XXxXXXX110.15672/hujms.xxArticle Type

<EMAIL_ADDRESS>(A. M. Mathai<EMAIL_ADDRESS>(M. Kumar)

# Optimal variable acceptance sampling plan for exponential distribution using
Bayesian estimate under Type I hybrid censoring

A. M. Mathai1, M. Kumar2 1,2Department of Mathematics, National Institute of
Technology, Calicut, Pin Code - 673601, India.

###### Abstract.

In this study, variable acceptance sampling plans under Type I hybrid
censoring is designed for a lot of independent and identical units with
exponential lifetimes using Bayesian estimate of the parameter $\vartheta$.
This approach is new from the conventional methods in acceptance sampling plan
which relay on maximum likelihood estimate and minimising of Bayes risk.
Bayesian estimate is obtained using squared error loss and Linex loss
functions. Optimisation problem is solved for minimising the testing cost
under each methods and optimal values of the plan parameters $n,t_{1}$ and
$t_{2}$ are calculated. The proposed plans are illustrated using various
examples and a real life case study is also conducted. Expected testing cost
of the sampling plan obtained using squared error loss function is much lower
than the cost of existing plans using maximum likelihood estimate.

###### Key words and phrases:

Acceptance sampling plan, Type I hybrid censoring, Exponential distribution,
Bayesian estimate, Squared error loss, Linex loss, Lindley’s approximation

DD.MM.YYYYDD.MM.YYYY

##  1\. Introduction

In the world of manufacturing and quality control, ensuring the consistent
quality of products is of paramount importance. However, inspecting each and
every item produced can be impractical, time-consuming, and costly. This is
where acceptance sampling plans come into play. An acceptance sampling plan
(ASP) is a statistical technique used to make informed decisions about
accepting or rejecting a batch or lot of items based on the inspection of a
sample.

Understanding and implementing a well-designed acceptance sampling plan is
crucial for manufacturers seeking to maintain consistent product quality,
optimize resources, and meet customer expectations. By employing statistically
sound sampling techniques, organizations can make reliable decisions regarding
batch acceptance, leading to improved efficiency, reduced costs, and enhanced
customer satisfaction.

Censoring schemes play a crucial role in ASP, and are designed to optimize the
efficiency and effectiveness of the sampling process. Type-I and Type-II
censoring schemes are widely used in statistical analysis, particularly in
reliability testing and acceptance sampling. In a Type-I censoring scheme, the
time of experiment is predetermined, but the total number of failures occurred
becomes a random variable. Conversely, in a Type-II censoring scheme, the
testing time is a random variable since the number of observed failures is
fixed. When a combination of both Type-I and Type-II censoring schemes is
employed, it is referred to as a hybrid censoring scheme (HCS).

Here, we consider an experiment where a sample of $n$ units are subjected to
testing. Also, assume that the sample units’ lifespans are independent random
variables and identically follows exponential distribution. The sample units
ordered lifetimes are represented as $X_{1,n},X_{2,n},...,X_{n,n}$. The test
is concluded either when a predetermined number, $\gamma$ (where $\gamma<n$),
of the $n$ items have failed, or when a predetermined time, $\mathcal{T}$, has
been reached. In other words, the life-test ends at a random time
$\mathcal{T}^{*}$, which is the minimum of either the ordered lifetime
($X_{\gamma,n}$) or the predetermined time $\mathcal{T}$. Moreover, a widely
accepted assumption is that the failed units in the experiment are not
substituted or replaced.

The Type-I HCS, proposed by Epstein [8], has found extensive application in
reliability acceptance testing, such as in MIL-STD-781-C (1977). Since its
introduction, extensive research has been conducted on hybrid censoring and
its various variations. Epstein initially introduced the Type-I HCS and
discussed estimation methods for exponential distribution. Additionally, a
two-sided confidence interval was put forth without a mathematical proof of
how it was obtained.

Subsequent researchers, such as Fairbanks et al.[9], Chen and Bhattacharyya
[5], Bartholomew [3], and Barlow et al. [2] made slight modifications to
Epstein’s proposition and worked on the derivation of the conditional moment
generating function approach to derive the exact distribution of the
conditional maximum likelihood estimator (MLE) of the mean life parameter .
Childs et al. [6] obtained a simplified but equivalent form of the exact
distribution of the MLE as derived by Chen and Bhattacharyya [5]. Draper and
Guttman [7] delved into the Bayesian inference within the Type-I HCS and
obtained Bayesian estimates and two-sided credible intervals using an inverted
gamma prior.

In [12], Kumar and Ramyamol, designed a cost efficient acceptance sampling
plans for exponentially distributed lifetime data under Type I censoring based
on MLE of the the mean life. Most of the works in acceptance sampling plans
are centered on MLE of the parameter for deciding the acceptance and rejection
of lot, even though Bayesian estimator is more reliable than MLE.

Bayesian acceptance sampling plans have gained considerable attention due to
their ability to incorporate prior information and make informed decisions
based on observed data. However, existing Bayesian sampling plans focus on
minimising Bayes risk, while the consideration of testing cost optimization
remains limited. Lam [22], Lam and Choy [23], Lin et al. [16], and Huang and
Lin [10, 11] focused on conventional Type I, Type II, and random censored
samples. Additionally, Chen et al. [4] examined Bayesian sampling plans for
exponential distributions based on Type I hybrid censored samples.

Recent contributions by Chen et al. [4], Lin et al. [15] and Prajapati et al.
[20] have specifically addressed Bayesian sampling plans for exponential
distributions by reducing Bayes risk utilizing Type I hybrid censored samples
. But these plans are not based on Bayesian estimator of mean life parameter
which is better than MLE ([14]).

In this work, we examine an acceptance sampling plan for a batch of units
where the failure time follows an exponential distribution with mean
$\vartheta$. Our aim is to design an optimal Bayesian acceptance sampling plan
under Type I HCS by using Bayesian estimator of the mean life parameter,
$\vartheta$, as the decision function. Our plan stands out from all other
existing Bayesian sampling plans by centering the decision of acceptance or
rejection of a lot on the Bayesian estimator of the mean lifetime parameter.
Also, here testing cost is minimized while maintaining stringent quality
standards. This departure from traditional Bayes risk-oriented approaches
allows us to focus explicitly on the optimization of testing costs, a critical
consideration in practical industrial settings.

The rest of the paper is organised as follows: Section 2 provides a detailed
information about Bayesian inference of exponential distributed life time data
under Type I HCS. Bayesian estimator of mean life parameter, $\vartheta$, is
considered using both squared error loss (SEL) function and Linex loss
function. Section 3 presents the formulation and optimization framework of the
proposed acceptance sampling plan. Here, the plan parameters ($n,t_{1},t_{2}$)
are obtained by minimising the expected testing cost (ETC) subject to required
probability conditions using both the loss functions namely, SEL and Linex
loss function. The distribution of the Bayesian estimator of $\vartheta$ under
Type I HCS for both loss functions are derived using delta method. In Section
4, we present the results of numerical computations of the proposed acceptance
sampling plan and a comparison with existing work in [12]. A real life case
study is done to illustrate the performance of the sampling plan presented
here. Finally, Section 5 concludes the paper with a summary of the
contributions and future research directions.

##  2\. Bayesian Estimation for Type I hybrid censoring scheme

Consider a lot of units, all of which experience failures over time following
an exponential distribution. The probability density function that
characterizes the failure time of each unit is given by:

$f(x,\vartheta)=\frac{1}{\vartheta}exp\left\\{-\frac{-x}{\vartheta}\right\\}\qquad
x\geq 0,\vartheta>0$ (1)

Here we conduct Type-I hybrid censored life testing, and we obtain one of the
following two forms of observations as our observed data:

$\displaystyle\text{Data type I}:$
$\displaystyle\left\\{X_{1,n}<X_{2,n}<\cdots<X_{\gamma,n}\right\\}\quad\text{if}\quad\mathcal{T}^{*}=X_{\gamma},$
(2) $\displaystyle\text{Data type II}:$
$\displaystyle\left\\{X_{1,n}<X_{2,n}<\cdots<X_{\mathcal{D},n}\right\\}\quad\text{if}\quad\mathcal{T}^{*}=\mathcal{T},$

where $\mathcal{D}$ denotes the number of failures observed before time
$\mathcal{T}^{*}$. Thus, $\mathcal{D}=\gamma$ when
$\mathcal{T}^{*}=X_{\gamma}$.

In this section, the method of obtaining Bayesian estimator of the unknown
parameter $\vartheta$ under the squared error loss and Linex loss functions
and their corresponding distributions are discussed. Bayesian estimator of the
parameter $\vartheta$ is used for the designing of acceptance sampling plan in
the next sections and this idea is new in acceptance sampling plans.

Under Type-I HCS, the likelihood function is obtained as

$L(x,\vartheta)=\frac{1}{\vartheta^{\gamma}}exp\left\\{-\frac{1}{\vartheta}\sum\limits_{i=1}^{\mathcal{D}}x_{i,n}-\frac{(n-\gamma)}{\vartheta}\mathcal{T}^{*}\right\\},$
(3)

and log-likelihood function as

$L(\vartheta,x)=-\gamma\operatorname{log}\vartheta-\frac{1}{\vartheta}\sum\limits_{i=1}^{\mathcal{D}}x_{i,n}-\frac{(n-\gamma)}{\vartheta}\mathcal{T}^{*},$

where $\mathcal{T}^{*}=\operatorname{Min}\\{\mathcal{T},X_{\gamma,n}\\}$ and
$\mathcal{D}$ = Number of units failed before $\mathcal{T}^{*}$.

Hence, by referring to the observed sample, the MLE of $\vartheta$ can be
determined as follows ([5]):

$\hat{\vartheta}_{MLE}=\begin{cases}\frac{1}{\mathcal{D}}\left[\sum\limits_{i=1}^{\mathcal{D}}x_{i,n}+(n-\mathcal{D})\mathcal{T}\right]&\text{if}\quad
1\leq\mathcal{D}\leq\gamma-1\\\
\frac{1}{\gamma}\left[\sum\limits_{i=1}^{\gamma}x_{i,n}+(n-\gamma)X_{\gamma,n}\right]&\text{if}\quad\mathcal{D}=\gamma\\\
n\mathcal{T}&\text{if}\quad\mathcal{D}=0.\end{cases}$ (4)

It is evident that $\widehat{\vartheta}_{MLE}$ is a conditioned MLE of
$\vartheta$, with the condition that at least one observed failure is present.

Here for Bayesian inference is done by using two kinds of loss functions,
namely, squared error loss (SEL) and Linex loss function. SEL is a symmetric
function and its symmetric nature is demonstrated as

$l_{1}(\eta,\widehat{\eta})=(\widehat{\eta}-\eta)^{2}$ (5)

where $\widehat{\eta}$ is the estimate of the unknown parameter $\eta$.

The Bayesian estimation of any function $g=g(\vartheta)$ under the SEL in (5),
is obtained by

$\widehat{g}_{s}=E(g\mid
Data)=\frac{\int_{\vartheta}g(\vartheta)L(x,\vartheta)\rho(\vartheta)d\vartheta}{\int_{\vartheta}L(x,\vartheta)\rho(\vartheta)d\vartheta}.$
(6)

where $\vartheta$ follows the prior distribution with PDF $\rho(\vartheta)$.

Using Type-I HCS, Draper and Guttman [7] investigated the application of
Bayesian inference to estimate the unknown parameter $\vartheta$ under squared
error loss function. They assumed in the study that $\vartheta$ follows an
inverted gamma prior distribution, with the probability density function (PDF)
given below

$\rho(\vartheta)=\frac{a^{b}}{\Gamma(b)}\vartheta^{-(b+1)}e^{-a/\vartheta},\quad\vartheta>0,\
a>0,\ \text{and}\ b>0.$ (7)

The prior in (7) becomes non-informative prior when the hyper-parameters
$a=b=0$.

The posterior density function of $\vartheta$, based on the knowledge from the
observed data and the inverted gamma prior given by (7), becomes

$h(\vartheta\mid Data)=\frac{\left(\mathcal{D}\
\hat{\vartheta}_{MLE}+a\right)^{(\mathcal{D}+b)}}{\Gamma(\mathcal{D}+b)}\vartheta^{-\left(\mathcal{D}+b+1\right)}\operatorname{exp}\left\\{\left(\mathcal{D}\
\hat{\vartheta}_{MLE}+a\right)/\vartheta\right\\}$ (8)

Considering that $\frac{\left(\mathcal{D}\
\hat{\vartheta}_{MLE}+a\right)}{\vartheta}$ is distributed as
$\chi^{2}_{2\left(b+a\right)}/2$ a posteriori, the posterior mean serves as
the Bayesian estimate of $\vartheta$ under the squared-error loss function. If
$\mathcal{D}+b>1$, this estimator is simply produced as

$\widehat{\vartheta_{s}}=\frac{\mathcal{D}\
\hat{\vartheta}_{MLE}+a}{\mathcal{D}+b-1}.$ (9)

When $a=b=0$, that is, when the prior becomes non-informative, it is evident
that the MLE of $\vartheta$ provided by (4) agrees with the Bayesian estimator
in (9).

The second is the Linex loss function, which is referred to as an asymmetric
function and is represented by the following formula:

$l_{2}(\eta,\widehat{\eta})\propto
e^{c(\widehat{\eta}-\eta)}-c(\widehat{\eta}-\eta)-1,\quad c\neq 0.$ (10)

The direction and magnitude of asymmetry are both determined by the parameter
$c$’s value. Bayesian estimation tends to overestimate when $c<0$, but it
favours underestimating when $c>0$. The Linex loss value resembles the squared
error loss when $c$ approaches 0, leading to an almost symmetric behaviour.
Applying the Linex loss function, we can derive the Bayesian estimation of
$g(\vartheta)$ as

$\widehat{g}_{l}=-\frac{1}{c}\ln E\left(e^{-cg}\mid Data\right),$ (11)

where

$E\left(e^{-cg}\mid
Data\right)=\frac{\int_{\vartheta}e^{-cg(\vartheta)}L(x,\vartheta)\rho(\vartheta)d\vartheta}{\int_{\vartheta}L(x,\vartheta)\rho(\vartheta)d\vartheta}.$
(12)

But in most of the cases, the ratio of integrals in (6) and (12) is not
possible to derive an analytical expression for it. As a result, Lindley [17]
developed approximate procedures for determining the ratio of two integrals,
such as (6) and (12). Several researchers ([13]) have used this to derive
approximate Bayesian estimates.

Thus, based on Linux loss functions, we get the Bayesian estimator of
$g(\vartheta)=\vartheta$ as

$\widehat{\vartheta_{l}}=-\frac{1}{c}\ln E\left(e^{-c\vartheta}\mid
Data\right).$ (13)

By applying Lindley’s approximate method in (13) for solving ratio of
integrals without a closed form, we get the Bayesian estimate of $\vartheta$
as following

$\widehat{\vartheta_{l}}=\widehat{\vartheta}_{MLE}-\frac{1}{c}\operatorname{ln}\left[1+\frac{c}{2\mathcal{D}}\left(c\widehat{\vartheta}_{MLE}^{2}-2a+2\widehat{\vartheta}_{MLE}(b-1)\right)\right].$
(14)

##  3\. Acceptance sampling plans under Type I hybrid censoring scheme

In this section, we design ASPs for a lot of units having exponential failure
time with PDF given by (1). For that, a sample comprising $n$ elements is
taken from the lot and is tested under Type I HCS.

