# Generalized harmonic numbers via poly-Bernoulli polynomials

Levent Kargın, Mehmet Cenkci Ayhan Dil and Mümün Can
Department of Mathematics, Akdeniz University, Antalya, Turkey
lkargin@akdeniz.edu.trcenkci@akdeniz.edu.tradil<EMAIL_ADDRESS>
###### Abstract

We present a relationship between the generalized hyperharmonic numbers and
the poly-Bernoulli polynomials, motivated from the connections between
harmonic and Bernoulli numbers. This relationship yields numerous identities
for the hyper-sums and several congruences.

MSC 2010: 11B75, 11B68, 11B73, 11A07

Keywords: Harmonic numbers, hyperharmonic numbers, generalized harmonic
numbers, poly-Bernoulli polynomials, Stirling numbers, hyper-sums,
congruences.

##  1 Introduction

The $n$th harmonic number is defined by
$H_{n}=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n},$ where $H_{0}$ is
conventionally understood to be zero. The harmonic numbers naturally find
places in mathematics and applications such as combinatorics, mathematical
analysis, number theory, computer sciences. Therefore, introducing new
representations and closed forms for the harmonic numbers and their
generalizations, relating harmonic numbers with other subjects are active
research areas (see, for example [4, 6, 10, 14, 21, 22, 32, 33, 35]).

Among many other generalizations of the harmonic numbers, we are interested in
a unified generalization, the generalized hyperharmonic numbers, defined as

$H_{n}^{\left(p,r\right)}=\sum_{k=1}^{n}H_{k}^{\left(p,r-1\right)},$

with $H_{k}^{\left(p,0\right)}=1/k^{p}$ (see [14]). These numbers extend two
famous generalizations of the harmonic numbers, namely the generalized
harmonic numbers
$H_{n}^{\left(p,1\right)}=H_{n}^{\left(p\right)}=\sum_{k=1}^{n}$ $1/k^{p}$ and
the hyperharmonic numbers
$H_{n}^{\left(1,r\right)}=h_{n}^{(r)}=\sum_{k=1}^{n}h_{k}^{\left(r-1\right)}$.

There is an extensive literature on the harmonic and hyperharmonic numbers.
Among which we emphasize the formulas

$\sum_{k=0}^{n}\left(-1\right)^{k}\genfrac{[}{]}{0.0pt}{}{n+1}{k+1}B_{k}=n!H_{n+1}$
(1)

(see [10] and see also [6, 21, 32]) and

$\sum_{k=0}^{n}\genfrac{[}{]}{0.0pt}{}{n+r}{k+r}_{r}B_{k}=n!h_{n+1}^{\left(r-1\right)}$
(2)

(see [6, 21]). Here $B_{n}$ is $n$th Bernoulli number defined by means of the
generating function

$\frac{t}{e^{t}-1}=\sum_{n=0}^{\infty}B_{n}\frac{t^{n}}{n!},$

$\genfrac{[}{]}{0.0pt}{}{n+r}{k+r}_{r}$ is the $r$-Stirling number of the
first kind defined by

$\left(x+r\right)^{\overline{n}}=\left(x+r\right)\left(x+r+1\right)\cdots\left(x+r+n-1\right)=\sum\limits_{k=0}^{n}\genfrac{[}{]}{0.0pt}{}{n+r}{k+r}_{r}x^{k}$

(see [5]), and
$\genfrac{[}{]}{0.0pt}{}{n+1}{k+1}_{1}=\genfrac{[}{]}{0.0pt}{}{n+1}{k+1}$ is
the ordinary Stirling number of the first kind.

Equations (1) and (2) represent harmonic and hyperharmonic numbers in terms of
the Stirling and $r$-Stirling numbers of the first kind, and Bernoulli
numbers. These then give rise to the natural question of representing
generalized harmonic and generalized hyperharmonic numbers as similar
formulas. An affirmative answer to this question is given by the following
theorem, which represents the generalized hyperharmonic numbers in terms of
the $r$-Stirling numbers and poly-Bernoulli polynomials
$B_{n}^{\left(p\right)}\left(x\right).$ The polynomials
$B_{n}^{\left(p\right)}\left(x\right)$ are defined by

$\sum_{n=0}^{\infty}B_{n}^{\left(p\right)}\left(x\right)\frac{t^{n}}{n!}=\frac{\mathrm{Li}_{p}\left(1-e^{-t}\right)}{1-e^{-t}}e^{xt}$
(3)

(see [2]), where
$\mathrm{Li}_{p}\left(t\right)=\sum_{n=1}^{\infty}t^{n}/n^{p}$ stands for the
polylogarithm function.

###### Theorem 1

For all non-negative integers $n$ and $r$, we have

$\sum_{k=0}^{n}\genfrac{[}{]}{0.0pt}{}{n+r}{k+r}_{r}B_{k}^{\left(p\right)}\left(q\right)=n!H_{n+1}^{\left(p,q+r\right)}.$
(4)

As a result, we deduce the following identity which answers the question of
which type of the Bernoulli numbers are related to the generalized harmonic
numbers $H_{n}^{\left(p\right)}$.

###### Corollary 2

For any non-negative integer $n$

$\sum_{k=0}^{n}\genfrac{[}{]}{0.0pt}{}{n+1}{k+1}B_{k}^{\left(p\right)}=n!H_{n+1}^{\left(p\right)},$
(5)

where $B_{k}^{\left(p\right)}=B_{k}^{\left(p\right)}\left(0\right)$ is the
$k$th poly-Bernoulli number (see [19]).

We also observe the following correspondence between the generalized
hyperharmonic numbers and the hyper-sums
$S_{p}^{\left(q\right)}\left(n\right)$:

$H_{n}^{\left(-p,q+1\right)}=S_{p}^{\left(q\right)}\left(n\right).$ (6)

The hyper-sum is introduced by Faulhaber as

$S_{p}^{\left(r\right)}\left(n\right)=\sum_{k=1}^{n}S_{p}^{(r-1)}\left(k\right),$

with
$S_{p}^{\left(0\right)}\left(n\right)=S_{p}\left(n\right)=1^{p}+2^{p}+\cdots+n^{p}$
(see [23]). The sums of powers of integers $S_{p}\left(n\right)$ have been
interested since the classical times, for details see [15, 16, 23, 26]. Some
recent studies on the hyper-sums include explicit formulas, connection with
the Bernoulli numbers, congruences, generating functions, and recurrence
formulas ([8, 9, 18, 25]). The surprising correspondence (6) gives rise to
numerous identities such as

$S_{p}^{\left(q\right)}\left(n\right)=\sum_{j=1}^{p}\left(-1\right)^{p+j}n\genfrac{\\{}{\\}}{0.0pt}{}{p}{j}\binom{q+n+j}{q+1+j}j!,$

and congruences like

$S_{p}^{\left(q\right)}\left(n\right)\equiv\binom{n+q+1}{q+2}\pmod{p},$

for a prime number $p$.

Apart from the identities (1) and (2), the harmonic numbers are related to the
Bernoulli numbers and polynomials via

$\sum_{k=1}^{n}\left(-1\right)^{k-1}\genfrac{[}{]}{0.0pt}{}{n}{k}kB_{k-1}=\frac{2n!}{n+1}H_{n}$
(7)

(see [22]) and

$\displaystyle\sum_{k=1}^{n}\genfrac{[}{]}{0.0pt}{}{n+m}{k+m}_{m}kB_{k-1}\left(r\right)\mathbf{=}n!\binom{n+r+m-1}{r+m-2}$
$\displaystyle\left\\{\left(H_{n+r+m-1}-H_{r+m-2}\right)^{2}\right.$
$\displaystyle\left.-H_{n+r+m-1}^{\left(2\right)}+H_{r+m-2}^{\left(2\right)}\right\\}$
(8)

(see [21]). Since $B_{k}^{\prime}\left(x\right)=kB_{k-1}\left(x\right)$, the
terms $kB_{k-1}$ and $kB_{k-1}\left(r\right)$ in the summands suggest whether

$\sum_{k=1}^{n}\genfrac{[}{]}{0.0pt}{}{n+m}{k+m}_{m}\left.\frac{d^{l}}{dx^{l}}B_{k}\left(x\right)\right|_{x=r}$

can be written in terms of the harmonic numbers of any kind. We deal with this
problem in Theorem 6. It is worthwhile to mention that in the proof of Theorem
6, the question raised in [13, 27] on the general form of higher order
derivatives of $h_{n}^{\left(x\right)}$ with respect to $x$ is answered in a
different sense.