Let $\vartheta_{A}$ be the acceptable quality level (AQL) and $\vartheta_{U}$
be the unacceptable quality level (UQL) of a unit item in the lot. The
acceptance and rejection of the lot depends on the probability conditions
given by:

$\displaystyle P\left(\text{Lot is
rejected}\mid\vartheta=\vartheta_{A}\right)$ $\displaystyle\leq\alpha,$ (15)
$\displaystyle P\left(\text{Lot is
accepted}\mid\vartheta=\vartheta_{U}\right)$ $\displaystyle\leq\beta,$

where $\beta$ is the consumer’s risk, $\alpha$ is the producer’s risk, and
$0<\alpha,\beta<1$.

###  3.1. Normal approximation of the Bayesian estimators of $\vartheta$ using
delta method

Here, the Bayesian estimator of the unknown parameter $\vartheta$ is used to
design the acceptance sampling plan. In order to proceed, it is necessary to
derive the distribution of the Bayesian estimator. Recall that both the
Bayesian estimators are derived in terms of $\hat{\vartheta}_{MLE}$, the MLE
of $\vartheta$. In [6], Childs et.al derived the exact distribution for the
conditional MLE of $\vartheta$, $\widehat{\vartheta}_{MLE}$ for
$\mathcal{D}\geq 1$ with PDF as

$f_{\widehat{\vartheta}_{MLE}}(x)=\left(1-v^{n}\right)^{-1}\left[\sum_{\mathcal{D}=1}^{\gamma-1}\sum_{k=0}^{\mathcal{D}}A_{k,\mathcal{D}}\
q\left(x-\mathcal{T}_{k,\mathcal{D}};\frac{\mathcal{D}}{\vartheta},\mathcal{D}\right)+q\left(x;\frac{\gamma}{\vartheta},\gamma\right)\right.\\\
+\left.\gamma\left(\begin{array}[]{l}n\\\
\gamma\end{array}\right)\sum_{k=1}^{\gamma}\frac{(-1)^{k}v^{n-\gamma+k}}{n-\gamma+k}\left(\begin{array}[]{c}\gamma-1\\\
k-1\end{array}\right)q\left(x-T_{k,\gamma};\frac{\gamma}{\vartheta},\gamma\right)\right],\
0<x<n\mathcal{T}$ (16)

where

$v=\operatorname{exp}\left\\{-\frac{\mathcal{T}}{\vartheta}\right\\},\quad
T_{k,\mathcal{D}}=\frac{(n-\mathcal{D}+k)\mathcal{T}}{\mathcal{D}},$
$A_{k,\mathcal{D}}=(-1)^{k}\left(\begin{array}[]{l}n\\\
\mathcal{D}\end{array}\right)\left(\begin{array}[]{l}\mathcal{D}\\\
k\end{array}\right)u^{n-\mathcal{D}+k},$

and

$q(x;p,t)=\begin{cases}\frac{p^{t}}{\Gamma(t)}x^{t-1}e^{-px},&x>0\\\ 0,&\text{
otherwise. }\end{cases}$

From (16), the expectation of $\widehat{\vartheta}_{MLE}$ and
$E(\widehat{\vartheta}_{MLE}^{2})$ are obtained respectively as

$E(\widehat{\vartheta}_{MLE})=\left(1-v^{n}\right)^{-1}{\left[\sum_{\mathcal{D}=1}^{\gamma-1}\sum_{k=0}^{\mathcal{D}}A_{k,\mathcal{D}}\left(\vartheta+T_{k,\mathcal{D}}\right)+\vartheta\right.}\\\
\left.+\gamma\left(\begin{array}[]{l}n\\\
\gamma\end{array}\right)\sum_{k=1}^{\gamma}\frac{(-1)^{k}v^{n-\gamma+k}}{n-\gamma+k}\left(\begin{array}[]{l}\gamma-1\\\
k-1\end{array}\right)\left(\vartheta+T_{k,\gamma}\right)\right],$ (17)

and

$E\left(\widehat{\vartheta}_{MLE}^{2}\right)=\left(1-v^{n}\right)^{-1}\left[\sum_{\mathcal{D}=1}^{\gamma-1}\sum_{k=0}^{\mathcal{D}}A_{k,\mathcal{D}}\left\\{\frac{\vartheta^{2}}{\mathcal{D}}(1+\mathcal{D})+2T_{k,\mathcal{D}}\vartheta+\left(T_{k,\mathcal{D}}\right)^{2}\right\\}\right.\\\
+\frac{\vartheta^{2}}{\gamma}(1+\gamma)+\gamma\left(\begin{array}[]{l}n\\\
\gamma\end{array}\right)\left.\cdot\sum_{k=1}^{\gamma}\frac{(-1)^{k}v^{n-\gamma+k}}{n-\gamma+k}\left(\begin{array}[]{l}\gamma-1\\\
k-1\end{array}\right)\left\\{\frac{\vartheta^{2}}{\gamma}(1+\gamma)+2T_{k,\gamma}\vartheta+\left(T_{k,\gamma}\right)^{2}\right\\}\right].$
(18)

It is important to note that the variance of $\widehat{\vartheta}_{MLE}$,
denoted as $\sigma^{2}\left({\widehat{\vartheta}_{MLE}}\right)$, can be
obtained by applying equations (17) and (18).

Define the Bayesian estimator of $\vartheta$ using SEL,
$\widehat{\vartheta_{s}}=U_{1}\left(\widehat{\vartheta}_{MLE}\right)$ then by
delta method (as in [1]), $\widehat{\vartheta_{s}}$ follows Normal
distribution with mean,

$E\left(\widehat{\vartheta_{s}}\right)=U_{1}\left(E\left(\widehat{\vartheta}_{MLE}\right)\right)=\frac{\mathcal{D}\
E\left(\hat{\vartheta}_{MLE}\right)+a}{\mathcal{D}+b-1},$ (19)

and variance,

$\displaystyle V\left(\widehat{\vartheta_{s}}\right)$
$\displaystyle=\left(U_{1}^{\prime}\left(E\left(\widehat{\vartheta}_{MLE}\right)\right)\right)^{2}\sigma^{2}\left({\widehat{\vartheta}_{MLE}}\right)$
(20)
$\displaystyle=\left(\frac{\mathcal{D}}{\mathcal{D}+b-1}\right)^{2}\sigma^{2}\left({\widehat{\vartheta}_{MLE}}\right).$

Similarly, from (14), we can define the Bayesian estimator of $\vartheta$
using Linex loss function, $\widehat{\vartheta_{l}}$ as
$U_{2}\left(\widehat{\vartheta}_{MLE}\right)$. Then by delta method,
$\widehat{\vartheta_{l}}$ follows Normal distribution with mean,

$\displaystyle E\left(\widehat{\vartheta_{l}}\right)$
$\displaystyle=U_{2}\left(E\left(\widehat{\vartheta}_{MLE}\right)\right)$ (21)
$\displaystyle=E\left(\widehat{\vartheta}_{MLE}\right)-\frac{1}{c}\operatorname{ln}\left[1+\frac{c}{2\mathcal{D}}\left(cE\left(\widehat{\vartheta}_{MLE}\right)^{2}-2a+2E\left(\widehat{\vartheta}_{MLE}\right)(b-1)\right)\right],$

and variance,

$\displaystyle V\left(\widehat{\vartheta_{l}}\right)$
$\displaystyle=\left(U_{2}^{\prime}\left(E\left(\widehat{\vartheta}_{MLE}\right)\right)\right)^{2}\sigma^{2}\left({\widehat{\vartheta}_{MLE}}\right)$
(22)
$\displaystyle=\left(1-\frac{2cE\left(\widehat{\vartheta}_{MLE}\right)+2b-2}{2\mathcal{D}+c\left(cE\left(\widehat{\vartheta}_{MLE}\right)^{2}-2a+2E\left(\widehat{\vartheta}_{MLE}\right)(b-1)\right)}\right)^{2}\
\sigma^{2}\left({\widehat{\vartheta}_{MLE}}\right).$

###  3.2. ASP with Bayesian estimator of $\vartheta$ using SEL

For this ASP, a sample consisting of $n$ items is taken from the lot and
examined using Type I hybrid censoring. The probability requirements stated in
(15) are used to determine whether to accept or reject a lot. Our ASP in this
situation is as follows:

  1. Step 1: A random sample of size $n$ is taken from the lot and is tested up to time $\mathcal{T}^{*}=\operatorname{Min}\\{\mathcal{T},X_{\gamma}\\}$ where $\mathcal{T}$ is the pre-fixed time and $\gamma<n$ is the pre-fixed number of failures.The number of failures that happened prior to the time $\mathcal{T}^{*}$ with associated lifetimes provided by (2), i.e, either data type I or type II, is denoted by $\mathcal{D}$. Also, fix the values of AQL, UQL, producer’s risk ($\alpha$), and consumer’s risk ($\beta$).

  2. Step 2: Calculate the Bayesian estimate of $\vartheta$ using SEL ($\widehat{\vartheta_{s}}$), which is given by

$\widehat{\vartheta_{s}}=\frac{\mathcal{D}\
\hat{\vartheta}_{MLE}+a}{\mathcal{D}+b-1}.$

  3. Step 3: Continue the testing process, by repeating Steps 1, and 2 if $t_{1}\leq\widehat{\vartheta_{s}}<t_{2}$. If not, go to Step 4.

  4. Step 4: Accept the lot if $\widehat{\vartheta_{s}}\geq t_{2}$ and reject the lot if $\widehat{\vartheta_{s}}<t_{1}.$

The key objective is to minimize the total testing cost subject to the
constrains given in (15) in order to obtain the optimal values of $n,t_{1}$
and $t_{2}$. The total testing cost is the product of the testing cost of an
item for unit time with total testing time. Based on the sampling plan
described above, the total testing time is the product of the time taken to
reach a decision for each sample, $\widehat{\vartheta_{s}}$, by the number of
samples tested. But, $\widehat{\vartheta_{s}}$ and number of samples are
random variables. So, here the expected testing cost is obtained.

Consider $p_{c}$, $p_{a}$, and $p_{r}$ as the probabilities of continuation of
testing, accepting the lot and rejecting the lot respectively. Then, the
corresponding long run probabilities of acceptance and rejection are given by
$P_{a}=\frac{p_{a}}{1-p_{c}}$ and $P_{r}=\frac{p_{r}}{1-p_{c}}$, and expected
number of items failed is given by $\frac{1}{1-p_{c}}$ (see [21]). Thus, the
expected testing cost for the ASP is obtained as

$\text{ETC}=\frac{C\ E\left(\widehat{\vartheta_{s}}\right)}{1-p_{c}},$

where $C$ is the testing cost of an item for unit time.

From equations (19) and (20), we get that, $\widehat{\vartheta_{s}}$ follows
normal distribution with mean,

$E\left(\widehat{\vartheta_{s}}\right)=\frac{\mathcal{D}\
E\left(\hat{\vartheta}_{MLE}\right)+a}{\mathcal{D}+b-1},$

and variance,

$V\left(\widehat{\vartheta_{s}}\right)=\left(\frac{\mathcal{D}}{\mathcal{D}+b-1}\right)^{2}\sigma^{2}\left({\widehat{\vartheta}_{MLE}}\right).$

Then, the required probabilities are given by,

$p_{a}=P\left(\widehat{\vartheta_{s}}\geq
t_{2}\right)=P\left[Z\geq\frac{t_{2}-E\left(\widehat{\vartheta_{s}}\right)}{\sqrt{V\left(\widehat{\vartheta_{s}}\right)}}\right],$
(23)
$p_{r}=P\left(\widehat{\vartheta_{s}}<t_{1}\right)=P\left[Z<\frac{t_{1}-E\left(\widehat{\vartheta_{s}}\right)}{\sqrt{V\left(\widehat{\vartheta_{s}}\right)}}\right],$
(24)
$p_{c}=P\left(t_{1}\leq\widehat{\vartheta_{s}}<t_{2}\right)=P\left[\frac{t_{1}-E\left(\widehat{\vartheta_{s}}\right)}{\sqrt{V\left(\widehat{\vartheta_{s}}\right)}}\leq
Z<\frac{t_{2}-E\left(\widehat{\vartheta_{s}}\right)}{\sqrt{V\left(\widehat{\vartheta_{s}}\right)}}\right].$
(25)

As the expected testing cost is a function of the unknown parameter
$\vartheta$, here we consider the expected testing cost at $\vartheta_{A}$ as
the objective of the optimisation problem. Thus, the required optimisation
problem, $P_{1}\left(n,t_{1},t_{2}\right)$, is obtained by minimising the
expected testing cost at $\vartheta_{A}$, subject to the probability
conditions in (15).

$\displaystyle P_{1}(n,t_{1},t_{2}):$
$\displaystyle\underset{\left(n,t_{1},t_{2}\right)}{\operatorname{Min}}\left(\frac{C\
E\left(\widehat{\vartheta_{s}}\right)}{1-p_{c}}\right)\text{at}\
\vartheta_{A}$ (26) subject to
$\displaystyle\left(P_{r}\mid\vartheta=\vartheta_{A}\right)\leq\alpha,$
$\displaystyle\left(P_{a}\mid\vartheta=\vartheta_{U}\right)\leq\beta,$

where $t_{1},t_{2}>0$ and $t_{2}>t_{1}$. That is,

$\displaystyle\underset{\left(n,t_{1},t_{2}\right)}{\operatorname{Min}}\left(\frac{C\
E\left(\widehat{\vartheta_{s}}\right)}{1-p_{c}}\right)\text{at}\
\vartheta_{A}$ (27) subject to
$\displaystyle\left(\frac{p_{r}}{1-p_{c}}\mid\vartheta=\vartheta_{A}\right)\leq\alpha,$
$\displaystyle\left(\frac{p_{a}}{1-p_{c}}\mid\vartheta=\vartheta_{U}\right)\leq\beta,$

where $t_{1},t_{2}>0$ and $t_{2}>t_{1}$. Thus, by substituting the expressions
of $p_{a},p_{r}$ and $p_{c}$ from equations (23), (24) and (25), the non-
linear optimization problem, $P_{1}(n,t_{1},t_{2})$, becomes:

$\displaystyle P_{1}(n,t_{1},t_{2}):$
$\displaystyle\underset{\left(n,t_{1},t_{2}\right)}{\operatorname{Min}}\left(\frac{C\
E\left(\widehat{\vartheta_{s}}\right)}{1-P\left[\frac{t_{1}-E\left(\widehat{\vartheta_{s}}\right)}{\sqrt{V\left(\widehat{\vartheta_{s}}\right)}}\leq
Z<\frac{t_{2}-E\left(\widehat{\vartheta_{s}}\right)}{\sqrt{V\left(\widehat{\vartheta_{s}}\right)}}\right]}\right)\text{at}\
\vartheta_{A}$ (28) subject to
$\displaystyle\left(\frac{P\left[Z<\frac{t_{1}-E\left(\widehat{\vartheta_{s}}\right)}{\sqrt{V\left(\widehat{\vartheta_{s}}\right)}}\right]}{1-P\left[\frac{t_{1}-E\left(\widehat{\vartheta_{s}}\right)}{\sqrt{V\left(\widehat{\vartheta_{s}}\right)}}\leq
Z<\frac{t_{2}-E\left(\widehat{\vartheta_{s}}\right)}{\sqrt{V\left(\widehat{\vartheta_{s}}\right)}}\right]}\mid\vartheta=\vartheta_{A}\right)\leq\alpha,$
$\displaystyle\left(\frac{P\left[Z\geq\frac{t_{2}-E\left(\widehat{\vartheta_{s}}\right)}{\sqrt{V\left(\widehat{\vartheta_{s}}\right)}}\right]}{1-P\left[\frac{t_{1}-E\left(\widehat{\vartheta_{s}}\right)}{\sqrt{V\left(\widehat{\vartheta_{s}}\right)}}\leq
Z<\frac{t_{2}-E\left(\widehat{\vartheta_{s}}\right)}{\sqrt{V\left(\widehat{\vartheta_{s}}\right)}}\right]}\mid\vartheta=\vartheta_{U}\right)\leq\beta,$

where $t_{1},t_{2}>0$ and $t_{2}>t_{1}$.

For $n,t_{1},$ and $t_{2}$, this non-linear optimisation problem $P_{1}$ can
be solved by the following steps:

  1. (1)

Compute the minimum value of $\gamma$ which satisfies the constrains of the
problem $P_{1}$.

  2. (2)

Utilising the obtained value of $\gamma$, from Step 1, solve $P_{1}$ for the
optimal values of $n$, $t_{1}$, and $t_{2}$.

The optimal values of $n$, $t_{1}$, and $t_{2}$ are obtained by solving the
above non-linear optimization problem using genetic algorithm solver in MATLAB
and are tabulated in Table 1.

###  3.3. ASP with Bayesian estimator of $\vartheta$ using Linex loss function

In this section, the Bayesian estimate of $\vartheta$ using Linex loss
function is applied in the ASP. For that, a sample of size $n$ is taken from
the lot. The testing is terminated at time $\mathcal{T}^{*}$ and the observed
lifetime data will be as given in (2). Here we define an ASP as follows:

  1. Step 1: Take a random sample of size $n$ from the lot. Fix the time $\mathcal{T}$, failure number, $\gamma<n$, AQL, UQL, producer’s risk ($\alpha$), and consumer’s risk ($\beta$). Testing is stopped at time, $\mathcal{T}^{*}=\operatorname{Min}\\{\mathcal{T},X_{\gamma}\\}$.

  2. Step 2: Calculate the Bayesian estimate of $\vartheta$ using Linex loss function ($\widehat{\vartheta_{l}}$), given by

$\widehat{\vartheta_{l}}=\widehat{\vartheta}_{MLE}-\frac{1}{c}\operatorname{ln}\left[1+\frac{c}{2\mathcal{D}}\left(c\widehat{\vartheta}_{MLE}^{2}-2a+2\widehat{\vartheta}_{MLE}(b-1)\right)\right].$

  3. Step 3: Continue the test, if $t_{1}\leq\widehat{\vartheta_{l}}<t_{2}$ and repeat Steps 1 and 2 .
Else, go to Step 4.

  4. Step 4: Accept the lot, if $\widehat{\vartheta_{l}}\geq t_{2}$ and reject the lot, if $\widehat{\vartheta_{l}}<t_{1}.$

The main problem here is determining the optimal values of $n$, $t_{1}$, and
$t_{2}$ while minimising the testing cost under probability constraints as
mentioned in (15). Total testing cost is a product of testing cost of a unit
item in unit time and total testing time. Total testing time is obtained by
multiplying the overall number of samples tested with decision making time
while testing a sample, given by $\widehat{\vartheta}_{l}$. But they are both
random variables. Thus, the total testing time becomes a random quantity with
expectation, $\frac{1}{1-p_{c}}\ E\left(\widehat{\vartheta_{l}}\right)$. As a
result, here we use the expected testing cost as the objective function and it
is given by

$\text{ETC}=\frac{C\ E\left(\widehat{\vartheta_{l}}\right)}{1-p_{c}},$

where $C$ is the testing cost of an item for unit time.

From equations (21) and (22), we get that, $\widehat{\vartheta_{l}}$ follows
Normal distribution with mean,

$E\left(\widehat{\vartheta_{l}}\right)=E\left(\widehat{\vartheta}_{MLE}\right)-\frac{1}{c}\operatorname{ln}\left[1+\frac{c}{2\mathcal{D}}\left(cE\left(\widehat{\vartheta}_{MLE}\right)^{2}-2a+2E\left(\widehat{\vartheta}_{MLE}\right)(b-1)\right)\right],$

and variance,

$V\left(\widehat{\vartheta_{l}}\right)=\left(1-\frac{2cE\left(\widehat{\vartheta}_{MLE}\right)+2b-2}{2\mathcal{D}+c\left(cE\left(\widehat{\vartheta}_{MLE}\right)^{2}-2a+2E\left(\widehat{\vartheta}_{MLE}\right)(b-1)\right)}\right)^{2}\
\sigma^{2}\left({\widehat{\vartheta}_{MLE}}\right).$

Using this information, we can evaluate the required probabilities as

$p_{a}=P\left(\widehat{\vartheta_{l}}\geq
t_{2}\right)=P\left[Z\geq\frac{t_{2}-E\left(\widehat{\vartheta_{l}}\right)}{\sqrt{V\left(\widehat{\vartheta_{l}}\right)}}\right],$
(29)
$p_{r}=P\left(\widehat{\vartheta_{l}}<t_{1}\right)=P\left[Z<\frac{t_{1}-E\left(\widehat{\vartheta_{l}}\right)}{\sqrt{V\left(\widehat{\vartheta_{l}}\right)}}\right],$
(30)
$p_{c}=P\left(t_{1}\leq\widehat{\vartheta_{l}}<t_{2}\right)=P\left[\frac{t_{1}-E\left(\widehat{\vartheta_{l}}\right)}{\sqrt{V\left(\widehat{\vartheta_{l}}\right)}}\leq
Z<\frac{t_{2}-E\left(\widehat{\vartheta_{l}}\right)}{\sqrt{V\left(\widehat{\vartheta_{l}}\right)}}\right].$
(31)

Here, we evaluate the objective function of the optimisation problem at
$\vartheta_{A}$ since the overall testing cost includes the unknown parameter
$\vartheta$. Thus, the expected testing cost at $\vartheta_{A}$ is minimised
by satisfying the probability conditions given in (15). Hence, the required
non-linear optimisation problem is formulated as

$\displaystyle P_{2}(n,t_{1},t_{2}):$
$\displaystyle\underset{\left(n,t_{1},t_{2}\right)}{\operatorname{Min}}\left(\frac{C\
E\left(\widehat{\vartheta_{l}}\right)}{1-p_{c}}\right)\text{at}\
\vartheta_{A}$ (32) subject to
$\displaystyle\left(P_{r}\mid\vartheta=\vartheta_{A}\right)\leq\alpha,$
$\displaystyle\left(P_{a}\mid\vartheta=\vartheta_{U}\right)\leq\beta,$

where $t_{1},t_{2}>0$ and $t_{2}>t_{1}$. That is,

$\displaystyle\underset{\left(n,t_{1},t_{2}\right)}{\operatorname{Min}}\left(\frac{C\
E\left(\widehat{\vartheta_{l}}\right)}{1-p_{c}}\right)\text{at}\
\vartheta_{A}$ (33) subject to
$\displaystyle\left(\frac{p_{r}}{1-p_{c}}\mid\vartheta=\vartheta_{A}\right)\leq\alpha,$
$\displaystyle\left(\frac{p_{a}}{1-p_{c}}\mid\vartheta=\vartheta_{U}\right)\leq\beta,$

where $t_{1},t_{2}>0$ and $t_{2}>t_{1}$. The non-linear optimisation problem,
$P_{2}(n,t_{1},t_{2})$, is modified by substituting the expressions for
$p_{a}$, $p_{r}$, and $p_{c}$ from equations (29), (30), and (31) as

$\displaystyle P_{2}(n,t_{1},t_{2}):$
$\displaystyle\underset{\left(n,t_{1},t_{2}\right)}{\operatorname{Min}}\left(\frac{C\
E\left(\widehat{\vartheta_{l}}\right)}{1-P\left[\frac{t_{1}-E\left(\widehat{\vartheta_{l}}\right)}{\sqrt{V\left(\widehat{\vartheta_{l}}\right)}}\leq
Z<\frac{t_{2}-E\left(\widehat{\vartheta_{l}}\right)}{\sqrt{V\left(\widehat{\vartheta_{l}}\right)}}\right]}\right)\text{at}\
\vartheta_{A}$ (34) subject to
$\displaystyle\left(\frac{P\left[Z<\frac{t_{1}-E\left(\widehat{\vartheta_{l}}\right)}{\sqrt{V\left(\widehat{\vartheta_{l}}\right)}}\right]}{1-P\left[\frac{t_{1}-E\left(\widehat{\vartheta_{l}}\right)}{\sqrt{V\left(\widehat{\vartheta_{l}}\right)}}\leq
Z<\frac{t_{2}-E\left(\widehat{\vartheta_{l}}\right)}{\sqrt{V\left(\widehat{\vartheta_{l}}\right)}}\right]}\mid\vartheta=\vartheta_{A}\right)\leq\alpha,$
$\displaystyle\left(\frac{P\left[Z\geq\frac{t_{2}-E\left(\widehat{\vartheta_{l}}\right)}{\sqrt{V\left(\widehat{\vartheta_{l}}\right)}}\right]}{1-P\left[\frac{t_{1}-E\left(\widehat{\vartheta_{l}}\right)}{\sqrt{V\left(\widehat{\vartheta_{l}}\right)}}\leq
Z<\frac{t_{2}-E\left(\widehat{\vartheta_{l}}\right)}{\sqrt{V\left(\widehat{\vartheta_{l}}\right)}}\right]}\mid\vartheta=\vartheta_{U}\right)\leq\beta,$

where $t_{1},t_{2}>0$ and $t_{2}>t_{1}$.

The following steps are used to resolve the above mentioned nonlinear
optimisation problem $P_{2}$ for ($n,t_{1},t_{2}$):

  1. (1)

First, determine the least value of $\gamma$ subjected to the constrains of
the problem $P_{2}$.

  2. (2)

Then solve $P_{2}$, using the value of $\gamma$ obtained in Step 1.

This non-linear optimisation problem is solved using genetic algorithm solver
in MATLAB and the obtained optimal values of $n$, $t_{1}$, and $t_{2}$ are
tabulated in Table 2 and Table 3 for $c=0.5$ and $c=-0.5$, respectively.

##  4\. Computational results, comparisons and real data case study

Numerical computations of all the above sections are discussed in this section
and are displayed in Table 1 to 3. The hyper parameters, $a=1.25$ and $b=2.5$
are used to compute the Bayesian estimator throughout this work since they
give more approximate Bayesian estimate for $\vartheta$ (see [15] and [22]).
ETC of the ASP under Type I hybrid censoring using Bayesian estimate of
$\vartheta$ subject to Linex loss function for both $c=0.5$ and $c=-0.5$ is
computed and is shown in Table 2 and Table 3, respectively. From that, we can
observe that, the ETC is less for $c=0.5$ than that of $c=-0.5$. For example,
with $C=1$, $\vartheta_{A}=200$, $\vartheta_{U}=100$, $\mathcal{T}=100$,
$\alpha=0.05$ and $\beta=0.05$, gives ETC of 127.2029 for $c=0.5$ and ETC is
increased to 228.1767 for $c=-0.5$. A comparison of testing costs obtained
from ASPs under Type I hybrid censoring using a Bayesian estimate of the
parameter $\vartheta$ using the SEL and Linex loss functions is also done
using the results in Tables 1, 2 and 3. Consider the following example for the
set of parameters: $C=1$, $\vartheta_{A}=500$, $\vartheta_{U}=200$,
$\mathcal{T}=50$, $\alpha=0.05$ and $\beta=0.05$, the ETC obtained for ASP
with SEL and Linex loss functions are 475.5810, 557.9282 ($c=0.5$) and
627.7466 ($c=-0.5$), respectively. These comparisons shows that the ASP has
the lowest test cost when Bayesian estimate of $\vartheta$ is computed using
SEL function.

Table 4 illustrates the comparison of ETC computed for the ASP under Type I
hybrid censoring with Bayesian estimate of $\vartheta$ using SEL function with
the ASP under Type I censoring scheme designed using MLE of $\vartheta$ in
[12]. It can be seen that, for parameters, $C=1$, $\vartheta_{A}=200$,
$\vartheta_{U}=100$, $\mathcal{T}=100$, $\alpha=0.05$ and $\beta=0.25$, the
ETC of ASP using SEL is 144.1837 and 156.53 for the ASP in [12]. Thus, one can
observe that cost of our plan is less than the one described in [12]. Also,
the proposed acceptance sampling plans are applied to a real life data.

$\vartheta_{A}$ | $\vartheta_{U}$ | $\mathcal{T}$ | $\alpha$ | $\beta$ | $\gamma$ | $t_{1}$ | $t_{2}$ | $n$ | ETC
---|---|---|---|---|---|---|---|---|---
200 | 100 | 100 | 0.05 | 0.05 | 26 | 162.3926 | 162.3957 | 31 | 123.4077
|  |  | 0.01 | 0.05 | 20 | 74.7316 | 74.7322 | 23 | 90.7417
|  |  | 0.01 | 0.01 | 25 | 146.3421 | 146.3394 | 26 | 125.6045
500 | 200 | 50 | 0.05 | 0.05 | 21 | 313.5638 | 313.5638 | 26 | 475.5810
|  |  | 0.01 | 0.05 | 22 | 310.5940 | 310.5948 | 27 | 480.9310
|  |  | 0.01 | 0.01 | 30 | 225.7734 | 225.7735 | 32 | 563.9248
500 | 200 | 100 | 0.05 | 0.05 | 26 | 313.3180 | 313.3188 | 37 | 509.9392
|  |  | 0.01 | 0.05 | 14 | 311.1179 | 311.1179 | 19 | 503.4323
|  |  | 0.01 | 0.01 | 24 | 390.4518 | 390.4526 | 38 | 566.6305
3000 | 1500 | 1000 | 0.05 | 0.05 | 16 | 2005.3416 | 2005.3425 | 62 | 475.5810
|  |  | 0.01 | 0.05 | 10 | 2000.6681 | 2000.6682 | 92 | 480.9310
|  |  | 0.01 | 0.01 | 15 | 2022.5679 | 2022.568 | 72 | 2684.0441
Table 1. ASP under Type I HCS using SEL function for C=1. $\vartheta_{A}$ | $\vartheta_{U}$ | $\mathcal{T}$ | $\alpha$ | $\beta$ | $\gamma$ | $t_{1}$ | $t_{2}$ | $n$ | ETC
---|---|---|---|---|---|---|---|---|---
200 | 100 | 100 | 0.05 | 0.05 | 21 | 63.1133 | 63.1134 | 35 | 127.2029
|  |  | 0.01 | 0.05 | 19 | 56.3674 | 56.3675 | 50 | 118.7482
|  |  | 0.01 | 0.01 | 28 | 140.9731 | 140.9733 | 30 | 143.1099
500 | 200 | 50 | 0.05 | 0.05 | 21 | 367.4209 | 367.421 | 34 | 557.9282
|  |  | 0.01 | 0.05 | 26 | 361.9554 | 361.9555 | 32 | 556.9578
|  |  | 0.01 | 0.01 | 28 | 371.9294 | 371.9295 | 36 | 595.2252
500 | 200 | 100 | 0.05 | 0.05 | 22 | 367.4918 | 367.4918 | 39 | 521.9267
|  |  | 0.01 | 0.05 | 27 | 354.8226 | 354.8229 | 37 | 513.5307
|  |  | 0.01 | 0.01 | 25 | 364.1113 | 364.1115 | 36 | 524.21
3000 | 1500 | 1000 | 0.05 | 0.05 | 15 | 2473.6173 | 2473.6174 | 24 | 3379.9555
|  |  | 0.01 | 0.05 | 11 | 2437.8132 | 2437.8133 | 33 | 3161.2578
|  |  | 0.01 | 0.01 | 6 | 2428.6943 | 2428.6944 | 14 | 3705.1855
Table 2. ASP under Type I HCS using Linex Loss for $c=0.5$ and C=1. $\vartheta_{A}$ | $\vartheta_{U}$ | $\mathcal{T}$ | $\alpha$ | $\beta$ | $\gamma$ | $t_{1}$ | $t_{2}$ | $n$ | ETC
---|---|---|---|---|---|---|---|---|---
200 | 100 | 100 | 0.05 | 0.05 | 28 | 186.2786 | 186.2793 | 34 | 228.1767
|  |  | 0.01 | 0.05 | 26 | 178.3335 | 178.3342 | 29 | 232.5816
|  |  | 0.01 | 0.01 | 23 | 177.5453 | 177.5455 | 45 | 220.3530
500 | 200 | 50 | 0.05 | 0.05 | 12 | 400.025 | 400.0248 | 25 | 627.7466
|  |  | 0.01 | 0.05 | 15 | 396.9158 | 396.9158 | 26 | 635.9504
|  |  | 0.01 | 0.01 | 14 | 405.41 | 405.4198 | 26 | 636.7619
500 | 200 | 100 | 0.05 | 0.05 | 18 | 423.5114 | 423.5116 | 43 | 600.3209
|  |  | 0.01 | 0.05 | 22 | 405.5332 | 405.5335 | 38 | 620.4932
|  |  | 0.01 | 0.01 | 24 | 411.5643 | 411.5644 | 37 | 623.4914
3000 | 1500 | 1000 | 0.05 | 0.05 | 19 | 3783.5928 | 3783.5929 | 29 | 3449.2903
|  |  | 0.01 | 0.05 | 11 | 2507.4709 | 2507.471 | 30 | 3411.8618
|  |  | 0.01 | 0.01 | 9 | 2557.26 | 2557.2699 | 27 | 3450.2116
Table 3. ASP under Type I HCS using Linex Loss for $c=-0.5$ and C=1. $\vartheta_{A}$ | $\vartheta_{U}$ | $\mathcal{T}$ | $\alpha$ | $\beta$ | $A$ | $B$ | $C$
---|---|---|---|---|---|---|---
200 | 100 | 100 | 0.05 | 0.25 | 123.4077 | 151.0479 | 156.53
3000 | 1500 | 1500 | 0.05 | 0.1 | 1423.5748 | 2256.9985 | 2290.35
Table 4.  Comparison of ETCs for ASP using two different loss functions and
testing cost under Type I Censoring. The ETCs are calculated using the SEL
function in column A and Linex loss with $c=0.5$ in column B. Column C
presents the expected testing cost obtained for ASP under Type I censoring, as
explained in Kumar et al.’s study in [12]. $C=1$