The organization of the paper is as follows. Section 2 is a preliminary
section in which we give notation and basic definitions needed in the paper.
In Section 3 we prove Theorems 1 and 6, and obtain some recurrence formulas
for the hyperharmonic and generalized harmonic numbers. In Section 4 utilizing
(6) we derive numerous formulas for the hyper-sums
$S_{p}^{\left(q\right)}\left(n\right)$. Moreover, new results for the sums of
powers of integers $S_{p}\left(n\right)$ are presented. We conclude the paper
with Section 5 where we present some congruences for the generalized
hyperharmonic numbers, hyperharmonic numbers and hyper-sums.

##  2 Preliminaries

The $r$-Stirling numbers, which are natural generalizations of the ordinary
Stirling numbers, may be defined either in a combinatorial or in an analytic
way. Analytic way includes the generating functions. The $r$-Stirling numbers
of the second kind, denoted by $\genfrac{\\{}{\\}}{0.0pt}{}{n+r}{k+r}_{r}$,
are defined by means of the exponential generating function

$\sum_{n=k}^{\infty}\genfrac{\\{}{\\}}{0.0pt}{}{n+r}{k+r}_{r}\frac{z^{n}}{n!}=\frac{\left(e^{z}-1\right)^{k}}{k!}e^{rz}$
(9)

(see [5]). The $r$-Stirling numbers of the first and the second kind are
related via the $r$-Stirling transform:

$b_{n}=\sum_{k=0}^{n}\genfrac{[}{]}{0.0pt}{}{n+r}{k+r}_{r}a_{k}\text{ if and
only if
}a_{n}=\sum_{k=0}^{n}\left(-1\right)^{n-k}\genfrac{\\{}{\\}}{0.0pt}{}{n+r}{k+r}_{r}b_{k}$

(see [5]). In particular
$\genfrac{\\{}{\\}}{0.0pt}{}{n}{k}_{0}=\genfrac{\\{}{\\}}{0.0pt}{}{n}{k},\text{
}\genfrac{\\{}{\\}}{0.0pt}{}{n+1}{k+1}_{1}=\genfrac{\\{}{\\}}{0.0pt}{}{n+1}{k+1},$
where $\genfrac{\\{}{\\}}{0.0pt}{}{n}{k}$ is the ordinary Stirling number of
the second kind,
and$\genfrac{[}{]}{0.0pt}{}{n}{k}_{0}=\genfrac{[}{]}{0.0pt}{}{n}{k},\text{
}\genfrac{[}{]}{0.0pt}{}{n+1}{k+1}_{1}=\genfrac{[}{]}{0.0pt}{}{n+1}{k+1},$
where $\genfrac{[}{]}{0.0pt}{}{n}{k}$ is the ordinary Stirling number of the
first kind.

We list some of the basic facts about the Stirling numbers in the following.

###### Lemma 3

We have

(1) $\genfrac{[}{]}{0.0pt}{}{0}{0}=1$, |  $\genfrac{[}{]}{0.0pt}{}{n}{0}=\genfrac{[}{]}{0.0pt}{}{0}{n}=0$, $n>0$, |  $\genfrac{[}{]}{0.0pt}{}{n}{1}=\left(n-1\right)!$, |  $\genfrac{[}{]}{0.0pt}{}{n}{n-1}=\frac{n\left(n-1\right)}{2}$, | $\genfrac{[}{]}{0.0pt}{}{n}{n}=1.\vspace{0.05in}$
---|---|---|---|---
(2) $\genfrac{\\{}{\\}}{0.0pt}{}{0}{0}=1$, |  $\genfrac{\\{}{\\}}{0.0pt}{}{n}{0}=\genfrac{\\{}{\\}}{0.0pt}{}{0}{n}=0$, $n>0$, |  $\genfrac{\\{}{\\}}{0.0pt}{}{n}{1}=\genfrac{\\{}{\\}}{0.0pt}{}{n}{n}=1$, | $\genfrac{\\{}{\\}}{0.0pt}{}{n}{n-1}=\frac{n\left(n-1\right)}{2}$ |  ([11]).

(3) $\genfrac{[}{]}{0.0pt}{}{n+r}{0+r}_{r}=r^{\overline{n}}$,
$\genfrac{[}{]}{0.0pt}{}{n+r}{1+r}_{r}=n!h_{n}^{\left(r\right)}$,
$\genfrac{[}{]}{0.0pt}{}{n+r}{n-1+r}_{r}=\frac{n\left(n-1\right)}{2}+nr$,
$\genfrac{[}{]}{0.0pt}{}{n+r}{n+r}_{r}=1$, and
$\genfrac{[}{]}{0.0pt}{}{n+n}{m+n}_{n}=\delta_{m,n}$, where $\delta_{m,n}$
stands for the Kronecker’s delta ([5]).

(4) For any $r$,
$\genfrac{\\{}{\\}}{0.0pt}{}{n+r}{k+r}_{r}\equiv\genfrac{[}{]}{0.0pt}{}{n+r}{k+r}_{r}\equiv
0\pmod{n}$ for a prime number $n$, provided that $k=2,3,\ldots,n-1$ ([17]).

The generating function of the generalized hyperharmonic numbers
$H_{n}^{\left(p,q\right)}$ is given by

$\sum_{n=0}^{\infty}H_{n}^{\left(p,q\right)}t^{n}=\frac{Li_{p}\left(t\right)}{\left(1-t\right)^{q}}$
(10)

(see [14]), which reduces to

$\sum_{n=0}^{\infty}h_{n}^{\left(q\right)}t^{n}=-\frac{\ln\left(1-t\right)}{\left(1-t\right)^{q}}$
(11)

(see [12]), the generating function of the hyperharmonic numbers
$h_{n}^{\left(q\right)}$, that is related to the harmonic numbers by

$h_{n}^{\left(q\right)}=\binom{n+q-1}{q-1}\left(H_{n+q-1}-H_{q-1}\right).$
(12)

The poly-Bernoulli polynomials $B_{n}^{\left(p\right)}\left(x\right)$, defined
in (3), can be expressed in terms of the poly-Bernoulli numbers
$B_{n}^{\left(p\right)}=B_{n}^{\left(p\right)}\left(0\right)$ as

$B_{n}^{\left(p\right)}\left(x\right)=\sum_{k=0}^{n}\binom{n}{k}B_{k}^{\left(p\right)}x^{n-k}.$

The poly-Bernoulli numbers can be also represented in terms of the Stirling
numbers of the second kind by

$B_{n}^{\left(p\right)}=\left(-1\right)^{n}\sum_{k=0}^{n}\genfrac{\\{}{\\}}{0.0pt}{}{n}{k}\frac{\left(-1\right)^{k}k!}{\left(k+1\right)^{p}}$
(13)

(see [19]). Hence

$B_{0}^{\left(p\right)}=1\text{,
}B_{1}^{\left(p\right)}=\frac{1}{2^{p}}\text{,
}B_{0}^{\left(p\right)}\left(x\right)=1\text{,
}B_{1}^{\left(p\right)}\left(x\right)=x+\frac{1}{2^{p}},$ (14)

etc. The poly-Bernoulli polynomials and numbers, which are studied recently in
different directions ([2, 7, 24, 34]), are generalizations of the Bernoulli
polynomials $B_{n}\left(x\right)$ and the Bernoulli numbers $B_{n}$ in that
$B_{n}^{\left(1\right)}\left(x-1\right)=B_{n}\left(x\right)$ and
$B_{n}^{\left(1\right)}=B_{n}$ with $B_{1}^{\left(1\right)}=-B_{1}$.

##  3 Generalized hyperharmonic numbers

We start this section by proving Theorem 1. Proof of Theorem 1. Using (3) and
(10) we have

$\displaystyle\sum_{k=0}^{\infty}B_{k}^{\left(p\right)}\left(q+1-r\right)\frac{t^{k}}{k!}$
$\displaystyle=\frac{\mathrm{Li}_{p}\left(1-e^{-t}\right)}{1-e^{-t}}e^{\left(q+1-r\right)t}$
$\displaystyle=\sum_{n=0}^{\infty}\left(-1\right)^{n}n!H_{n+1}^{\left(p,q+1\right)}\frac{\left(e^{-t}-1\right)^{n}}{n!}e^{-rt}.$
(15)

Here by considering (9) we obtain

$\sum_{k=0}^{\infty}B_{k}^{\left(p\right)}\left(q+1-r\right)\frac{t^{k}}{k!}=\sum_{k=0}^{\infty}\left(\sum_{n=0}^{k}\left(-1\right)^{k-n}\genfrac{\\{}{\\}}{0.0pt}{}{k+r}{n+r}_{r}n!H_{n+1}^{\left(p,q+1\right)}\right)\frac{t^{k}}{k!}.$

Then, comparing the coefficients of $\frac{t^{k}}{k!}$ in the both sides of
the above equation gives

$B_{n}^{\left(p\right)}\left(q+1-r\right)=\sum_{k=0}^{n}\left(-1\right)^{n-k}\genfrac{\\{}{\\}}{0.0pt}{}{n+r}{k+r}_{r}k!H_{k+1}^{\left(p,q+1\right)}.$
(16)

Applying the $r$-Stirling transform to the above equation, we obtain the
desired result (4).