###  4.1. Case Study

In this section, the acceptance sampling plans presented in this work are
demonstrated using a real-world example from [18]. The failure rates for 36
appliances that underwent an automated life test are listed below; the
lifetime shown here refers to the number of cycles of use that the appliances
can withstand before failing.

$\begin{aligned}
&11,35,49,170,329,381,708,958,1062,1167,1594,1925,1990,2223,2327,2400,2451,2471,2551,2565,2568,\\\
&2694,2702,2761,2831,3034,3059,3112,3214,3478,3504,4329,6367,6976,7846,13403.\end{aligned}$

The above data follows exponential distribution. In order to apply the
proposed sampling plans in Sections 3 and 4, we take our plan parameters as
$\vartheta_{A}=3000$, $\vartheta_{U}=600$, $\mathcal{T}=2000$, $\alpha=0.1$
and $\beta=0.2$ as in [12]. Now, we can solve the nonlinear optimisation
problems $P_{1}$ and $P_{2}$ to obtain the optimal values of plan parameters
$t_{1}$ and $t_{2}$. Note that, $t_{1}$ and $t_{2}$ are real numbers but for
the application of our acceptance sampling plans we need them to be integers.
So we first solve $P_{1}$ and $P_{2}$ , if we get integer values for $t_{1}$
and $t_{2}$, then we can look for the acceptance and rejection of lot with
appropriate conditions. If $t_{1}$ and $t_{2}$ are non-integers then we
restrict them to be integers. Hence our optimisation problems $P_{1}$ and
$P_{2}$ becomes non-linear integer programming problem with integer solutions.

For the above mentioned data, the different sampling plans proposed in our
work yield the following outcomes:

  1. (1)

ASP under Type I HCS using SEL function gives the parameters as $\gamma=9$,
$t_{1}=2064$, $t_{2}=2065$, and $n=31$. Next, we consider the first $31$
samples from the above data. The samples are tested upto time
$\mathcal{T}^{*}=\operatorname{Min}\\{\mathcal{T},X_{9}\\}=1062$ since
$X_{9}=1062$. Now we compute the Bayesian estimate of $\vartheta$ with SEL
function as $\widehat{\vartheta_{s}}=2577.9286$ and it is greater than
$t_{2}=2065$. So, we can accept the lot with ETC = 2405.

  2. (2)

Based on ASP under Type I hybrid using Linex loss function for $c=0.5$, we get
the optimal values for test parameters as $\gamma=11$, $t_{1}=2156$,
$t_{2}=2157$, and $n=27$. Then, the first $27$ samples are taken from the data
and the testing is stopped at
$\mathcal{T}^{*}=\operatorname{Min}\\{\mathcal{T},X_{11}\\}=1594$ as
$X_{11}=1594$. Next, we calculate the Bayesian estimate of $\vartheta$ using
Linex loss function and get $\widehat{\vartheta_{l}}=2883.2339$. Hence, we
accept the lot as $\widehat{\vartheta_{l}}>t_{2}$ and the ETC = 2909.

##  5\. Conclusion

In conclusion, a novel approach for Type I hybrid censoring in acceptance
sampling plans is proposed here by utilizing Bayesian inference. Unlike
conventional methods that rely on maximum likelihood estimation (MLE) of
$\vartheta$ and minimizing Bayes risk, this study considered the Bayesian
estimators of $\vartheta$, for designing the ASP and taking into account the
expected testing cost. The squared error loss and linex loss were employed to
compute the Bayesian estimators of $\vartheta$.

The aim of all ASPs was to compute the optimal values of $n$, $t_{1}$ and
$t_{2}$ by solving nonlinear optimisation problems of minimising ETC. A
comparative study of ETC involved in ASPs using SEL and Linex loss functions
is illustrated here using Table 1 to 3 for different values of AQL
($\vartheta_{A}$), UQL ($\vartheta_{U}$), producer’s risk ($\alpha$),
consumer’s risk ($\beta$), $\mathcal{T}$ and $\gamma$. The lower expected
testing cost for SEL compared to Linex loss functions signifies that SEL
provides a more cost-effective approach for our plan. This insight can have
substantial implications for resource allocation, budget optimization, and
overall project efficiency. ETC obtained for ASP using Linex loss function is
computed for both $c=0.5$ and $c=-0.5$ and is tabulated in Table 2 and Table
3, respectively. From that, we get the ETC for $c=-0.5$ is more than ETC
obtained for $c=0.5$. This is mainly because the Bayesian estimator of
$\vartheta$ is overestimated for $c=-0.5$ using Linex loss function (see
[19]).

Table 4 shows the comparison of expected costs for our plan using Bayesian
inference and the ASP under Type I censoring in [12] which uses MLE of
$\vartheta$. The comparison results has revealed the importance of utilizing
Bayesian inference to minimize costs as the ETC involved in ASP under Type I
censoring in [12] is more than the ASPs discussed in this work. Thus, by
adopting Bayesian inference and incorporating SEL as the loss function, we can
better manage costs and improve the overall performance of our plan. Moreover,
a real life case study is conducted to explain the practical application of
the various ASPs discussed here. In terms of future research possibilities,
the application of Bayesian inference can be extended to the design of ASPs,
incorporating accelerated life testing, which helps to reduce the testing
time.

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# Forbidden conductors of $L$-functions and continued fractions of particular
form

J.KACZOROWSKI, A.PERELLI and M.Radziejewski

In memory of Professor Andrzej Schinzel

Abstract. In this paper we study the forbidden values of the conductor $q$ of
the $L$-functions of degree 2 in the extended Selberg class by a novel
technique, linking the problem to certain continued fractions and to their
weight $w_{q}$. Our basic result states that if an $L$ function with conductor
$q$ exists, then the weight $w_{q}$ is unique in a suitable sense. From this
we deduce several results, both of theoretical and computational nature.

Mathematics Subject Classification (2010): 11M41, 11A55

Keywords: Selberg class; converse theorems; continued fractions.

##  1\. Introduction

It is expected that the conductor $q$ of an $L$-function from the extended
Selberg class ${\mathcal{S}}^{\sharp}$, see Section 2 for definitions, can
attain only certain special values. For example, it is expected that the
$L$-functions in ${\mathcal{S}}^{\sharp}$ with degree 2 cannot have conductor
$q<1$, and that the $L$-functions in the Selberg class ${\mathcal{S}}$ always
have $q\in\mathbb{N}$. Both such expectations are far from being proved at
present.

In this paper we focus on $L$-functions of degree 2 in
${\mathcal{S}}^{\sharp}$ and investigate the problem of the admissible values
of their conductor $q$ via an unexpected link with certain continued fractions
$c(q,{\bf m})$, which we now define. Let $q>0$ be given. For a vector ${\bf
m}=(m_{0},\dots,m_{k})\in\mathbb{Z}^{k+1}$ with some $k\geq 0$ we set

$c(q,{\bf
m})=m_{k}+\cfrac{1}{qm_{k-1}+\cfrac{q}{qm_{k-2}+\cfrac{q}{\ddots\,+\cfrac{q}{qm_{0}}}}}.$
(1.1)

Here we assume that all denominators in (1.1) are non-zero. Such a vector $\bf
m$ is called a path for $q$, or simply a path.

Of course, (1.1) can be translated to the standard continued fraction notation
where all numerators are one and indices are in increasing order

$[a_{0},\dotsc,a_{k}]=a_{0}+\cfrac{1}{a_{1}+\cfrac{1}{\ddots\,+\cfrac{1}{a_{k}}}}\
\ \ ,\ \ \ a_{j}\in\begin{cases}\mathbb{Z},&2\mid j,\\\ q\mathbb{Z},&2\nmid
j.\end{cases}$

In this paper we use the notation $c(q,{\bf m})$, as it is a better fit to our
transformation formula for $L$-functions.

The fraction $c(q,{\bf m})$ and the path ${\bf m}$ are called proper if all
$m_{j}$, $j=0,\dots,k-1$, are non-zero. The proper fractions are those arising
naturally in connection with $L$-functions. The integer $k$ is the length of
the path, and clearly $c(q,{\bf m})=m_{0}$ for a path of length 0. The weight
$w_{q}({\bf m})$ of $c(q,{\bf m})$ is defined for a path of length $k\geq 1$
as

$\begin{split}w_{q}({\bf m})&=q^{k/2}\prod_{j=0}^{k-1}|c(q,{\bf m}_{j})|\\\
&=q^{k/2}\left|m_{k-1}+\cfrac{1}{qm_{k-2}+\cfrac{q}{\ddots\,+\cfrac{q}{qm_{0}}}}\right|\left|m_{k-2}+\cfrac{1}{\ddots\,+\cfrac{q}{qm_{0}}}\right|\cdots\left|m_{0}\right|,\end{split}$
(1.2)

where ${\bf m}_{j}=(m_{0},\dots,m_{j})$ for $0\leq j\leq k$. If $k=0$ we
simply write $w_{q}({\bf m})=1$. Note that the weight $w_{q}({\bf m})$ does
not depend on the last entry $m_{k}$, and that always $w_{q}({\bf m})>0$.
Moreover, we say that the weight $w_{q}$ is unique if $w_{q}({\bf
m})=w_{q}({\bf n})$ whenever $c(q,{\bf m})=c(q,{\bf n})$.

The main result of this paper reads as follows.

######  Theorem 1.

If there exists $F\in{\mathcal{S}}^{\sharp}$ of degree $2$ and conductor $q$,
then the weight $w_{q}$ is unique.

The proof of Theorem 1 is based on the properties of certain nonlinear twists
of $L$-functions and is given in Section 3.

We shall also prove, see Lemma 4 in Section 3, that the weight $w_{q}$ is
unique if and only if $w(q,{\bf m})=1$ for all proper fractions of type (1.1)
representing 0, i.e. such that $c(q,{\bf m})=0$. A fraction $c(q,{\bf m})$
representing 0 is called a loop; in such a case, the path ${\bf m}$ is also
called a loop. The loop $c(q,(0))$ is the trivial loop.

Examples. 1. Let $q=2/3$. Then one easily checks that the fraction
$c(2/3,(1,-1,-3))$, which has $k=2$, satisfies $c(2/3,(1,-1,-3))=0$. Moreover,
using the definition (1.2) we see that $w_{2/3}((1,-1,-3))=1/3\neq 1$. Hence,
in view of Theorem 1, there are no functions of degree 2 in
${\mathcal{S}}^{\sharp}$ with conductor $q=2/3$.

2. Sometimes loops can be quite long. For $q=7/2$ the sequence

$\mathbf{m}=(2,-5,-1,1,-1,1,-1,1,-1,1,2)$

is a loop with $k=10$ and $w_{7/2}(\mathbf{m})=8$. By solving the Diophantine
equation

$c(7/2,(m_{0},\dotsc,m_{k}))=0$

for $k<10$ we can see that there are no shorter loops of weight $\neq 1$ for
$q=7/2$. As before, we conclude that there are no functions of degree 2 in
${\mathcal{S}}^{\sharp}$ with conductor $q=7/2$.

3. Choose now $q=\sqrt{3}$ and ${\bf m}=(1,1,1,-1,1)$. A simple computation shows that

$c(\sqrt{3},{\bf m})=0\qquad\text{and}\qquad w_{\sqrt{3}}({\bf
m})=\sqrt{3}+2\neq 1,$

hence, again, there are no functions of degree 2 in ${\mathcal{S}}^{\sharp}$
with conductor $q=\sqrt{3}$.

4. Finally, let $q=2$ and ${\bf m}=(1,-1,1)$. In this case we have

$c(2,{\bf m})=0\qquad\text{and}\qquad w_{2}({\bf m})=1.$

The last equality is not surprising, as it is well known that there exist
$L$-functions in ${\mathcal{S}}^{\sharp}$ of degree 2 and conductor $q=2$, cf.
Lemma 6 with $m=4$. In fact, Theorem 1 tells us that $w_{2}({\bf m})=1$ not
only for ${\bf m}=(1,-1,1)$ but for every loop ${\bf
m}=(m_{0},\dots,m_{k})\in\mathbb{Z}^{k+1}$. ∎

As is clear from the above examples, Theorem 1 and Lemma 4 enable to prove
non-existence of $L$-functions of degree 2 with a given conductor $q$ by
producing a proper loop $c(q,{\bf m})$ with weight $w_{q}({\bf m})\neq 1$.
This problem is suitable for computations and, for example, in that way we
obtain the following result; see Section 4.

######  Corollary 1.

There exist no $L$-functions of degree $2$ in ${\mathcal{S}}^{\sharp}$ with
conductor of the form

$q=\frac{a}{nb}\ \text{with}\ (a,b)=1,\ 2\leq b\leq 300,\ n\geq 1$

if at least one of the following conditions holds

  * •

$b\leq 292$ and $a/b<1$,

  * •

$b\leq 150$ and $a/b<3/2$,

  * •

$b\leq 100$ and $a/b<2$,

  * •

$b\leq 30$ and $a/b<3$,

  * •

$b\leq 9$ and $a/b<4$,

  * •

$a\leq 25$ and $a/b<4$.

Remark. Actually, our computations were performed for $q=a/b$ with $a$, $b$
satisfying one of the conditions specified. The result in Corollary 1 follows
from these computations observing that if $q$ is a forbidden conductor then
$q/n$ is also forbidden for every integer $n\geq 1$. Suppose indeed that there
exists $F\in{\mathcal{S}}^{\sharp}$ of degree 2 with conductor $q/n$. Then
$FG$ has degree 2 and conductor $q$ for any $G\in{\mathcal{S}}^{\sharp}$ of
degree 0 and conductor $n$, a contradiction since such functions $G$ actually
exist; see [3]. Alternatively, Proposition 1 in Section 3 shows the link
between properties of fractions $c(q,{\mathbf{m}})$ and $c(q/n,{\mathbf{m}})$
without reference to $L$-functions. The second assertion follows from the same
computations and properties of loops described in Section 4. ∎

Theorem 1 justifies a closer study of continued fractions of type (1.1). There
are some natural questions to be asked about them. For instance, we would like
to know for which values of $q$ the representation of a real number $a$ in the
form $a=c(q,{\bf m})$, with $c(q,{\bf m})$ proper, is unique. For such, our
method of detecting forbidden conductors does not work, because, as Lemma 4
shows, the weight is necessarily unique. So an even more interesting problem
is to find all $q$ without the above uniqueness property. Among them, there
are $q$’s such that $w(q)$ is not unique, so Theorem 1 applies. Hence the
basic open question in this direction is to describe the set of such $q$
explicitly.

######  Theorem 2.

If $q>0$ is transcendental or $q\geq 4$, then every real number $a$ can be
represented as $a=c(q,{\bf m})$ with a proper fraction $c(q,{\bf m})$ at most
in one way. In particular, in this case the weight $w_{q}$ is unique.

Let $L(q)$ denote the set of loops for a given $q$.

######  Corollary 2.

If $q>0$ and $L(q)$ contains a loop of odd length, then $1/q$ is an algebraic
integer.

The proofs of these results do not lie particularly deep. In particular, they
do not depend on the theory of $L$-functions. Moreover, it shows that the
problem of the uniqueness of the weight $w_{q}$ is non-trivial for algebraic
$q<4$ only. The latter case is far more subtle. In contrast to the proof of
Theorem 2, our proof of the following result heavily depends on $L$-functions,
in particular on Theorem 1 and the Hecke theory of modular forms for the
triangle groups $G(\lambda)$. In passing we remark that Hecke’s theory shows
the existence of $L$-functions of degree 2 for every conductor $q\geq 4$, see
Lemma 6. Thus, although our method cannot detect forbidden conductors among
the values $q$ in Theorem 2, actually there are no forbidden conductors $q\geq
4$.

######  Theorem 3.

Let $q\in{\mathbb{R}}$ be a positive algebraic number. The weight $w_{q}$ is
unique in each of the following two cases:

i) $q$ has a Galois conjugate which is greater or equal to $4$; in other
words, for a certain $\sigma\in$ Gal$(\overline{\mathbb{Q}}/{\mathbb{Q}})$ we
have $q^{\sigma}\geq 4$;

ii) $q$ is a totally positive algebraic integer.

Remarks. 1. As remarked before, our proof of Theorem 3 depends on
$L$-functions, but its formulation does not. The case of $q^{\sigma}\geq 4$
also follows from Theorem 2. A natural problem is to give a proof of the
second case, independent of the theory of $L$-functions.

2. From Theorem 3, we know that the weight $w_{q}$ is unique for the following pair of Galois conjugated algebraic integers $q_{\pm}=(3\pm\sqrt{5})/2$. From Lemma 6 applied with $m=5$ we know that there exists an $L$-function in ${\mathcal{S}}^{\sharp}_{2}$ with conductor $q_{+}$. So, in that case, the uniqueness of $w_{q_{+}}$ follows from Theorem 1. In contrast, no $L$-function $F\in{\mathcal{S}}^{\sharp}_{2}$ with $q_{F}=q_{-}$ is known at present, and it is not clear if it exists at all. Analyzing the proof of Theorem 1, we see that the uniqueness of $w_{q}$ for $q=q_{F}$, $F\in{\mathcal{S}}^{\sharp}_{2}$, follows from consistency conditions imposed by the basic transformation formula, see Lemma 2, and hence implicitly by the functional equation of $F$. Thus the uniqueness of $w_{q_{-}}$ can be interpreted as the lack of obstacles for the existence of $F\in{\mathcal{S}}^{\sharp}_{2}$ with $q_{F}=q_{-}$.

3. All algebraic integers of the form

$q=4\cos^{2}(\pi\ell/m)\qquad(m\geq 3,1\leq\ell<m,(\ell,m)=1)$

are totally positive. Thus for such $q$’s the weight $w_{q}$ is unique. In
particular it shows that the set of algebraic $q$ for which $w_{q}$ is unique
is dense in the interval $(0,4)$. ∎

In the opposite direction we have the following theorem.

######  Theorem 4.

The weight $w_{q}$ is not unique for

$q=\frac{4}{n}\cos^{2}(\pi\ell/(2k+1)),\qquad k\geq
1,\;1\leq\ell<2k+1,\;(\ell,2k+1)=1,\;n\geq 2.$

In particular, there are no functions of degree $2$ in
${\mathcal{S}}^{\sharp}$ with such conductors.

We conclude with some open problems.

1. Construct an $L$-function $F\in{\mathcal{S}}_{2}^{\sharp}$ with conductor $q_{F}=(3-\sqrt{5})/2$ or show that it does not exist. Show that there exists a real number $q>0$ such that $w(q)$ is unique but there is no $F\in{\mathcal{S}}^{\sharp}_{2}$ with conductor $q$.

2. Show that the set of $q$ for which the weight $w_{q}$ is not unique is also dense in the interval $(0,4)$.

3. It follows from Theorems 2 and 3 that for every $q>0$ there exists a positive integer $n$ such that $w_{nq}$ is unique. The last question is if for every algebraic $q>0$ there exists a positive integer $n$ such that $w_{q/n}$ is not unique.

Acknowledgements. This research was partially supported by the Istituto
Nazionale di Alta Matematica, by the MIUR grant PRIN-2017 “Geometric,
algebraic and analytic methods in arithmetic” and by grant 2021/41/BST1/00241
“Analytic methods in number theory” from the National Science Centre, Poland.

##  2\. Definitions and basic requisites

Throughout the paper we write $s=\sigma+it$ and $\overline{f}(s)$ for
$\overline{f(\overline{s})}$. The extended Selberg class
${\mathcal{S}}^{\sharp}$ consists of non identically vanishing Dirichlet
series

$F(s)=\sum_{n=1}^{\infty}\frac{a(n)}{n^{s}},$

absolutely convergent for $\sigma>1$, such that $(s-1)^{m}F(s)$ is entire of
finite order for some integer $m\geq 0$, and satisfying a functional equation
of type

$F(s)\gamma(s)=\omega\overline{\gamma}(1-s)\overline{F}(1-s),$

where $|\omega|=1$ and the $\gamma$-factor

$\gamma(s)=Q^{s}\prod_{j=1}^{r}\Gamma(\lambda_{j}s+\mu_{j})$

has $Q>0$, $r\geq 0$, $\lambda_{j}>0$ and $\Re(\mu_{j})\geq 0$. The Selberg
class ${\mathcal{S}}$ is, roughly, the subclass of ${\mathcal{S}}^{\sharp}$ of
the functions having, in addition, an Euler product representation and
satisfying the Ramanujan conjecture. Note that the conjugate function
$\overline{F}$ has conjugated coefficients $\overline{a(n)}$, and clearly
$\overline{F}\in{\mathcal{S}}^{\sharp}$. We refer to the survey papers
[2],[4],[9],[10],[11],[12] for further definitions, examples and the basic
theory of the classes ${\mathcal{S}}^{\sharp}$ and ${\mathcal{S}}$.

Degree $d$, conductor $q$ and $\xi$-invariant $\xi_{F}$ of
$F\in{\mathcal{S}}^{\sharp}$ are defined as

$d=2\sum_{j=1}^{r}\lambda_{j},\qquad
q=(2\pi)^{d}Q^{2}\prod_{j=1}^{r}\lambda_{j}^{2\lambda_{j}},\qquad\xi_{F}=2\sum_{j=1}^{r}(\mu_{j}-1/2):=\eta_{F}+id\theta_{F}$

with $\eta_{F},\theta_{F}\in\mathbb{R}$. In this paper we deal mainly with
functions in ${\mathcal{S}}^{\sharp}$ of degree $d=2$; the subclass of such
functions is denoted by ${\mathcal{S}}^{\sharp}_{2}$.

For $\sigma>1$ and $F\in{\mathcal{S}}^{\sharp}$ with degree 2 and conductor
$q$ we consider the nonlinear twist

$F(s,\alpha,\beta)=\sum_{n=1}^{\infty}\frac{a(n)}{n^{s}}e(-\alpha
n-\beta\sqrt{n}),$ (2.1)

where $\alpha,\beta\in\mathbb{R}$ and $e(x)=e^{2\pi ix}$. Note that, according
to our notation above, we have

$\overline{F}(s,\alpha,\beta)=\overline{F(\overline{s},\alpha,\beta)}=\sum_{n=1}^{\infty}\frac{\overline{a(n)}}{n^{s}}e(\alpha
n+\beta\sqrt{n}).$

To avoid ambiguities, we also use the following notation when we consider a
nonlinear twist of the conjugate function $\overline{F}$

$(\overline{F})(s,\alpha,\beta)=\sum_{n=1}^{\infty}\frac{\overline{a(n)}}{n^{s}}e(-\alpha
n-\beta\sqrt{n}).$

Thanks to the periodicity of the complex exponential, for
$\alpha\in\mathbb{Z}$, the twist in (2.1) reduces to the standard twist.

$F(s,\beta)=\sum_{n=1}^{\infty}\frac{a(n)}{n^{s}}e(-\beta\sqrt{n}),$

and for $m\in\mathbb{Z}$, we have

$F(s,\alpha+m,\beta)=F(s,\alpha,\beta).$ (2.2)

Writing

$n_{\beta}=q\beta^{2}/4\quad\text{and}\quad a(n_{\beta})=0\ \text{if}\
n_{\beta}\not\in\mathbb{N},$

the spectrum of $F$ is defined as

$\text{Spec}(F):=\\{\beta>0:a(n_{\beta})\neq
0\\}=\Big{\\{}2\sqrt{m/q}:m\in\mathbb{N}\ \text{with}\ a(m)\neq 0\Big{\\}}.$
(2.3)

Moreover, for $\ell=0,1,\dots$ we write

$s_{\ell}=\frac{3}{4}-\frac{\ell}{2}\quad\text{and}\quad
s^{*}_{\ell}=s_{\ell}-i\theta_{F}.$

######  Lemma 1.

Let $\beta\neq 0$. Then the standard twist $F(s,\beta)$ is entire if
$|\beta|\not\in Spec(F)$, while for $|\beta|\in Spec(F)$ it is meromorphic on
$\mathbb{C}$ with at most simple poles at the points $s_{\ell}^{*}$. Moreover,
when $|\beta|\in Spec(F)$ the residue of $F(s,\beta)$ at $s=s_{0}^{*}$ does
not vanish.

We refer to [5], and [6] for this and other results on the standard twist.
Clearly

Spec$(\overline{F})$ = Spec$(F)$

and, since $\theta_{\overline{F}}=-\theta_{F}$, the possible poles of
$(\overline{F})(s,\beta)$ are at the points
$\overline{s_{\ell}^{*}}=s_{\ell}+i\theta_{F}$, and $\overline{s_{0}^{*}}$ is
again a simple pole.

######  Lemma 2.

Let $F\in{\mathcal{S}}^{\sharp}$ be of degree $2$ and conductor $q$, and let
$\alpha>0$ and $\beta\in\mathbb{R}$. Then

$F(s,\alpha,\beta)=e^{as+b}\
\overline{F}\left(s+2i\theta_{F},\frac{1}{q\alpha},-\frac{\beta}{\sqrt{q}\alpha}\right)+h(s)$
(2.4)

with certain $a\in\mathbb{R}$ and $b\in\mathbb{C}$, where $h(s)$ is
holomorphic for $\sigma>1/2$.

Since the explicit values of $a$ and $b$ are not specified, this is a less
precise form of the Lemma in [7], in the case where $F$ is suitably
normalized. Moreover, Lemma 2 follows by similar but more straightforward
arguments in the more general case stated in Lemma 2, where $\theta_{F}$ is
not necessarily vanishing. We will also need an analogous expression for
negative values of the first parameter in $F(s,\alpha,\beta)$, thus for
$\alpha>0$, we consider the twist $F(s,-\alpha,\beta)$ and note that.

$F(s,-\alpha,\beta)=\overline{(\overline{F})(\overline{s},\alpha,-\beta)}.$

Since the conductors of $\overline{F}$ and $F$ are equal, from Lemma 2 we
finally obtain that for $\alpha>0$

$F(s,-\alpha,\beta)=e^{as+b}\
(\overline{F})\left(s-2i\theta_{F},\frac{1}{q\alpha},\frac{\beta}{\sqrt{q}\alpha}\right)+h(s)$
(2.5)

with certain $a\in\mathbb{R}$ and $b\in\mathbb{C}$ and a function $h(s)$
holomorphic for $\sigma>1/2$.

In the next section, we shall use an argument based on repeated applications
of (2.4) and (2.5). Since what really matters in such an argument is only the
value of $1/(q\alpha)$ and $|\beta/(\sqrt{q}\alpha)|$, to simplify notation,
we denote by

$\widetilde{F}\left(s\pm
2i\theta_{F},\frac{1}{q\alpha},\pm\left|\frac{\beta}{\sqrt{q}\alpha}\right|\right)$
(2.6)

the right hand side of both (2.4) and (2.5). Clearly, for $\sigma>1/2$ the
function in (2.6) has the same singularities as the functions $\overline{F}$
or $(\overline{F})$ on the right hand side of (2.4) or (2.5).

##  3\. Proof of the theorems

3.1. Some properties of fractions and weights. We start with some initial
properties of the fractions and weights in (1.1) and (1.2), and we refer to
Section 4 for further discussions on this subject. Directly from the
definitions, we see that.