It worths to note recent works [29] and [30], which relate the Stirling
numbers of the first kind and poly-Bernoulli numbers in different manner.

The following interesting alternating sum

$-H_{1}+H_{2}-H_{3}+\cdots-H_{2n-1}+H_{2n}=\frac{1}{2}H_{n}$

motivates the next result, which follows from the relation

$\mathrm{Li}_{p}\left(-t\right)+\mathrm{Li}_{p}\left(t\right)=2^{1-p}\mathrm{Li}_{p}\left(t^{2}\right).$

###### Proposition 4

We have

$\sum_{k=0}^{2n}\left(-1\right)^{k}\binom{q+k-1}{k}H_{2n-k}^{\left(p,q\right)}=\frac{1}{2^{p}}H_{n}^{\left(p,q\right)}.$

In particular,

$\sum_{k=0}^{2n}\left(-1\right)^{k}H_{2n-k}^{\left(p\right)}=\frac{1}{2^{p}}H_{n}^{\left(p\right)}\text{
\ and \
}\sum_{k=0}^{2n}\left(-1\right)^{k}\binom{q+k-1}{k}h_{2n-k}^{\left(q\right)}=\frac{1}{2}h_{n}^{\left(q\right)}.$

###### Proposition 5

We have

$H_{n}^{\left(p,q+1\right)}=\sum_{k=0}^{n}\left(-1\right)^{k}\binom{p-q}{k}H_{n-k}^{\left(p,p+1\right)}.$

In particular,

$H_{n}^{\left(p\right)}=\sum_{k=0}^{n}\left(-1\right)^{k}\binom{p}{k}H_{n-k}^{\left(p,p+1\right)}.$

Proof. We have

$\displaystyle\sum_{n=0}^{\infty}H_{n}^{\left(p,q+1\right)}t^{n}$
$\displaystyle=\frac{\mathrm{Li}_{p}\left(t\right)}{\left(1-t\right)^{p+1}}\left(1-t\right)^{p-q}$
$\displaystyle=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}H_{k}^{\left(p,p+1\right)}\binom{p-q}{n-k}\left(-1\right)^{n-k}\right)t^{n},$

from which the desired result follows.

We now turn our attention to the sum

$\sum_{k=1}^{n}\genfrac{[}{]}{0.0pt}{}{n+r}{k+r}_{r}\left.\frac{d^{l}}{dx^{l}}B_{k}\left(x\right)\right|_{x=r},$

which is a more general form of (1), (7) and (8). To evaluate this sum we
first recall that

$\sum_{k=0}^{\infty}\binom{m+k}{m}P\left(r,m+k,m\right)t^{k}=\frac{\left(-\ln\left(1-t\right)\right)^{r}}{\left(1-t\right)^{m+1}}$
(17)

(see [35]), where

$P\left(r,m+k,m\right)=P_{r}\left(H_{m+k}^{\left(1\right)}-H_{m}^{\left(1\right)},H_{m+k}^{\left(2\right)}-H_{m}^{\left(2\right)},\ldots,H_{m+k}^{\left(r\right)}-H_{m}^{\left(r\right)}\right),$

and the polynomial $P_{n}\left(x_{1},x_{2},\ldots,x_{n}\right)$ is defined by
$P_{0}=1$ and

$P_{n}\left(x_{1},x_{2},\ldots,x_{n}\right)=\left(-1\right)^{n}Y_{n}\left(-0!x_{1},-1!x_{2},\ldots,-\left(n-1\right)!x_{n}\right),$

where $Y_{n}$ is the exponential Bell polynomial [11]. A first few of them may
be listed as $P_{1}\left(x_{1}\right)=x_{1},\text{
}P_{2}\left(x_{1},x_{2}\right)=x_{1}^{2}-x_{2},\text{
}P_{3}\left(x_{1},x_{2},x_{3}\right)=x_{1}^{3}-3x_{1}x_{2}+2x_{3}.$

###### Theorem 6

For nonnegative integers $q,r,$ and $n$, we have

$\displaystyle\sum_{k=l}^{n}\genfrac{[}{]}{0.0pt}{}{n+r}{k+r}_{r}k\left(k-1\right)\cdots\left(k-l+1\right)B_{k-l}\left(q\right)$
$\displaystyle\ =n!\binom{n+q+r-1}{q+r-2}P\left(l+1,n+q+r-1,q+r-2\right).$
(18)

Proof. Differentiating both sides of (11) with respect to $x$ $l$ times gives

$\sum_{n=0}^{\infty}\frac{d^{l}}{dx^{l}}h_{n}^{\left(x+1\right)}t^{n}=\frac{\left(-\ln\left(1-t\right)\right)^{l+1}}{\left(1-t\right)^{x+1}}.$

Setting $x=q$ in the above equation and using (17), we see that

$\left.\frac{d^{l}}{dx^{l}}h_{n}^{\left(x+1\right)}\right|_{x=q}=\binom{q+n}{q}P\left(l+1,q+n,q\right).$
(19)

For $p=1$ and $q\rightarrow x-1,$ with the use of
$B_{k}^{\left(1\right)}\left(x-1\right)=B_{k}\left(x\right),$ (4) turns into

$\frac{1}{n!}\sum_{k=0}^{n}\genfrac{[}{]}{0.0pt}{}{n+r}{k+r}_{r}B_{k}\left(x\right)=h_{n+1}^{\left(x+r-1\right)}.$

Iterating $B_{k}^{\prime}\left(x\right)=kB_{k-1}\left(x\right)$ $l$ times on
the left-hand side and observing (19) for the right-hand side yield the closed
formula (18).

It is seen that (18) reduces to (1) for $r=q=1$ and $l=0$, and to (8) for
$l=1$. Another demonstration of (18) is the following example corresponding to
$l=2$:

$\displaystyle\frac{1}{n!}\sum_{k=2}^{n}\genfrac{[}{]}{0.0pt}{}{n+r}{k+r}_{r}k\left(k-1\right)B_{k-2}\left(q\right)$
$\displaystyle=\binom{n+q+r-1}{q+r-2}\left\\{\left(H_{n+q+r-1}-H_{q+r-2}\right)^{3}+2\left(H_{n+q+r-1}^{\left(3\right)}-H_{q+r-2}^{\left(3\right)}\right)\right.$
$\displaystyle\
\left.\qquad\qquad\qquad\qquad\quad-3\left(H_{n+q+r-1}-H_{q+r-2}\right)\left(H_{n+q+r-1}^{\left(2\right)}-H_{q+r-2}^{\left(2\right)}\right)\right\\},$

in particular for $r=q=1$

$\frac{1}{n!}\sum_{k=2}^{n}\left(-1\right)^{k}\genfrac{[}{]}{0.0pt}{}{n+1}{k+1}k\left(k-1\right)B_{k-2}=\left(H_{n+1}\right)^{3}-3H_{n+1}H_{n+1}^{\left(2\right)}+2H_{n+1}^{\left(3\right)}.$

###### Remark 1

We note that (19) gives an answer to the question raised in [13, 27] on the
general form of higher order derivatives of $h_{n}^{\left(x\right)}$ with
respect to $x$.