$c(q,{\bf m})=m_{k}+\frac{1}{qc(q,{\bf m}_{k-1})}\quad\text{and}\quad
w_{q}({\bf m})=\sqrt{q}|c(q,{\bf m}_{k-1})|w_{q}({\bf m}_{k-1}).$ (3.1)

Moreover, it is easy to check that for $k\geq 2$, we also have

$c(q,(m_{0},\dots,m_{j-1},0,m_{j+1},\dots,m_{k}))=c(q,(m_{0},\dots,m_{j-2},m_{j-1}+m_{j+1},m_{j+2},\dots,m_{k})).$
(3.2)

Thus, by repeated applications of (3.2) we can transform a path ${\bf m}$ to a
proper path ${\bf m^{*}}$ in such a way that $c(q,{\bf m})=c(q,{\bf m^{*}})$.
This process is called zero-skipping, and the notation ${\bf m^{*}}$ will be
used also later on. For example,

$c(q,(1,0,1))=c(q,(2))=2\quad\text{and}\quad c(q,(1,0,-1))=c(q,(0))=0.$

We also note that the zero-skipping process preserves the weight, namely
$w_{q}({\bf m})=w_{q}({\bf m^{*}})$ if ${\bf m}$ and ${\bf m^{*}}$ are as
above, since

$q^{3/2}\left|m_{j-1}+\frac{1}{0+\frac{q}{qm_{j+1}+\frac{q}{*}}}\right|\left|0+\frac{1}{qm_{j+1}+\frac{q}{*}}\right|\left|m_{j+1}+\frac{1}{*}\right|=\sqrt{q}\left|m_{j-1}+m_{j+1}+\frac{1}{*}\right|.$

Further, given two loops ${\bf m}=(m_{0},\dots,m_{k})$ and ${\bf
n}=(n_{0},\dots,n_{\ell})$ we define the composition of ${\bf m}$ and ${\bf
n}$ as

${\bf mn}=(m_{0},\dots,m_{k}+n_{0},\dots,n_{\ell}).$

Note that ${\bf mn}$ is a path and moreover

$c(q,(m_{0},\dots,m_{k}+n_{0},\dots,n_{j}))=c(q,{\bf
n_{j}})\quad\text{for}\quad j=0,\dots,\ell.$ (3.3)

Indeed, since ${\bf m}$ is a loop, we have
$c(q,(m_{0},\dots,m_{k}+n_{0}))=c(q,(n_{0}))$, hence (3.3) follows. In
particular we have that

$c(q,{\bf mn})=c(q,{\bf n}).$ (3.4)

As a consequence, we have that

if $c(q,{\bf m})$ and $c(q,{\bf n})$ are two loops then $c(q,{\bf mn})$ is
also a loop. (3.5)

Clearly, by zero-skipping, we may transform $c(q,{\bf mn})$ to a proper loop
with the same weight. Finally, we have that the weight is multiplicative with
respect to the composition of loops, namely, if $c(q,{\bf m})$ and $c(q,{\bf
n})$ are two loops, then

$w_{q}({\bf mn})=w_{q}({\bf m})w_{q}({\bf n}).$ (3.6)

Indeed, thanks to (1.2) and (3.3) we have

$\begin{split}w_{q}({\bf mn})&=q^{(k+\ell)/2}\prod_{j=0}^{k-1}|c(q,{\bf
m}_{j})|\prod_{j=0}^{\ell-1}|c(q,(m_{0},\dots,m_{k}+n_{0},\dots,n_{j}))|\\\
&=q^{k/2}\prod_{j=0}^{k-1}|c(q,{\bf
m}_{j})|q^{\ell/2}\prod_{j=0}^{\ell-1}|c(q,{\bf n}_{j})|=w_{q}({\bf
m})w_{q}({\bf n}).\end{split}$

It can be shown that proper loops form a group under composition (with zero-
skipping) and thus $w_{q}$ is a group homomorphism. The inverse of a loop
$\mathbf{m}=(m_{0},\dotsc,m_{k})$ is $(-m_{k},\dotsc,-m_{0})$.

######  Lemma 3.

Let $q>0$. If two proper fractions satisfy $c(q,{\bf m})=c(q,{\bf n})$, then
there is a proper loop $\bf u$ such that ${\bf m}=({\bf u}{\bf n})^{*}$. The
loop $\bf u$ is non-zero if and only if ${\bf m}\neq{\bf n}$.

Proof. Let ${\bf m}=(m_{0},\dots,m_{k})$ and ${\bf n}=(n_{0},\dots,n_{\ell})$
be such that

$c(q,{\bf m})=c(q,{\bf n}).$

If ${\bf m}={\bf n}$, the assertions are clear. Otherwise, we note that ${\bf
u^{\prime}}=(m_{0},\dots,m_{k}-n_{\ell},-n_{\ell-1},\dotsc,-n_{0})$ is a loop.
Indeed,

$c(q,(m_{0},\dots,m_{k-1},m_{k}-n_{\ell},-n_{\ell-1},\dotsc,-n_{j+1}))=c(q,{\bf
n}_{j}),\qquad j=\ell-1,\dotsc,0,$

by induction, and finally
$c(q,(m_{0},\dots,m_{k-1},m_{k}-n_{\ell},-n_{\ell-1},\dotsc,-n_{0}))=0$. Let
${\bf u}={\bf u^{\prime*}}$, where $*$ denotes zero-skipping, and let $j$ be
the number of times (3.2) was applied in this operation. i.e. the largest
integer such that

$m_{k-i}=n_{\ell-i},\qquad i=0,\dotsc,j-1.$

We have $j<\min(k+1,\ell+1)$, otherwise ${\bf u}$ would be a loop with the
first or last entry equal to zero, which is impossible. Hence

$\begin{split}({\bf u}{\bf
n})^{*}&=((m_{0},\dots,m_{k-j-1},m_{k-j}-n_{\ell-j},-n_{\ell-j-1},\dotsc,-n_{0})(n_{0},\dots,n_{\ell}))^{*}\\\
&=((m_{0},\dots,m_{k-j-1},m_{k-j},n_{\ell-j+1},\dots,n_{\ell}))^{*}\\\ &=(\bf
m)^{*}={\bf m}.\end{split}$

∎

######  Lemma 4.

Let $q>0$. The following statements are equivalent

i) the weight $w_{q}$ is unique;

ii) $w_{q}({\bf m})=1$ for every proper loop ${\bf m}$;

iii) $w_{q}({\bf m})=w_{q}({\bf n})$ for every non-trivial proper loops ${\bf
m}$ and $\bf n$.

Proof. The implication i) $\Rightarrow$ ii) follows from the convention that
$w_{q}((0))=1$ and $c(q,{\bf m})=0=c(q,(0))$ for every loop ${\bf m}$. The
implication ii) $\Rightarrow$ iii) is trivial. Now we assume iii) and prove
ii) first and then i). Let $c(q,{\bf m})$ be a non-trivial proper loop.
Consider the composition ${\bf mm}$ and recall that, thanks to (3.5),
$c(q,{\bf mm})$ is also a loop. Therefore by (3.4) and (3.6) applied with
${\bf n}={\bf m}$ we get

$w_{q}({\bf m})=w_{q}({\bf mm})=w_{q}({\bf m})^{2},$

thus $w_{q}({\bf m})=1$. This implies ii). The equality of fractions $c(q,{\bf
m})=c(q,{\bf n})$ implies ${\bf m}={\bf u}{\bf n}$ for some proper loop $\bf
u$, by Lemma 3. Hence

$w_{q}({\bf m})=w_{q}({\bf u})w_{q}({\bf n})=w_{q}({\bf n})$

by (3.6) and ii). This proves i). ∎

Recall that $L(q)$ denotes the set of loops for a given $q$.

######  Proposition 1.

Let $q>0$. Let $\mathbf{m}=(m_{0},\dotsc,m_{k})\in L(q)$ and $r,r^{\prime}$ be
rational numbers with $rr^{\prime}>0$ such that

$m^{\prime}_{j}=\begin{cases}rm_{j},&j\equiv k\pmod{2}\\\
r^{\prime}m_{j},&j\not\equiv k\pmod{2}\end{cases}$

are all integers. Moreover, let
$\mathbf{m^{\prime}}=(m^{\prime}_{0},\dotsc,m^{\prime}_{k})$. Then for
$q^{\prime}=\frac{q}{rr^{\prime}}$ we have $\mathbf{m^{\prime}}\in
L(q^{\prime})$ and

$w_{q^{\prime}}(\mathbf{m^{\prime}})=\begin{cases}w_{q}(\mathbf{m}),&2\mid
k,\\\ \sqrt{\frac{r}{r^{\prime}}}w_{q}(\mathbf{m}),&2\nmid k.\end{cases}$

Proof. This follows from the relation

$c(q^{\prime},\mathbf{m^{\prime}}_{j})=\begin{cases}rc(q,\mathbf{m}_{j}),&j\equiv
k\pmod{2},\\\ r^{\prime}c(q,\mathbf{m}_{j}),&j\not\equiv
k\pmod{2},\end{cases}$

for $j=0,\dotsc,k$. ∎

######  Corollary 3.

If $L(q)$ contains a loop $\mathbf{m}$ of odd length, then the weight
$w_{q/n}$ is not unique for any integer $n\geq 2$.

Proof. Let $q^{\prime}=\frac{q}{n}$. By Proposition 1 there exist
$\mathbf{m^{\prime}},\mathbf{m^{\prime\prime}}\in L(q^{\prime})$ with

$w_{q^{\prime}}(\mathbf{m^{\prime}})=\sqrt{n}w_{q}(\mathbf{m})\qquad\text{and}\qquad
w_{q^{\prime}}(\mathbf{m^{\prime\prime}})=\sqrt{\frac{1}{n}}w_{q}(\mathbf{m}),$

so at least one of $w_{q^{\prime}}(\mathbf{m^{\prime}})$,
$w_{q^{\prime}}(\mathbf{m^{\prime\prime}})$ is different from 1. The corollary
follows from Lemma 4. ∎

3.2. Proof of Theorem 1. Theorem 1 is an immediate consequence of Lemma 4 and
the following lemma.

######  Lemma 5.

Let $F\in{\mathcal{S}}^{\sharp}$ be of degree $2$ and conductor $q$. Then
$w_{q}({\bf m})=1$ for every proper loop $c(q,{\bf m})$.

Proof. Let $\beta\in$ Spec$(F)$; we use the notation in (1.1),(1.2) and (2.6).
By repeated applications of (2.2),(2.4) and (2.5) we obtain

$\begin{split}F(s,\beta)&=F(s,m_{0},\beta)\\\ &=\widetilde{F}\left(s\pm
2i\theta_{F},\frac{1}{qm_{0}},\pm\frac{\beta}{\sqrt{q}|m_{0}|}\right)\\\
&=\widetilde{F}\left(s\pm
2i\theta_{F},m_{1}+\frac{1}{qm_{0}},\pm\frac{\beta}{\sqrt{q}|m_{0}|}\right)\\\
&=\widetilde{F}\left(s+2n_{1}i\theta_{F},\frac{1}{qm_{1}+\frac{q}{qm_{0}}},\pm\frac{\beta}{q|(m_{1}+\frac{1}{qm_{0}})m_{0}|}\right)\\\
&=\widetilde{F}\left(s+2n_{1}i\theta_{F},m_{2}+\frac{1}{qm_{1}+\frac{q}{qm_{0}}},\pm\frac{\beta}{q|(m_{1}+\frac{1}{qm_{0}})m_{0}|}\right)\\\
&=\dots\\\ &=\widetilde{F}\left(s+2n_{k-1}i\theta_{F},c(q,{\bf
m}),\pm\frac{\beta}{w_{q}({\bf m})}\right)\end{split}$

where ${\bf m}=(m_{0},\dots,m_{k})$ and the $n_{j}$’s, $j=1,\dots,k-1$, are
certain integers.

If $c(q,{\bf m})$ is a non-trivial proper loop, then the above equation
reduces, essentially, to the equality of two standard twists; more precisely,
it becomes

$F(s,\beta)=\widetilde{F}\left(s+2n_{k-1}i\theta_{F},\pm\frac{\beta}{w_{q}({\bf
m})}\right).$

Since $\beta\in$ Spec$(F)$, both sides must have a simple pole at
$s=s_{0}^{*}$ and hence by Lemma 1 we have that $\beta/w_{q}({\bf m})\in$
Spec$(F)$ as well. Moreover, Lemma 1 implies that
$s_{0}^{*}+2n_{k-1}i\theta_{F}$ must be either $s_{0}^{*}$ or
$\overline{s_{0}^{*}}$. Thus the opposite implication holds as well, namely if
$\beta/w_{q}({\bf m})\in$ Spec$(F)$ then $\beta\in$ Spec$(F)$. Therefore we
have that

$\text{$\beta\in$ Spec$(F)$ $\Longleftrightarrow$ $\beta/w_{q}({\bf m})\in$
Spec$(F)$}.$

This, in view of the shape of Spec$(F)$ in (2.3), implies that $w_{q}({\bf
m})=1$, and the lemma follows. ∎

3.3. Proof of Theorem 2 and its corollary. For a proper path ${\bf
m}=(m_{0},\ldots,m_{k})$ we define the rational function

$R(x,{\bf m})=m_{k}+\cfrac{1}{xm_{k-1}+\cfrac{x}{\ddots+\cfrac{x}{xm_{0}}}}.$
(3.7)

Moreover, we define the polynomials $P_{\ell}(x,{\bf m})$ and $Q_{\ell}(x,{\bf
m})$, $0\leq\ell\leq k$, inductively as

$P_{0}(x,{\bf m})\equiv m_{0},\qquad Q_{0}(x,{\bf m})\equiv 1$ (3.8)

and for $1\leq\ell\leq k$

$P_{\ell}(x,{\bf m})=m_{\ell}xP_{\ell-1}(x,{\bf m})+Q_{\ell-1}(x,{\bf
m}),\qquad Q_{\ell}(x,{\bf m})=xP_{\ell-1}(x,{\bf m}).$ (3.9)

Then we have

$R(x,{\bf m})=\frac{P_{k}(x,{\bf m})}{Q_{k}(x,{\bf m})}.$ (3.10)

By a trivial induction we show that

$\deg P_{\ell}=\deg Q_{\ell},\qquad 0\leq\ell<k.$ (3.11)

Now we show that two rational functions of the above type, say $R(x,{\bf m})$
and $R(x,{\bf n})$ with ${\bf m}=(m_{0},\ldots,m_{k})$ and ${\bf
n}=(n_{0},\ldots,n_{\ell})$, coincide if and only if ${\bf m}={\bf n}$.
Sufficiency is trivial, and so is necessity for $k=\ell=0$. Suppose first that
$k,\ell>0$. Then

$R(x,{\bf m})=m_{k}+\frac{1}{xR(x,{\bf m}_{k-1})}\quad\text{and}\quad R(x,{\bf
n})=n_{l}+\frac{1}{xR(x,{\bf n}_{\ell-1})}.$ (3.12)

By (3.11) we have $R(x,{\bf m})\asymp 1$ and $R(x,{\bf n})\asymp 1$ as
$|x|\to\infty$. Thus (3.12) gives

$m_{k}=n_{\ell}+O\left(\frac{1}{|x|}\right)$

as $|x|\to\infty$ and hence $m_{k}=n_{\ell}$. Again by (3.12) we have
$R(x,{\bf m}_{k-1})=R(x,{\bf n}_{\ell-1})$, therefore by induction we conclude
that $R(x,{\bf m})=R(x,{\bf n})$ implies ${\bf m}={\bf n}$. Finally, if $k>0$
and $\ell=0$ (or viceversa) then

$R(x,{\bf m})=m_{k}+\frac{1}{xR(x,{\bf m}_{k-1})}=n_{0},$

a contradiction proving our assertion in this case as well.

After this preparation we can conclude the proof. Suppose that a real number
$a$ has two different representations as a proper fraction with transcendental
parameter $q$. Then

$a=R(q,{\bf m})=R(q,{\bf n})$

for two different proper paths ${\bf m}=(m_{0},\ldots,m_{k})$ and ${\bf
n}=(n_{0},\ldots,n_{\ell})$. Since the rational functions $R(x,{\bf m})$ and
$R(x,{\bf n})$ are distinct, we deduce that the polynomial

$H(x):=P_{k}(x,{\bf m})Q_{l}(x,{\bf n})-P_{\ell}(x,{\bf n})Q_{k}(x,{\bf
m})\in{\mathbb{Z}}[x]$

is not identically vanishing and moreover $H(q)=0$. This is impossible if $q$
is transcendental.

Suppose now $q\geq 4$ and suppose ${\bf m}=(m_{0},\ldots,m_{k})$ is the
shortest non-zero proper path such that $\left|c(q,{\bf m})\right|\leq 1/2$.
We have $\left|m_{0}\right|\geq 1$, so $k\neq 0$. Since $\left|c(q,{\bf
m}_{k-1})\right|>1/2$, we have

$\left|\frac{1}{qc(q,{\bf m}_{k-1})}\right|<\frac{1}{2}.$

From $m_{k}=c(q,{\bf m})-\frac{1}{qc(q,{\bf m}_{k-1})}$ we obtain
$\left|m_{k}\right|<1$, contradicting $m_{k}\neq 0$. Therefore there is no
non-zero proper path ${\bf m}$ such that $\left|c(q,{\bf m})\right|\leq 1/2$.
In particular, there is no non-zero proper loop. The assertion follows from
Lemma 3.∎

To show the corollary we note that, by Theorem 2, the number $q$ is algebraic.
Let

$a_{n}x^{n}+\dotsc+a_{1}x+a_{0},$

be its minimal polynomial, where $a_{0},\dotsc,a_{n}$ are integers with
$\gcd(a_{0},\dotsc,a_{n})=1$. For $k=2l+1$ we have
$P_{2l+1}(x,\mathbf{m})=x^{l}(1+xH(x))$, where $H(x)$ is some polynomial with
integer coefficients dependent on $\mathbf{m}$. If $L(q)$ contains a loop of
length $k$, we have $P_{2l+1}(x,\mathbf{m})=0$. This implies that the minimal
polynomial of $q$ divides $1+xH(x)$. Hence $\left|a_{0}\right|=1$. ∎

3.4. Proof of Theorem 3. We need two further lemmas.

######  Lemma 6.

Let $q\in\mathbb{R}$ be either of the form $q=4\cos^{2}(\pi/m)$ with $m\geq 3$
or $q\geq 4$. Then there exists an $L$-function in ${\mathcal{S}}^{\sharp}$
with degree $2$ and conductor $q$.

Proof. From the classical Hecke theory we know that there are non-trivial
automorphic forms $f$ for the Hecke trangle group $G(\lambda)$ if $\lambda\geq
2$ or $\lambda=2\cos(\pi/m)$ with integer $m\geq 3$, see e.g. [1]. The
corresponding normalized $L$-function $L_{f}$ satisfies the functional
equation

$\left(\frac{\lambda}{2\pi}\right)^{s}\Gamma(s+\frac{k-1}{2})L_{f}(s)=\omega\left(\frac{\lambda}{2\pi}\right)^{1-s}\Gamma(1-s+\frac{k-1}{2})L_{f}(1-s),$

where $\omega=\pm 1$. We cannot claim that $L_{f}$ belongs to
${\mathcal{S}}^{\sharp}$ because conjugation of $L_{f}(1-s)$ is missing in the
above functional equation. This however can easily be repaired. Without loss
of generality we may assume that $L_{f}(s)$ has at least one coefficient with
a non-zero real part, otherwise we consider $iL_{f}(s)$. Then
$F(s):=L_{f}(s)+\overline{L_{f}}(s)$ has real coefficients, satisfies a
functional equation of the right type and is not identically zero. Thus it
belongs to ${\mathcal{S}}^{\sharp}$, has degree 2 and its conductor equals
$\lambda^{2}$; therefore the lemma follows. ∎

######  Lemma 7.

Let $\alpha$ be a totally positive algebraic integer with all conjugates
smaller than $4$. Then there exist positive integers $m$ and $\ell$ satisfying
$m\geq 3$, $1\leq\ell<m$, $(\ell,m)=1$ such that
$\alpha=4\cos^{2}(\pi\ell/m)$.

Proof. Let $\beta:=\sqrt{\alpha}$. Then $\beta$ is a totally real algebraic
integer with all Galois conjugates in absolute value smaller than $2$. By the
Kronecker theorem, $\beta=2\cos(\pi\ell/m)$ for certain positive coprime
integers $\ell$ and $m$; see Theorem 2.5 of [8]. Thus
$\alpha=4\cos^{2}(\pi\ell/m)$. Since $0<\alpha<4$, we have $m\geq 3$.
Moreover, by the periodicity of $\cos^{2}x$, we can assume that $1\leq\ell<m$,
and the lemma follows. ∎

Let $q\in\overline{\mathbb{Q}}$ be a positive algebraic number and suppose
that ${\bf m}=(m_{0},\ldots,m_{k})\in{\mathbb{Z}}^{k+1}$ is a non-trivial loop
for $q$, namely

$m_{k}+\cfrac{1}{qm_{k-1}+\cfrac{q}{\ddots+\cfrac{q}{qm_{0}}}}=0.$

Then for every $\sigma\in$ Gal$(\overline{\mathbb{Q}}/{\mathbb{Q}})$ we have

$m_{k}+\cfrac{1}{q^{\sigma}m_{k-1}+\cfrac{q^{\sigma}}{\ddots+\cfrac{q^{\sigma}}{q^{\sigma}m_{0}}}}=0$

so that $\bf m$ is a loop for $q^{\sigma}$ as well. Moreover, it is easy to
check that $w_{q}({\bf m})=1$ if and only if $w_{q^{\sigma}}({\bf m})=1$.
Hence we conclude that the weights $w_{q}$ and $w_{q^{\sigma}}$ are
simultaneously unique or not.

We can now complete the proof of Theorem 3. If $q^{\sigma}\geq 4$ for a
certain $\sigma\in$ Gal$(\overline{\mathbb{Q}}/{\mathbb{Q}})$ then by Lemma 6
there exists an $L$-function in ${\mathcal{S}}_{2}^{\sharp}$ with conductor
$q^{\sigma}$, and the weight $w_{q^{\sigma}}$ is unique by Theorem 1.
Consequently $w_{q}$ is unique as well, thus proving i). To show ii) we assume
that $q$ is a totally positive algebraic integer with all conjugates in the
interval $(0,4)$. By Lemma 7 this means that $q=4\cos^{2}(\pi\ell/m)$ for
certain $m\geq 3$ and $1\leq\ell<m$, $(m,\ell)=1$. Let $\sigma\in$
Gal$(\overline{\mathbb{Q}}/{\mathbb{Q}})$ be such that $\sigma:\exp(2\pi
i\ell/m)\mapsto\exp(2\pi i/m)$. Then $q^{\sigma}=4\cos^{2}(\pi/m)$. According
to Lemma 6 there exists an $L$-function in ${\mathcal{S}}^{\sharp}_{2}$ with
such a conductor, thus $w_{q^{\sigma}}$ is unique according to Theorem 1.
Consequently $w_{q}$ is unique as well, and the proof is complete. ∎

3.5. Proof of Theorem 4. We need the following explicit expression for the
polynomials $P_{\ell}(x,{\bf m})$ defined in the proof of Theorem 2. The proof
of such an expression consists of a straightforward induction, which we omit.

######  Lemma 8.

Let $k\geq 1$ and $\mathbf{m}=(m_{0},\dotsc,m_{2k})$. Then

$P_{2k-1}(x,\mathbf{m})=x^{k-1}\sum_{j=0}^{k}\left(\sum_{A\in
I(2k-1,2j-1)}\prod_{i\in A}m_{i}\right)x^{j}$

and

$P_{2k}(x,\mathbf{m})=x^{k}\sum_{j=0}^{k}\left(\sum_{A\in I(2k,2j)}\prod_{i\in
A}m_{i}\right)x^{j},$

where $I(h,j)$ denotes the set of subsets
$\left\\{a_{0},\dotsc,a_{j}\right\\}$ of $\left\\{0,\dotsc,h\right\\}$ such
that $a_{0}<\dotsc<a_{j}$ and $a_{i}\equiv i\pmod{2}$ for every
$i=0,\dotsc,j$.

Let $k$ and $\ell$ be as in Theorem 4 and

$q=4\cos^{2}(\pi\ell/(2k+1))=\left(e\left(\frac{\ell}{4k+2}\right)+e\left(-\frac{\ell}{4k+2}\right)\right)^{2}.$
(3.13)

By Corollary 3 it suffices to show that $L(q)$ contains a loop of odd length.
Let

$\mathbf{m}=(m_{0},\dotsc,m_{2k-1}),\qquad m_{j}=(-1)^{j},\ j=0,\dotsc,2k-1.$

By Lemma 8 we have

$\displaystyle q^{-k+1}P_{2k-1}(q,\mathbf{m})$
$\displaystyle=\sum_{j=0}^{k}\left(\sum_{A\in I(2k-1,2j-1)}\prod_{i\in
A}(-1)^{i}\right)\left(e\left(\frac{\ell}{4k+2}\right)+e\left(-\frac{\ell}{4k+2}\right)\right)^{2j}$
$\displaystyle=\sum_{j=0}^{k}(-1)^{j}\left|I(2k-1,2j-1)\right|\sum_{m=-j}^{j}\binom{2j}{j+m}e\left(\frac{\ell
m}{2k+1}\right)$ $\displaystyle=\sum_{m=-k}^{k}e\left(\frac{\ell
m}{2k+1}\right)\sum_{j=\left|m\right|}^{k}(-1)^{j}\left|I(2k-1,2j-1)\right|\binom{2j}{j+m}.$

The subsets $\left\\{a_{0},\dotsc,a_{2j-1}\right\\}\in I(2k-1,2j-1)$
correspond one-to-one to the subsets

$\left\\{b_{0},\dotsc,b_{2j-1}\right\\}\subseteq\left\\{0,\dotsc,k+j-1\right\\}$

by the mapping $b_{i}=(a_{i}+i)/2$, so
$\left|I(2k-1,2j-1)\right|=\binom{k+j}{2j}$. Hence from the identity

$\sum_{i=0}^{m}(-1)^{i}\binom{n-i}{n-2i}\binom{n-2i}{m-i}=1,\qquad n\geq 0,\
0\leq 2m\leq n,$

which can be shown by induction, it follows that

$\displaystyle q^{-k+1}P_{2k-1}(q,\mathbf{m})$
$\displaystyle=\sum_{m=-k}^{k}e\left(\frac{\ell
m}{2k+1}\right)\sum_{j=\left|m\right|}^{k}(-1)^{j}\binom{k+j}{2j}\binom{2j}{j+m}$
$\displaystyle=\sum_{m=-k}^{k}e\left(\frac{\ell
m}{2k+1}\right)\sum_{j=0}^{k-\left|m\right|}(-1)^{k-j}\binom{2k-j}{2k-2j}\binom{2k-2j}{k-m-j}$
$\displaystyle=(-1)^{k}\sum_{m=-k}^{k}e\left(\frac{\ell m}{2k+1}\right)=0.$

Thus for $q$ as in (3.13) there exists an integer vector ${\bf m}$ of odd
length $2k-1$ with $P_{2k-1}(q,{\bf m})=0$; however, ${\bf m}$ may not be a
path for $q$.

Now let $\mathbf{n}=(n_{0},\dotsc,n_{2j-1})\in\mathbf{Z}^{2j}$ be such that
$P_{2j-1}(q,\mathbf{n})=0$ with the smallest possible $j$. In view of (3.7)
and (3.10), see also Proposition 2 in the next section, if $\mathbf{n}$ is not
a path for $q$ we have $P_{i}(q,\mathbf{n})=0$ for some $i<2j-1$, and $i$ is
even by the minimality of $j$. It follows from (3.8) and (3.9) that
$Q_{i}(q,\mathbf{n})\neq 0$, and moreover that
$P_{i+1}(q,\mathbf{n})=Q_{i}(q,\mathbf{n})$. Hence $i+1<2j-1$, so $i<2j-3$.
Further we have

$Q_{i+1}(q,\mathbf{n})=0,\quad
P_{i+2}(q,\mathbf{n})=n_{i+2}qQ_{i}(q,\mathbf{n})=cP_{0}(q,\mathbf{n^{\prime}}),\quad
Q_{i+2}(q,\mathbf{n})=qQ_{i}(q,\mathbf{n})=cQ_{0}(q,\mathbf{n^{\prime}}),$

where $\mathbf{n^{\prime}}=(n_{i+2},\dotsc,n_{2j-1})$ and
$c=qQ_{i}(q,\mathbf{n})\neq 0$. Consequently, using (3.9) again, we have

$P_{i+2+h}(q,\mathbf{n})=cP_{h}(q,\mathbf{n^{\prime}})\quad\text{for}\
h=1,\dotsc,2j-i-3,$

and in particular

$P_{2j-1}(q,\mathbf{n})=cP_{2j-i-3}(q,\mathbf{n^{\prime}}).$

Hence $P_{2j-i-3}(q,\mathbf{n^{\prime}})=0$, contrary to the minimality of
$j$. Therefore ${\bf n}$ is a path, and also a loop, of odd length; the
theorem now follows. ∎

##  4\. Computations

With the aid of machine computations we have been able to find loops of weight
$\neq 1$ for rational $q=\frac{a}{b}$, $(a,b)=1$, $0<q<4$, in each of the
following cases:

  * •

$a\leq 25$, arbitrary $b$;

  * •

$b\leq 292$ and $q<1$;

  * •

$b\leq 150$ and $q<3/2$;

  * •

$b\leq 100$ and $q<2$;

  * •

$b\leq 30$ and $q<3$;

  * •

$b\leq 9$.

An excerpt from the results is shown in Tables 1 and 2. Complete results and
the Python scripts needed to reproduce them are available online at
https://maciejr.web.amu.edu.pl/computations/conductors

$q$ | $\mathbf{m}$ | $w_{q}(\mathbf{m})$
---|---|---
$\frac{1}{2}$ | $(1,-2)$ | $\sqrt{\frac{1}{2}}$
$\frac{3}{2}$ | $(-1,1,-2)$ | $\frac{1}{2}$
$\frac{5}{2}$ | $(-1,1,-1,1,2)$ | $\frac{1}{4}$
$\frac{7}{2}$ | $(2,1,-1,1,-1,1,-1,1,-1,-5,2)$ | $\frac{1}{8}$
$\frac{1}{3}$ | $(1,-3)$ | $\sqrt{\frac{1}{3}}$
$\frac{2}{3}$ | $(1,-1,-3)$ | $\frac{1}{3}$
$\frac{4}{3}$ | $(-1,1,-3)$ | $\frac{1}{3}$
$\frac{5}{3}$ | $(-1,1,-1,-1,-3)$ | $\frac{1}{9}$
$\frac{7}{3}$ | $(-1,1,-1,2,-1,-1,3)$ | $\frac{1}{27}$
$\frac{8}{3}$ | $(1,-1,1,-1,6)$ | $\frac{1}{9}$
$\frac{10}{3}$ | $(2,-1,1,-1,1,-1,1,-5,15)$ | $\frac{1}{81}$
$\frac{11}{3}$ | $(-1,-1,-1,1,-1,1,-1,1,-1,1,-1,1,-30,1,-8)$ | $\frac{1}{243}$
$\frac{1}{4}$ | $(1,-4)$ | $\sqrt{\frac{1}{4}}$
$\frac{3}{4}$ | $(-1,1,4)$ | $\frac{1}{4}$
$\frac{5}{4}$ | $(-1,1,-4)$ | $\frac{1}{4}$
$\frac{7}{4}$ | $(1,-1,1,2,-2)$ | $\frac{1}{8}$
$\frac{9}{4}$ | $(-1,1,-1,2,2)$ | $\frac{1}{8}$
$\frac{11}{4}$ | $(-1,1,-1,1,-2,-1,4)$ | $\frac{1}{32}$
$\frac{13}{4}$ | $(1,-1,1,-1,1,-1,36)$ | $\frac{1}{64}$
$\frac{15}{4}$ | $(-2,1,-1,1,-1,1,-1,1,-1,1,-1,1,-6,11,-1,8,-1)$ | $\frac{1}{4096}$
Table 1. Examples of loops for $q=a/b$ with $b\leq 4$. $a$ | $b$ | $N$ | $b^{\prime\prime}$ | $\bf m$ | $\bf n$
---|---|---|---|---|---
$3$ | $1$ | $3$ | $1$ | $(-1,0,2)$ | $(1)$
$4$ | $1$ | $4$ | $1$ | $(-1,0,2)$ | $(1)$
$5$ | $1$ | $5$ | $1$ | $(-1,0,2)$ | $(1)$
$5$ | $2$ | $10$ | $2$ | $(-2,0,3)$ | $(1)$
$5$ | $3$ | $10$ | none | $(-1,1,-1,-1,-3)$ | $(0)$
Table 2. Examples of families of loops of weight $\neq 1$ for
$q^{\prime}=a/b^{\prime}$ where $b^{\prime}\equiv\pm b$ (mod $N$), $b\neq
b^{\prime\prime}$, and $N=na$. For each family we list the data necessary to
apply Corollary 6: $a$, the residue $b$, the modulus $N$, and the paths
(denoted $\bf m$ and $\bf n$ in the corollary). The possible exceptional value
of $b^{\prime}$ in (4.1) is denoted as $b^{\prime\prime}$ here.