##  4 Hyper-sums of powers

The hyper-sums of powers of integers $S_{p}^{\left(q\right)}\left(n\right)$
are defined recursively by

$S_{p}^{\left(q\right)}\left(n\right)=\sum_{k=1}^{n}S_{p}^{(q-1)}\left(k\right),$

with the initial condition
$S_{p}^{\left(0\right)}\left(n\right)=S_{p}\left(n\right)$. We note that
$S_{p}^{\left(q\right)}\left(n\right)$ satisfies the relation

$S_{p}^{\left(q\right)}\left(n\right)=\sum_{k=1}^{n}\binom{n+q-k}{q}k^{p}$

(see [25]). We also observe that

$H_{n}^{\left(p,q+1\right)}=\sum_{k=1}^{n}\binom{n+q-k}{q}\frac{1}{k^{p}}$

(see [14, p. 1648]). These simply imply the relation

$H_{n}^{\left(-p,q+1\right)}=S_{p}^{\left(q\right)}\left(n\right).$ (20)

Therefore, we can translate the results given for the generalized
hyperharmonic numbers $H_{n}^{\left(p,q\right)}$ to the hyper-sums
$S_{p}^{\left(q\right)}\left(n\right)$. For instance,

$\displaystyle S_{p}^{\left(q\right)}\left(n\right)$
$\displaystyle=\frac{1}{2^{p}}\sum_{k=0}^{2n}\binom{q+2n-k}{2n-k}\left(-1\right)^{k}S_{p}^{\left(q\right)}\left(k\right),$
$\displaystyle S_{p}^{\left(q\right)}\left(n\right)$
$\displaystyle=\sum_{k=0}^{n}\binom{n-k+p+q-1}{n-k}S_{p}^{\left(p\right)}\left(k\right),$

which follow from Proposition 4. In particular, we have

$S_{p}\left(n\right)=\frac{1}{2^{p}}\sum_{k=0}^{2n}\left(-1\right)^{k}S_{p}\left(k\right)\text{\
and
}S_{p}\left(n\right)=\sum_{k=0}^{n}\binom{n-k+p-1}{n-k}S_{p}^{\left(p\right)}\left(k\right).$

Moreover, using the duality theorem
$B_{n}^{\left(-p\right)}=B_{p}^{\left(-n\right)}$ for the poly-Bernoulli
numbers [19, Theorem 2] in (5), we obtain an evaluation formula for sums of
powers of integers:

$\frac{1}{n!}\sum_{k=0}^{n}\genfrac{[}{]}{0.0pt}{}{n+1}{k+1}B_{p}^{\left(-k\right)}=S_{p}\left(n+1\right).$

Our next result utilizes the following lemma, which is a polynomial extension
of the Arakawa-Kaneko formula

$B_{n}^{\left(-p\right)}=\sum_{j=0}^{\min\left\\{n,p\right\\}}\left(j!\right)^{2}\genfrac{\\{}{\\}}{0.0pt}{}{p+1}{j+1}\genfrac{\\{}{\\}}{0.0pt}{}{n+1}{j+1}$

(see [1, Theorem 2]).

###### Lemma 7

We have

$B_{n}^{\left(-p\right)}\left(q\right)=\sum_{j=0}^{\min\left\\{n,p\right\\}}\left(j!\right)^{2}\genfrac{\\{}{\\}}{0.0pt}{}{p+1}{j+1}\genfrac{\\{}{\\}}{0.0pt}{}{n+q+1}{j+q+1}_{q+1}.$
(21)

Proof. We can write (10) in the form

$\sum_{p=0}^{\infty}\sum_{k=1}^{\infty}\left(-1\right)^{k}H_{k}^{\left(-p,q+1\right)}\left(e^{-t}-1\right)^{k}\frac{y^{p}}{p!}=\left(1-e^{-t}\right)\frac{e^{\left(q+2\right)t+y}}{1-\left(1-e^{t}\right)\left(1-e^{y}\right)}$

by setting $t\rightarrow 1-e^{-t}$. On the other hand, we equivalently have

$\sum_{p=0}^{\infty}\sum_{k=0}^{\infty}\left(-1\right)^{k}H_{k+1}^{\left(-p,q+1\right)}\frac{\left(e^{-t}-1\right)^{k}}{e^{rt}}\frac{y^{p}}{p!}=\sum_{j=0}^{\infty}\left(j!\right)^{2}\frac{\left(1-e^{t}\right)^{j}}{j!}e^{\left(q+2-r\right)t}\frac{\left(1-e^{y}\right)^{j}}{j!}e^{y}.$

Last equality, (15) and (9) give the desired result.

Now, taking $-p$ in (4) and using (21) yield

$n!H_{n+1}^{\left(-p,q+r\right)}=\sum_{j=0}^{n}\left(j!\right)^{2}\genfrac{\\{}{\\}}{0.0pt}{}{p+1}{j+1}\sum_{k=j}^{n}\genfrac{[}{]}{0.0pt}{}{n+r}{k+r}_{r}\genfrac{\\{}{\\}}{0.0pt}{}{k+q+1}{j+q+1}_{q+1}.$

Utilizing the formula

$\sum_{k=j}^{n}\genfrac{[}{]}{0.0pt}{}{n+r}{k+r}_{r}\genfrac{\\{}{\\}}{0.0pt}{}{k+s}{j+s}_{s}=\frac{n!}{j!}\binom{n+r+s-1}{j+r+s-1}$

(see [28, Theorems 3.7 and 3.11]), we obtain the following.

###### Theorem 8

For non-negative integers $p$, $q$, and $n$, we have

$S_{p}^{\left(q\right)}\left(n\right)=\sum_{j=0}^{p}j!\genfrac{\\{}{\\}}{0.0pt}{}{p+1}{j+1}\binom{n+q}{j+q+1}.$
(22)

###### Remark 2

We note that a slightly different form of (22) can be found in [8, 9].

Equation (22) provides a natural extension of the expression

$S_{p}\left(n\right)=\sum_{j=0}^{p}j!\genfrac{\\{}{\\}}{0.0pt}{}{p}{j}\binom{n+1}{j+1}$

(see [23, p. 285]) to $S_{p}^{\left(q\right)}\left(n\right)$.

Next result is an alternative representation for
$S_{p}^{\left(q\right)}\left(n\right)$, which extends

$S_{p}\left(n\right)=\sum_{j=1}^{p}\left(-1\right)^{p+j}\genfrac{\\{}{\\}}{0.0pt}{}{p}{j}\binom{n+j}{j+1}j!$
(23)

(see [23, p. 285]).

###### Theorem 9

We have

$S_{p}^{\left(q\right)}\left(n\right)=\sum_{j=1}^{p}\left(-1\right)^{p+j}\genfrac{\\{}{\\}}{0.0pt}{}{p}{j}\binom{n+q+j}{q+j+1}j!.$
(24)

Proof. We set $-p$ for $p$ in (10). Then

$\sum_{n=0}^{\infty}H_{n}^{\left(-p,q\right)}t^{n}=\frac{1}{\left(1-t\right)^{q}}\sum_{k=1}^{\infty}k^{p}t^{k}=\frac{1}{\left(1-t\right)^{q+1}}w_{p}\left(\frac{t}{1-t}\right),$

where $w_{n}\left(x\right)$ is the $n$th geometric polynomial, defined by

$\sum_{k=1}^{\infty}k^{n}t^{k}=\frac{1}{1-x}w_{n}\left(\frac{x}{1-x}\right)$
(25)

(see [3]). Using (25), the relation
$\left(1+x\right)w_{n}\left(x\right)=x\left(-1\right)^{n}w_{n}\left(-x-1\right),$
$n>0$, (see [20, Eq. (22)]) and

$w_{p}\left(x\right)=\sum_{j=0}^{p}\genfrac{\\{}{\\}}{0.0pt}{}{p}{j}j!x^{j}$

(see [3]), we reach at

$\sum_{n=0}^{\infty}H_{n}^{\left(-p,q\right)}t^{n}=\sum_{n=1}^{\infty}t^{n}\sum_{j=0}^{p}\genfrac{\\{}{\\}}{0.0pt}{}{p}{j}\binom{n+k+q-1}{n-1}\left(-1\right)^{j+p}j!,$

which implies the result.

###### Remark 3

A slightly different form of (24) may be written as

$S_{p}^{\left(q\right)}\left(n\right)=\sum_{j=0}^{p}\left(-1\right)^{p+j}\genfrac{\\{}{\\}}{0.0pt}{}{p+1}{j+1}\binom{n+q+j+1}{q+j+1}j!,$

which can be obtained by utilizing

$\mathrm{Li}_{-p}\left(t\right)=\left(-1\right)^{p+1}\sum_{k=0}^{p}k!\genfrac{\\{}{\\}}{0.0pt}{}{p+1}{k+1}\left(\frac{-1}{1-t}\right)^{k+1}$

in the generating function (10).

Taking into account (6), it can be seen from (16) and (24) that

$B_{n}^{\left(-p\right)}\left(q\right)=\sum_{j=1}^{p}\genfrac{\\{}{\\}}{0.0pt}{}{p}{j}j!\left(-1\right)^{p+j}\sum_{k=0}^{n}\left(-1\right)^{n-k}\genfrac{\\{}{\\}}{0.0pt}{}{n+1}{k+1}\left(j+q+2\right)^{\overline{k}}.$

This reduces to

$B_{n}^{\left(-p\right)}\left(q\right)=\sum_{j=1}^{p}\genfrac{\\{}{\\}}{0.0pt}{}{p}{j}\left(-1\right)^{p+j}j!\left(j+q+1\right)^{n}$
(26)

utilizing the formula

$\left(x-r\right)^{n}=\sum_{n=k}\left(-1\right)^{n-k}\genfrac{\\{}{\\}}{0.0pt}{}{n+r}{k+r}_{r}x^{\overline{k}}$

(see [5]). (26) stands for a polynomial extension of the Arakawa-Kaneko
formula

$B_{n}^{\left(-p\right)}=\sum_{j=1}^{p}\genfrac{\\{}{\\}}{0.0pt}{}{p}{j}\left(-1\right)^{p+j}j!\left(j+1\right)^{n}$

(cf. [1]).