In our computations we make use of some observations that we state here
without complete proofs.

######  Proposition 2.

A sequence $\mathbf{m}\in\mathbf{Z}^{k+1}$ is a path for a given $q$ if and
only if $P_{l}(q,\mathbf{m})\neq 0$, $0\leq l<k$. In that case $\mathbf{m}$ is
a loop if and only if $P_{k}(q,\mathbf{m})=0$. Moreover,
$w_{q}(\mathbf{m})=\left|q^{-\frac{k}{2}}Q_{k}(q,\mathbf{m})\right|$.

######  Corollary 4.

For $q>0$ the set $L(q)$ contains a loop of length $1$ if and only if
$q=\frac{1}{b}$ for some positive integer $b$. In that case the weight $w_{q}$
is not unique unless $q=1$.

###### Proof.

Loops of length $1$ are solutions of $P_{1}(q,(m_{0},m_{1}))=0$, i.e.
$m_{0}m_{1}q+1=0$, so $q$ needs to be of the above form. In that case $(b,-1)$
is a loop of weight $\sqrt{b}$. ∎

######  Corollary 5.

For $q>0$ the set $L(q)$ contains a loop of length $2$ if and only if
$q=\frac{1}{u}+\frac{1}{v}$ for some non-zero integers $u,v$, with $u\neq-v$.
In that case the weight $w_{q}$ is not unique unless $q=1$ or $q=2$. In
particular, the weight is not unique whenever $q=\frac{2}{b}$ for some integer
$b\geq 3$.

###### Proof.

Loops of length $2$ are solutions of
$P_{2}(q,(m_{0},m_{1},m_{2}))=m_{0}m_{1}m_{2}q^{2}+(m_{0}+m_{2})q=0$, which
implies that $q$ is of the required form. Conversely, for
$q=\frac{1}{u}+\frac{1}{v}$ the sequence $\mathbf{m}=(u,-1,v)$ is always a
loop and $w_{q}(\mathbf{m})=\left|\frac{u}{v}\right|$, which is $\neq 1$
unless $v=u$. Suppose $v=u$, so $q=\frac{2}{u}$. If $2\mid u$, the assertion
follows from Corollary 4. Otherwise, unless $q=2$, we have $u=2l+1$ for some
positive integer $l$ and there is a loop $\mathbf{m^{\prime}}=(1,-l,-2l-1)$ of
weight $\frac{1}{2l+1}$. ∎

######  Proposition 3.

Let $\mathbf{m}=(m_{0},\dotsc,m_{k})$ be a path for some $q=\frac{a}{b}$,
where $a$ and $b$ are coprime positive integers. Let

$\frac{u_{j}}{v_{j}}=ac(q,\mathbf{m}_{j}),\qquad j=0,\dotsc,k$

be reduced fractions, in particular $u_{k}=0$ and $v_{k}=1$ if $\mathbf{m}\in
L(q)$. Then for arbitrary $\varepsilon_{j}=\pm 1$, $j=0,\dotsc,k$ and for
every positive $b^{\prime}$ satisfying
$b^{\prime}\equiv\varepsilon_{j}\varepsilon_{j+1}b\pmod{u_{j}}$,
$j=0,\dotsc,k-1$, the sequence
$\mathbf{m^{\prime}}=(m^{\prime}_{0},\dotsc,m^{\prime}_{k})$, where
$m_{0}^{\prime}=\varepsilon_{0}m_{0}$ and

$m_{j+1}^{\prime}=\varepsilon_{j+1}m_{j+1}+\frac{\varepsilon_{j+1}b-\varepsilon_{j}b^{\prime}}{u_{j}}v_{j},\qquad
j=0,\dotsc,k-1,$

satisfies $c(q^{\prime},\mathbf{m^{\prime}})=c(q,\mathbf{m})$ and
$w_{q^{\prime}}(\mathbf{m^{\prime}})=\left(\frac{b}{b^{\prime}}\right)^{k/2}w_{q}(\mathbf{m})$,
where $q^{\prime}=\frac{a}{b^{\prime}}$.

###### Proof.

It suffices to show that
$ac(q^{\prime},\mathbf{m^{\prime}}_{j})=\frac{\varepsilon_{j}u_{j}}{v_{j}}$,
$j=0,\dotsc,k$. Indeed, we have
$\frac{u_{0}}{v_{0}}=am_{0}=\varepsilon_{0}am_{0}^{\prime}$ and

$\displaystyle\frac{u_{j+1}}{v_{j+1}}$
$\displaystyle=\frac{am_{j+1}u_{j}+abv_{j}}{u_{j}}$
$\displaystyle=\pm\frac{am_{j+1}^{\prime}\varepsilon_{0}u_{j}+ab^{\prime}v_{j}}{\varepsilon_{0}u_{j}}.$

for $j=0,\dotsc,k-1$. ∎

######  Corollary 6.

Let $\mathbf{m}=(m_{0},\dotsc,m_{k})$ and $\mathbf{n}=(n_{0},\dotsc,n_{l})$ be
such that

$c(q,\mathbf{m})=c(q,\mathbf{n}),\qquad\text{for some $q=\frac{a}{b}$, where
$(a,b)=1$.}$

Let $\frac{u_{j}}{v_{j}}=ac(q,\mathbf{m}_{j})$, $j=0,\dotsc,k-1$,
$\frac{x_{j}}{y_{j}}=ac(q,\mathbf{n}_{j})$, $j=0,\dotsc,l-1$, be reduced
fractions and let $N$ be a positive integer such that

$u_{j}\mid N,\quad j=0,\dotsc,k-1,\qquad\text{and}\qquad x_{j}\mid N,\quad
j=0,\dotsc,l-1.$

If $k\neq l$, then $w_{q^{\prime}}$ is non-unique for every
$q^{\prime}=\frac{a}{b^{\prime}}$ such that

$b^{\prime}\equiv\pm b\pmod{N},\qquad
b^{\prime}\neq\left(\frac{w_{q}(\mathbf{m})}{w_{q}(\mathbf{n})}\right)^{2/(k-l)}b.$
(4.1)

If $k=l$ and $w_{q}(\mathbf{m})\neq w_{q}(\mathbf{n})$, then $w_{q^{\prime}}$
is non-unique for every $q^{\prime}=\frac{a}{b^{\prime}}$ such that

$b^{\prime}\equiv\pm b\pmod{N}.$

It also follows from Proposition 1 with $r=r^{\prime}=-1$ and the fact that
$w_{q}$ is a group homomorphism that $\mathbf{m}=(m_{0},\dotsc,m_{k})$ is a
loop with weight $\neq 1$ if and only if $(-m_{0},\dotsc,-m_{k})$ and
$(m_{k},\dotsc,m_{0})$ are such loops.

The computations for $q=a/b$ employed several methods, depending on the case
being considered; below is the list of the methods. In the complete table
available online, each example of loop is labelled with the number of the
method by which it was obtained. This number also corresponds to the script
number.

  1. (1)

The case $q=\frac{1}{b}$ was handled using Corollary 4.

  2. (2)

For $a=3,\dotsc,25$ and $b$ relatively prime to $a$ in all possible congruence
classes mod $na$, starting with $n=1$, a search for paths satisfying the
assumptions of Corollary 6, with $N=an$, was performed. (The cases $a=1$ and
$a=2$ follow from Corollaries 4 and 5 respectively). Examples of paths and
congruence classes that we have found are shown in Table 2. Exceptions
(asserted in the corollary) where noted and later handled by subsequent
methods, with the results stored in the full version of Table 1 (online).
Whenever appropriate paths were not found, classes mod a higher modulus had to
be considered, either by incrementing $n$, or by splitting the current class
mod $an$ to classes mod $ann^{\prime}$ with the smallest possible
$n^{\prime}$. The decision to increment $n$ or split the class was based on
how many residue classes mod $an$ were already successfully handled. Covering
the next case, $a=26$, with the current algorithm, would probably require
around 2 months of machine time.

  3. (3)

For a given $q=a/b$ all possible loops of a given length may be found by
solving the Diophantine equation $P_{k}(q,(m_{0},\dotsc,m_{k}))=0$ in non-
zero, integer $m_{j}$. This was mainly done recursively by

     * •

finding an upper bound $M$ for $\min\left|m_{j}\right|$,

     * •

checking all possible cases of $i=0,\dotsc,k$,
$|m_{i}|=\min\left|m_{j}\right|\leq M$,

     * •

substituting possible values $|m_{i}|\leq M$ and then solving each case.

For example, for $q=26/23$ and length $6$ the equation to solve is

$2^{3}\cdot 13^{3}m_{0}m_{1}m_{2}m_{3}m_{4}m_{5}m_{6}+2^{2}\cdot 13^{2}\cdot
23(m_{0}m_{1}m_{2}m_{3}m_{4}+m_{0}m_{1}m_{2}m_{3}m_{6}\\\
+m_{0}m_{1}m_{2}m_{5}m_{6}+m_{0}m_{1}m_{4}m_{5}m_{6}+m_{0}m_{3}m_{4}m_{5}m_{6}+m_{2}m_{3}m_{4}m_{5}m_{6})\\\
+2\cdot 13\cdot
23^{2}(m_{0}m_{1}m_{2}+m_{0}m_{1}m_{4}+m_{0}m_{1}m_{6}+m_{0}m_{3}m_{4}+m_{0}m_{3}m_{6}+m_{0}m_{5}m_{6}\\\
+m_{2}m_{3}m_{4}+m_{2}m_{3}m_{6}+m_{2}m_{5}m_{6}+m_{4}m_{5}m_{6})+23^{3}(m_{0}+m_{2}+m_{4}+m_{6})=0.$

This implies $\min|m_{j}|\leq 2$ which reduces to $m_{j}\in\\{1,2\\}$ for at
least one $j$ equal to $0$, $1$, $2$ or $3$. Each of these 8 substitutions
produces an equation in $6$ variables (of degree $6$) which can be solved by
applying the same method recursively. Ultimately we reach 3384779 subcases
involving a quadratic equation in $2$ variables, for which a dedicated
algorithm was used. Searching for longer loops requires solving much more
complex equations, e.g., those corresponding to length $10$ would not fit on
one page.

The computational complexity of this method is hard to estimate. Roughly, it
behaves like $(1+q^{-1})^{k!}$, but it is much less regular. The unique
feature of this method is that it also allows us to prove that loops of a
given length do not exist, and thus that a loop of the next possible length
(e.g., found with another method) has the smallest possible length.

This method was applied for $3\leq a\leq b\leq 300$ and loops of length $2$,
$4$ and, sometimes, longer loops. Results for a given $q$ also showed the
existence of loops of weight $\neq 1$ for other $q$, in accordance with
Propositions 1 and 3, allowing us to avoid the direct application of the
method for many of the smaller $q$, where the computation time would be
particularly long.

  4. (4)

Where previous methods were unsuccessful, loops of weight $\neq 1$ were found
by examining sequences constructed using the following simple heuristic:

     * •

start with $m_{0}=1$ or $m_{0}=b$;

     * •

given $m_{0},\dotsc,m_{k}$, consider several possible values of $m_{k+1}$ such
that

$\left|c(q,(m_{0},\dotsc,m_{k+1}))\right|<C$

for some fixed $C>0$;

     * •

if more than $N$ paths were generated (where $N$ is around $10^{7}$) discard
those with largest numerators.

Direct application of Method 3 allows us to exclude the existence of loops of
a given length and weight $\neq 1$ for a given $q$. This way we were able to
check that many of the results in the full table available online are optimal,
in the sense that there are no shorter loops of weight $\neq 1$ for such $q$.

## References

  * [1] E.Hecke - Lectures on Dirichlet Series, Modular Functions and Quadratic Forms \- Vanderhoeck $\&$ Ruprecht 1983.
  * [2] J.Kaczorowski - Axiomatic theory of $L$-functions: the Selberg class \- In Analytic Number Theory, C.I.M.E. Summer School, Cetraro (Italy) 2002, ed. by A.Perelli and C.Viola, 133–209, Springer L.N. 1891, 2006.
  * [3] J.Kaczorowski, A.Perelli - On the structure of the Selberg class, I: $0\leq d\leq 1$ \- Acta Math. 182 (1999), 207–241.
  * [4] J.Kaczorowski, A.Perelli - The Selberg class: a survey \- In Number Theory in Progress, Proc. Conf. in Honor of A.Schinzel, ed. by K.Györy et al., 953–992, de Gruyter 1999.
  * [5] J.Kaczorowski, A.Perelli - On the structure of the Selberg class, VI: non-linear twists \- Acta Arith. 116 (2005), 315–341.
  * [6] J.Kaczorowski, A.Perelli - Twists and resonance of $L$-functions, I \- J. European Math. Soc. 18 (2016), 1349–1389.
  * [7] J.Kaczorowski, A.Perelli - A weak converse theorem for degree $2$ $L$-functions with conductor $1$ \- Proc. Res. Inst. Math. Sci. Kyoto 53 (2017), 337–347.
  * [8] W.Narkiewicz - Elementary and analytic theory of algebraic numbers, third edition - Springer Verlag 2004.
  * [9] A.Perelli - A survey of the Selberg class of $L$-functions, part I \- Milan J. Math. 73 (2005), 19–52.
  * [10] A.Perelli - A survey of the Selberg class of $L$-functions, part II \- Riv. Mat. Univ. Parma (7) 3* (2004), 83–118.
  * [11] A.Perelli - Non-linear twists of $L$-functions: a survey \- Milan J. Math. 78 (2010), 117–134.
  * [12] A.Perelli - Converse theorems: from the Riemann zeta function to the Selberg class \- Boll. U.M.I. 10 (2017), 29–53.

Jerzy Kaczorowski, Faculty of Mathematics and Computer Science, Adam
Mickiewicz University, Poznań, 61-614 Poznań, Poland and Institute of
Mathematics of the Polish Academy of Sciences, 00-956 Warsaw, Poland. e-mail:
<EMAIL_ADDRESS>
Alberto Perelli, Dipartimento di Matematica, Università di Genova, via
Dodecaneso 35, 16146 Genova, Italy. e-mail<EMAIL_ADDRESS>
Maciej Radziejewski, Faculty of Mathematics and Computer Science, Adam
Mickiewicz University, Poznań, 61-614 Poznań, Poland. e-mail
<EMAIL_ADDRESS>