We conclude this section by stating further results for hyper-sums and sums of
powers, and some of their consequences.

###### Theorem 10

We have

$\displaystyle S_{p}^{\left(q\right)}\left(n\right)$
$\displaystyle=\frac{1}{q!}\sum_{k=0}^{q}\left(-1\right)^{k}\genfrac{[}{]}{0.0pt}{}{q+n+1}{k+n+1}_{n+1}S_{p+k}\left(n\right),$
(27) $\displaystyle S_{q}\left(n\right)$
$\displaystyle=\sum_{k=0}^{q}\left(-1\right)^{k}\genfrac{\\{}{\\}}{0.0pt}{}{q+n+1}{k+n+1}_{n+1}\binom{k+n}{k+1}k!,$
(28) $\displaystyle S_{q-1}\left(n\right)$
$\displaystyle=\sum_{k=0}^{q}\left(-1\right)^{k}\genfrac{\\{}{\\}}{0.0pt}{}{q+n+1}{k+n+1}_{n+1}k!h_{n}^{\left(k+1\right)}.$
(29)

Proof. For (27), we appeal to the generating function of
$S_{p}^{\left(k\right)}\left(n\right)$ depending on the index $k$

$\sum_{k=0}^{\infty}S_{p}^{\left(k\right)}\left(n\right)z^{k}=\frac{1}{\left(1-z\right)^{n+1}}\sum_{j=1}^{n}j^{p}\left(1-z\right)^{j}$

(cf. [25]). We set $z\rightarrow 1-e^{-t}$ and see that

$\sum_{k=0}^{\infty}\left(-1\right)^{k}k!S_{p}^{\left(k\right)}\left(n\right)\frac{\left(e^{-t}-1\right)^{k}}{k!}e^{-\left(n+1\right)t}=\sum_{j=1}^{n}j^{p}e^{-jt}.$

Hence, the proof follows from (9) and $r$-Stirling transform.

(29) follows similarly by using the generating function

$\sum_{k=0}^{\infty}h_{n}^{\left(k+1\right)}z^{k}=\frac{1}{\left(1-z\right)^{n+1}}\sum_{j=1}^{n}\frac{\left(1-z\right)^{j}}{j}$

(see [21]).

Utilizing the identity

$\genfrac{\\{}{\\}}{0.0pt}{}{q+n}{k+n}_{n}=\sum_{j=k}^{q}\binom{q}{j}\genfrac{\\{}{\\}}{0.0pt}{}{j}{k}n^{q-j}$

(cf. [5]) and (23) we find that

$\displaystyle\sum_{k=0}^{q}\left(-1\right)^{k}k!\genfrac{\\{}{\\}}{0.0pt}{}{q+n+1}{k+n+1}_{n+1}\binom{k+n}{k+1}$
$\displaystyle=\sum_{j=0}^{q}\left(-1\right)^{j}\binom{q}{j}\left(n+1\right)^{q-j}S_{j}\left(q\right)$
$\displaystyle=\left(n+1\right)^{q}\sum_{m=0}^{q}\left(\sum_{j=0}^{q}\left(-1\right)^{j}\binom{q}{j}m^{j}\left(n+1\right)^{-j}\right)=S_{q}\left(n\right),$

which is (28).

Equations (22) and (27) yield the following interesting identity.

###### Proposition 11

$\sum_{j=1}^{q}\frac{\left(-1\right)^{j}}{j}\binom{n}{j}\binom{n+q-j}{n}=\binom{n+q}{q}\left(H_{n+q}-H_{q}-H_{n}\right).$

Proof. Taking $p=-1$ in (27) gives

$q!h_{n}^{\left(q+1\right)}=\sum_{k=1}^{q}\left(-1\right)^{k}\genfrac{[}{]}{0.0pt}{}{q+n+1}{k+n+1}_{n+1}S_{k-1}\left(n\right)+q!\binom{n+q}{q}H_{n},$

by Lemma 3 (3). We now make use of (22) with $q=0$ and deduce that

$\displaystyle\sum_{k=1}^{q}\left(-1\right)^{k}\genfrac{[}{]}{0.0pt}{}{q+n+1}{k+n+1}_{n+1}S_{k-1}\left(n\right)$
$\displaystyle=\sum_{j=1}^{q}\left(j-1\right)!\binom{n}{j}\sum_{k=j}^{q}\left(-1\right)^{k}\genfrac{[}{]}{0.0pt}{}{q+n+1}{k+n+1}_{n+1}\genfrac{\\{}{\\}}{0.0pt}{}{k}{j}.$

Here we appeal to the equation

$\sum_{k=j}^{q}\left(-1\right)^{k-j}\genfrac{[}{]}{0.0pt}{}{q+r}{k+r}_{r}\genfrac{\\{}{\\}}{0.0pt}{}{k+s}{j+s}_{s}=\binom{q}{j}\left(r-s\right)^{\overline{q-j}}$

(see [5, 28]). Hence,

$\displaystyle\sum_{k=1}^{q}\left(-1\right)^{k}\genfrac{[}{]}{0.0pt}{}{q+n+1}{k+n+1}_{n+1}S_{k-1}\left(n\right)$
$\displaystyle=\sum_{j=1}^{q}\left(-1\right)^{j}\left(j-1\right)!\binom{n}{j}\binom{q}{j}\left(n+1\right)^{\overline{q-j}}$
$\displaystyle=q!\sum_{j=1}^{q}\frac{\left(-1\right)^{j}}{j}\binom{n}{j}\binom{n+q-j}{n}.$

Thus, the proof follows from (12).

##  5 Congruences

In this section we present several congruences for the hyperharmonic numbers,
the generalized hyperharmonic numbers, and the hyper-sums. The motivation
rises from the formulas established in Sections 3 and 4, where hyperharmonic
numbers and hyper-sums are formulated in terms of the Stirling numbers.

Throughout by a congruence $x\equiv y\pmod{m}$, we mean that the rational
numbers $x=\frac{a}{b}$ and $y=\frac{c}{d}$ satisfy $x\equiv y\pmod{m}$ if and
only if $m|\left(ad-bc\right)$. We also call a rational number as an
$m$-integer whenever its denominator is not divisible by $m$.

We start by reviewing some basic facts about Bernoulli numbers and
polynomials.

###### Lemma 12

We have

(1) (von Staudt-Clausen) For $n=1$ and for any even integer $n\geq 2$, we have

$B_{n}=A_{n}-\sum_{\left(p-1\right)|n}\frac{1}{p},$

where $A_{n}$ is an integer and $p$ is a prime number. Equivalently, we have

$pB_{n}\equiv\left\\{\begin{array}[c]{ll}-1\pmod{p},&\text{if }p-1\text{
divides }n,\\\ 0\pmod{p},&\text{if }p-1\text{ does not divide
}n.\end{array}\right.$

(2) If $q$ is an integer, then $\frac{B_{n}\left(q\right)-B_{n}}{n}$ is an
integer ([31, p. 6]).

The following result is about a divisibility property for generalized
hyperharmonic numbers.

###### Theorem 13

Let $n$ be an odd prime. For positive integers $p$ and $q$, we have

$n^{p}H_{n+1}^{\left(p,q\right)}\equiv q\pmod{n}\text{.}$

Proof. Let $n$ be an odd prime and $p\geq 2$. Setting $r=0$ in (4),
multiplying both sides by $n^{p-1}$ and separating out the terms with
$k=0,k=1,k=n-1$, and $k=n$, we have

$\displaystyle n^{p-1}n!H_{n+1}^{\left(p,q\right)}$
$\displaystyle=n^{p-1}\genfrac{[}{]}{0.0pt}{}{n}{0}B_{0}^{\left(p\right)}\left(q\right)+n^{p-1}\genfrac{[}{]}{0.0pt}{}{n}{1}B_{1}^{\left(p\right)}\left(q\right)$
$\displaystyle+n^{p-1}\genfrac{[}{]}{0.0pt}{}{n}{n-1}B_{n-1}^{\left(p\right)}\left(q\right)+n^{p-1}\genfrac{[}{]}{0.0pt}{}{n}{n}B_{n}^{\left(p\right)}\left(q\right)+\sum_{k=2}^{n-2}\genfrac{[}{]}{0.0pt}{}{n}{k}B_{k}^{\left(p\right)}\left(q\right).$

We consider each of these terms separately.

By Lemma 3 (1), we have
$n^{p-1}\genfrac{[}{]}{0.0pt}{}{n}{0}B_{0}^{\left(p\right)}\left(q\right)=0$.

By Lemma 3 (1) and (14), we have

$n^{p-1}\genfrac{[}{]}{0.0pt}{}{n}{1}B_{1}^{\left(p\right)}\left(q\right)=n^{p-1}\left(n-1\right)!\left(q+\frac{1}{2^{p}}\right).$

We write

$B_{n-1}^{\left(p\right)}\left(q\right)=B_{n-1}^{\left(p\right)}+\sum_{m=0}^{n-2}\binom{n-1}{m}B_{m}^{\left(p\right)}q^{n-1-m}.$

Now, from (13) we observe that $B_{m}^{\left(p\right)}$ is an $n$-integer for
$m=0,1,\ldots,n-2$. On the other hand, $n^{p}B_{n-1}^{\left(p\right)}$ is an
$n$-integer satisfying $n^{p}B_{n-1}^{\left(p\right)}\equiv-1\pmod{n}.$ Thus,
by Lemma 3 (1), we conclude that

$\displaystyle
n^{p-1}\genfrac{[}{]}{0.0pt}{}{n}{n-1}B_{n-1}^{\left(p\right)}\left(q\right)$
$\displaystyle=n^{p}B_{n-1}^{\left(p\right)}\frac{n-1}{2}+n^{p}\frac{n-1}{2}\sum_{m=0}^{n-2}\binom{n-1}{m}B_{m}^{\left(p\right)}q^{n-1-m}$
$\displaystyle\equiv-\frac{n-1}{2}\pmod{n}.$

Since

$B_{n}^{\left(p\right)}\left(q\right)=B_{n}^{\left(p\right)}+nB_{n-1}^{\left(p\right)}q+\sum_{m=0}^{n-2}\binom{n}{m}B_{m}^{\left(p\right)}q^{n-m},$

we have

$n^{p-1}\genfrac{[}{]}{0.0pt}{}{n}{n}B_{n}^{\left(p\right)}\left(q\right)=n^{p-1}B_{n}^{\left(p\right)}+n^{p}B_{n-1}^{\left(p\right)}q+n^{p-1}\sum_{m=0}^{n-2}\binom{n}{m}B_{m}^{\left(p\right)}q^{n-m}.$

Since $p\geq 2$ and $B_{m}^{\left(p\right)}$ is an $n$-integer for
$m=0,1,\ldots,n-2$, we obtain that

$n^{p-1}\sum_{m=0}^{n-2}\binom{n}{m}B_{m}^{\left(p\right)}q^{n-m}\equiv
0\pmod{n}.$

We also have $n^{p}B_{n-1}^{\left(p\right)}q\equiv-q\pmod{n}$. Now, in [1,
Theorem 1] it has been also proved that $n^{p-1}B_{n}^{\left(p\right)}$ is an
$n$-integer with

$n^{p-1}B_{n}^{\left(p\right)}\equiv\frac{1}{n}\genfrac{\\{}{\\}}{0.0pt}{}{n}{n-1}-\frac{n}{2^{p}}\pmod{n}.$

Since $\genfrac{\\{}{\\}}{0.0pt}{}{n}{n-1}=\frac{n\left(n-1\right)}{2}$, we
then have $n^{p-1}B_{n}^{\left(p\right)}\equiv\frac{n-1}{2}\pmod{n}$. Thus, we
arrive at

$n^{p-1}\genfrac{[}{]}{0.0pt}{}{n}{n}B_{n}^{\left(p\right)}\left(q\right)\equiv\frac{n-1}{2}-q\pmod{n}.$

Finally we write

$n^{p-1}\sum_{k=2}^{n-2}\genfrac{[}{]}{0.0pt}{}{n}{k}B_{k}^{\left(p\right)}\left(q\right)=n^{p-1}\sum_{k=2}^{n-2}\genfrac{[}{]}{0.0pt}{}{n}{k}\sum_{m=0}^{k}\binom{k}{m}B_{m}^{\left(p\right)}q^{k-m}.$

By Lemma 3 (4) and the fact that $B_{m}^{\left(p\right)}$ is an $n$-integer
for $m=0,1,\ldots,k$, where $k=2,3,\ldots,n-2$, we find that

$n^{p-1}\sum_{k=2}^{n-2}\genfrac{[}{]}{0.0pt}{}{n}{k}B_{k}^{\left(p\right)}\left(q\right)\equiv
0\pmod{n}.$

Combining these results we conclude that

$\displaystyle n^{p-1}n!H_{n+1}^{\left(p,q\right)}$ $\displaystyle\equiv
n^{p-1}\left(n-1\right)!\left(q+\frac{1}{2^{p}}\right)-\frac{n-1}{2}+\frac{n-1}{2}-q$
$\displaystyle\equiv-q\pmod{n}.$

The result follows from Wilson’s theorem.

Now for an odd prime $n$, let $p=1$ and $r=0$ in (4). Since
$B_{k}^{\left(1\right)}\left(q\right)=B_{k}\left(q+1\right)$, with
$B_{1}\left(q+1\right)=q+\frac{1}{2}$, we have

$\displaystyle n!H_{n+1}^{\left(1,q\right)}$
$\displaystyle=\genfrac{[}{]}{0.0pt}{}{n}{0}B_{0}\left(q+1\right)+\genfrac{[}{]}{0.0pt}{}{n}{1}B_{1}\left(q+1\right)+\genfrac{[}{]}{0.0pt}{}{n}{n-1}B_{n-1}\left(q+1\right)$
$\displaystyle\quad+\genfrac{[}{]}{0.0pt}{}{n}{n}B_{n}\left(q+1\right)+\sum_{k=2}^{n-2}\genfrac{[}{]}{0.0pt}{}{n}{k}B_{k}\left(q+1\right)$
$\displaystyle=\left(n-1\right)!\left(q+\frac{1}{2}\right)+\frac{n\left(n-1\right)^{2}}{2}\frac{B_{n-1}\left(q+1\right)-B_{n-1}}{n-1}+\frac{n-1}{2}nB_{n-1}$
$\displaystyle\quad+n\frac{B_{n}\left(q+1\right)-B_{n}}{n}+B_{n}+\sum_{k=2}^{n-2}k\genfrac{[}{]}{0.0pt}{}{n}{k}\frac{B_{k}\left(q+1\right)-B_{k}}{k}+\sum_{k=2}^{n-2}\genfrac{[}{]}{0.0pt}{}{n}{k}B_{k}.$

By Lemma 12 (2),

$\frac{n\left(n-1\right)^{2}}{2}\frac{B_{n-1}\left(q+1\right)-B_{n-1}}{n-1}\equiv
n\frac{B_{n}\left(q+1\right)-B_{n}}{n}\equiv 0\pmod{n},$

and $B_{n}=0$ since $n$ is an odd prime. On the other hand, by Lemma 12 (1),
$nB_{n-1}\equiv-1\pmod{n},$ and $B_{k}$ is an $n$-integer for
$k=2,3,\ldots,n-2$. Then, by Lemma 3 (4), we have

$n!H_{n+1}^{\left(1,q\right)}\equiv\left(n-1\right)!\left(q+\frac{1}{2}\right)-\frac{n-1}{2}\equiv\left(n-1\right)!\left(q+\frac{1}{2}+\frac{n-1}{2}\right)\pmod{n}$

by Wilson’s theorem. Canceling $\left(n-1\right)!$ gives the desired result.

Next we present a symmetric-type congruence for hyperharmonic numbers.

###### Theorem 14

For a prime $q$ and an integer $n$ with $1\leq n\leq q-1$, we have

$q\left(h_{n}^{\left(q+1\right)}+nh_{q}^{\left(n+1\right)}\right)\equiv
n\pmod{q}.$

Proof. Let $1\leq n\leq q-1$ and $q$ be a prime. Applying the $r$-Stirling
transform in (29) we find that

$\displaystyle q!h_{n}^{\left(q+1\right)}$
$\displaystyle=\sum_{k=0}^{q}\left(-1\right)^{k}\genfrac{[}{]}{0.0pt}{}{q+n+1}{k+n+1}_{n+1}S_{k-1}\left(n\right)$
$\displaystyle=\genfrac{[}{]}{0.0pt}{}{q+n+1}{0+n+1}_{n+1}H_{n}-\genfrac{[}{]}{0.0pt}{}{q+n+1}{1+n+1}_{n+1}S_{0}\left(n\right)+\left(-1\right)^{q}S_{q-1}\left(n\right)$
$\displaystyle+\sum_{k=2}^{q-1}\left(-1\right)^{k}\genfrac{[}{]}{0.0pt}{}{q+n+1}{k+n+1}_{n+1}S_{k-1}\left(n\right)$
$\displaystyle\equiv\left(n+1\right)\cdots\left(n+q\right)H_{n}-nq!h_{q}^{\left(n+1\right)}+\left(-1\right)^{q}S_{q-1}\left(n\right)\pmod{q}$

by Lemma 3 (3) and (4). Now, $1\leq n\leq q-1$ implies that

$\left(n+1\right)\left(n+2\right)\cdots\left(n+q\right)H_{n}\equiv 0\pmod{q}$

and

$S_{q-1}\left(n\right)=1^{q-1}+2^{q-1}+\cdots+n^{q-1}\equiv n\pmod{q}.$

Thus,

$q!\left(h_{n}^{\left(q+1\right)}+nh_{q}^{\left(n+1\right)}\right)\equiv\left(-1\right)^{q}n\pmod{q}.$

If $q=2$, then $n=1$, and we have

$2\left(h_{1}^{\left(3\right)}+h_{2}^{\left(2\right)}\right)=2\left(1+\frac{5}{2}\right)\equiv
1\pmod{2}.$

Otherwise,
$q!\left(h_{n}^{\left(q+1\right)}+nh_{q}^{\left(n+1\right)}\right)\equiv-n\pmod{q}$
and the result follows from Wilson’s theorem.

We conclude this section by stating two congruences about
$S_{p}^{\left(q\right)}\left(n\right)$. In [25] congruences for
$S_{p}^{\left(q\right)}\left(n\right)$ when $n$ is a prime number were given.
Here we give results modulo a prime number $p$.

###### Theorem 15

For a prime number $p$, we have

$S_{p}^{\left(q\right)}\left(n\right)\equiv\binom{n+q+1}{q+2}\pmod{p}.$

Proof. Let $p$ be a prime in (24). Then

$\displaystyle S_{p}^{\left(q\right)}\left(n\right)$
$\displaystyle=\left(-1\right)^{p+1}\genfrac{\\{}{\\}}{0.0pt}{}{p}{1}\binom{n+q+1}{q+2}+\genfrac{\\{}{\\}}{0.0pt}{}{p}{p}\binom{n+q+p}{q+p+1}p!$
$\displaystyle+\sum_{j=2}^{p-1}\left(-1\right)^{p+j}\genfrac{\\{}{\\}}{0.0pt}{}{p}{j}\binom{n+q+j}{q+j+1}j!.$

By Lemma 3 (2) and (4), we obtain the desired result.

Next result seems obvious, but we record it here as an application of the
formula for the hyper-sums in terms of the poly-Bernoulli polynomials with
negative index. For this we need the following result about the poly-Bernoulli
polynomials, which follows from (26), Lemma 3 (2) and (4).

###### Lemma 16

Given a prime number $p$, a positive integer $n$, and a nonnegative integer
$q$, we have

$B_{n}^{\left(-p\right)}\left(q\right)\equiv\left(q+2\right)^{n}\pmod{p}.$

###### Proposition 17

Given a prime number $p$ and a nonnegative integer $q$, we have

$pS_{p}^{\left(q\right)}\left(p+1\right)\equiv 0\pmod{p}.$

Proof. With the use of (20) in Theorem 1 for $r=0$, we find that

$n!S_{p}^{\left(q\right)}\left(n+1\right)=\sum_{k=0}^{n}\genfrac{[}{]}{0.0pt}{}{n}{k}B_{k}^{\left(-p\right)}\left(q\right).$

Now let $p$ be a prime and set $n=p$. By Lemma 3 (1) and (4),

$\displaystyle p!S_{p}^{\left(q\right)}\left(p+1\right)$
$\displaystyle=\genfrac{[}{]}{0.0pt}{}{p}{0}B_{0}^{\left(-p\right)}\left(q\right)+\genfrac{[}{]}{0.0pt}{}{p}{1}B_{1}^{\left(-p\right)}\left(q\right)+\genfrac{[}{]}{0.0pt}{}{p}{p}B_{p}^{\left(-p\right)}\left(q\right)+\sum_{k=2}^{p-1}\genfrac{[}{]}{0.0pt}{}{p}{k}B_{k}^{\left(-p\right)}\left(q\right)$
$\displaystyle\equiv\left(p-1\right)!B_{1}^{\left(-p\right)}\left(q\right)+B_{p}^{\left(-p\right)}\left(q\right)\pmod{p}.$

By Lemma 16 and Wilson’s theorem, we conclude that

$pS_{p}^{\left(q\right)}\left(n+1\right)\equiv-\left(q+2\right)\left[\left(q+2\right)^{p-1}-1\right]\pmod{p},$

and the result follows from Fermat’s little theorem.

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A Heavy Ion Fireball freeze-out Dipion Cocktail for Au-Au Collisions at
$\sqrt{s_{NN}}$=200 GeV (Part 1).
R.S. Longacrea
aBrookhaven National Laboratory, Upton, NY 11973, USA

###### Abstract

In this paper we develop all the ingredients that come into play in the
freeze-out of the heavy ion fireball for Au-Au collisions at
$\sqrt{s_{NN}}$=200 GeV into dipions. The resonance production of particles
that decay into dipions plus minijets that also decay into dipions are
explored. The final state re-scattering of the pions from minijets play an
important role in the mass spectrum of the cocktail. Mass shifts due to
minijet interference which depend on the volume of the re-scattering pions is
presented. The effective mass balance function is explained. The dipion mass
spectrum within a $p_{t}$ range(intermediate $p_{t}$) is fitted using thermal
and minijet amplitudes.

##  1 Introduction

The ultra-relativistic heavy ion collision starts out as a state of high
density nuclear matter called the Quark Gluon Plasma(QGP) and expands rapidly
to freeze-out. During the freeze-out phase quarks and gluons form a system of
strongly interacting hadrons. These hadrons continue to expand in a thermal
manner until no further scattering is possible because the system becomes to
dilute. However this transition from quarks and gluons(partons) into hadrons
is not a smooth affair. The expansion is very rapid and some faster or hard
scattered partons fragment directly into hadron through a minijet[1] process.
Thus we have thermal and minijet hadrons present in the last scattering of the
hadrons. This mixture of sources is considered in this paper and applied to
the dipion mass spectrum of the heavy ion fireball formed in Au-Au collisions
at $\sqrt{s_{NN}}$=200.

The paper is organized in the following manner:

Sec. 1 is the introduction to the cross sections for $\pi$ $\pi$ scattering.
Sec. 2 develops a two component model with direct production and decay plus a
background $\pi$ $\pi$ component which must re-scatter in order to satisfy
unitarity. Sec. 3 consider two channel unitary scattering $\pi$ $\pi$ and $K$
$\overline{K}$ for $J^{PC}$ = $0^{++}$ and shows what re-scattering would look
like in the $\pi$ $\pi$ channel. Sec. 4 introduces the balance function for
dipion data and applies it to the two component model. Sec. 5 applies the two
component model to dipion data within a $p_{t}$ range. Sec. 6 presents the
summary and discussion. Finally there are two appendices. Appendix A show the
mathematical details of how the two component model is worked out. Appendix B
determines the maximum value of the $\alpha$ parameter using photo-production
data reported in Ref.[2].

###  1.1 $\pi$ $\pi$ scattering cross section

For the first part of this story we will define what a scattering cross
section is. We will first only consider elastic scattering of pions. Two pions
can scatter at a certain energy which we will call $M_{\pi\pi}$. The
differential cross section $\sigma$ at a given $M_{\pi\pi}$ is

$\frac{d\sigma}{d\phi
d\theta}=\frac{1}{K^{2}}\left|\sum_{\ell}(2\ell+1)T_{\ell}P_{\ell}(cos\theta)\right|^{2}$
(1)

where $\phi$ and $\theta$ are the azimuthal and scattering angles,
respectively. $T_{\ell}$ is a complex scattering amplitude and $\ell$ is the
angular momentum. $P_{\ell}$ is the Legendre polynomial, which is a function
of $cos\theta$. $K$ is the flux factor equal to the pion momentum in the
center of mass. The $T_{\ell}$ elastic scattering amplitudes are complex
amplitudes described by one real number which is in units of angles. The form
of the amplitude is

$T_{\ell}=e^{i\delta_{\ell}}sin\delta_{\ell}$ (2)

We note that $\delta$ depends on the value of $\ell$ and $M_{\pi\pi}$. We will
use the phase shifts given in Ref[3].

##  2 The two component model

Let us consider two pions scattering in the final state of the heavy ion
collision. The scattering will be in some $\ell$ partial wave. The
$M_{\pi\pi}$ of the scattering dipion system will depend on the probability of
the phase space of the overlapping pions. The pions emerge from a close
encounter in a defined quantum state with a random phase. We will call this
amplitude $A$ and note that the absolute value squared of the amplitude is
proportional to the phase space overlap. The emerging pions can re-scatter
through the quantum state of the pions, which is a partial wave or a phase
shift. We have amplitude $A$ plus $A$ times the re-scattering of pions through
the phase shift consistent quantum state of $A$. The correct unitary way to
describe this process is given by Ref[4] equation(4.5)

$T=\frac{V_{1}U_{1}}{D_{1}}+\frac{\left(V_{2}+\frac{D_{12}V_{1}}{D_{1}}\right)\left(U_{2}+\frac{D_{12}U_{1}}{D_{1}}\right)}{D_{2}-\frac{D_{12}^{2}}{D_{1}}}$
(3)

In the above equation we have two terms, 1 and 2. The first term denoted by 1
is the $\pi\pi$ scattering through $\ell$-wave which will become the amplitude
$A$ mentioned above, where $V$ is the incoming and $U$ is the outgoing
$\pi\pi$ system. The second term denoted by 2 is the direct production of the
$\pi\pi$ system in the $\ell$-wave with $V$ being the production, the
propagation being $D$ and the decay being $U$. We see that there are terms
$D_{12}$ which involves a loop of pions between scattering pions and the
formation of a resonance($\ell$ =1 would be a $\rho$) by the pions.

###  2.1 Final equation for the two component model

The complete derivation is in Appendix A. From the appendix we get two terms,
one being the direct production of the resonance or $\pi\pi$ $\ell$-wave phase
shift and the second being the resonance from re-scattering. The final
equation 6 has two important factors, one is two-body phase space and the
other is a coefficient $\alpha$. This coefficient is related to the real part
of the $\pi\pi$ re-scattering loop and is given by equation 4. When the pions
re-scatter or interact at a close distance or a point the real part of
$\alpha$ has its maximum value of $\alpha_{0}$. While if the pions re-scatter
or interact at a distance determined by the diffractive limit the value of
$\alpha$ is zero.

In the equation 6 $|T|^{2}$ is the cross section for $\ell$ partial wave
produced, where $D$ is the direct production amplitude and $A$ is the
amplitude introduced above for the re-scattering pions into the $\ell$-wave
with $\delta_{\ell}$ the $\pi\pi$ phase shift[3]. The $q$ is the $\pi\pi$
center of mass momentum. At a given $p_{t}$ and $y$ bin, $D$ will have a
thermal factor as a function of $M_{\pi\pi}$. The $\alpha$ which is the real
part of the re-scattering factor has a simple form given by

$\alpha=(1.0-\frac{r^{2}}{r_{0}^{2}})\alpha_{0}$ (4)

where $r$ is the radius of re-scattering in fm’s and $r_{0}$ is 1.0 fm or the
limiting range of the strong interaction ranging to $r$ = 0.0 for point like
interactions.

The dependence of $A$ is calculated by the phase space overlap of dipions
added as four vectors and corrected for proper time, with the sum having the
correct $p_{t}$ , $y$ and phase space weighting for $\ell^{th}$ partial wave
for a given $M_{\pi\pi}$.

Finally we must use the correct two body phase space. For a two body system of
pions, phase space goes to an constant as $M_{\pi\pi}$ goes to infinity. Let
us choose this constant to be unity. Phase space which is denoted by PS is
equal to

$PS=\frac{2qB_{\ell}(q/q_{s})}{M_{\pi\pi}}$ (5)

where $B_{\ell}$ is a Blatt-Weisskopf-barrier factor[5] for $\ell$ angular
momentum quantum number. The $q_{s}$ is the momentum related to the range of
interaction of the $\pi\pi$ scattering. 1 fm is the usual interaction distance
which implies that $q_{s}$ is .200 GeV/c. For the $\rho$ meson $\ell$=1 the
barrier factor is $B_{1}$ = $\frac{(q/q_{s})^{2}}{(1+(q/q_{s})^{2})}$. The
phase space factor PS as a function of $q$ near the $\pi\pi$ threshold is
given by $q^{2\ell+1}$. Thus in the appendix we use $q^{2\ell+1}$ for the
factor PS except for equation 6 which is the final equation.

$|T|^{2}=|D|^{2}\frac{sin^{2}\delta_{\ell}}{PS}+\frac{|A|^{2}}{PS}\left|\alpha
sin\delta_{\ell}+PScos\delta_{\ell}\right|^{2}$ (6)

##  3 Re-scattering through the Swave $\pi\pi$ is a two channel problem

In Sec. 2 we derived equation 6 considering only elastic scattering of the
$\pi\pi$ system. If we consider the Dwave it couples to the $f_{2}(1270)$
($J^{PC}$ = $2^{++}$) with 85% of the cross section in the $\pi\pi$ channel.
The Pwave couples to the $\rho(770)$ ($J^{PC}$ = $1^{--}$) where 100% is in
the $\pi\pi$ channel. The Swave $\pi\pi$ ($J^{PC}$ = $0^{++}$) couples to two
resonances the $\sigma$ and the $f_{0}(980)$. The $\sigma$ is purely elastic
while the $f_{0}(980)$ is split between the $\pi$ $\pi$ and $K$ $\overline{K}$
channels. These two channels plus two resonances gives an additional
complexity to the re-scattering problem.

In order to handle the $\pi\pi$ and $K$ $\overline{K}$ channels we will use
the K-matrix approach. When we are below the $K$ $\overline{K}$ threshold the
system is only a one channel problem and the K-matrix is only a single term of
the matrix

$K_{11}=tan\delta_{0}.$ (7)

We see that when $\delta_{0}$ = $90^{\circ}$ that there will be a pole in the
K-matrix. It is standard to expand the K-matrix as a sum of poles.

$K_{11}=\sum_{i}{\frac{2\gamma_{i}^{2}q_{\pi\pi}}{M_{\pi\pi}}\over{(M_{i}^{2}-M_{\pi\pi}^{2})}}$
(8)

Where $\gamma_{i}$ is the coupling of the pole($i^{th}$) to the $\pi\pi$
channel, $q_{\pi\pi}$ is the center of mass momentum of the $\pi\pi$ channel
and $M_{i}$ is the mass of the pole($i^{th}$). The T-matrix is given by

$T_{11}=e^{i\delta_{0}}sin\delta_{0}={K_{11}\over{(1-iK_{11})}}=\,(1-iK)^{-1}K$
(9)

When both channels are open $j=1$ $\pi$ $\pi$ and $j=2$ $K$ $\overline{K}$,
the K-matrix is given by

$K_{11}=\sum_{i}{\frac{2\gamma_{i1}^{2}q_{1}}{M_{1}}\over{(M_{i}^{2}-M_{1}^{2})}},K_{21}=\sum_{i}{\frac{2\gamma_{i2}\sqrt{q_{2}}\gamma_{i1}\sqrt{q_{1}}}{M_{1}}\over{(M_{i}^{2}-M_{1}^{2})}},K_{12}=\sum_{i}{\frac{2\gamma_{i1}\sqrt{q_{1}}\gamma_{i2}\sqrt{q_{2}}}{M_{1}}\over{(M_{i}^{2}-M_{1}^{2})}},K_{22}=\sum_{i}{\frac{2\gamma_{i2}^{2}q_{2}}{M_{1}}\over{(M_{i}^{2}-M_{1}^{2})}}.$
(10)

The T-matrix is given by

$T=\,(\delta-iK)^{-1}K.$ (11)

We fit the Swave ($J^{PC}$ = $0^{++}$) $\pi\pi$ of Ref[3] using three poles
for the $\sigma$, $f_{0}$ and some background from higher mass poles. The
$T_{11}$ amplitude is shown in Figure 1.

Figure 1: The $T_{11}$ amplitude for the Swave ($J^{PC}$ = $0^{++}$) $\pi\pi$
comes from a fit to of Ref[3] using three K-matrix poles for the $\sigma$,
$f_{0}$ and some background from higher mass poles.

Two pions scattering in the final state of the heavy ion collision in a Swave
will be our amplitude $A$, where the emerging pions can re-scatter through the
Swave. We have amplitude $A$ plus $A$ times the re-scattering of pions through
the Swave phase shift.

The term of equation 6 $PSe^{i\delta_{0}}cos\delta_{0}$ is equal to
$PS(1+ie^{i\delta_{0}}sin\delta_{0})$. $T_{11}$ which is equal to
$e^{i\delta_{0}}sin\delta_{0}$ for the one channel case becomes $T_{11}$ =
$\eta e^{i\delta_{0}}sin\delta_{0}$ for the two channel case. Thus the re-
scattering term becomes $PS(1+i\eta e^{i\delta_{0}}sin\delta_{0})$ or
$PS(1+iT_{11})$. Using our K-matrix fit to the Swave one obtain the re-
scattering term plotted in Figure 2. In the lower mass we see a shift of the
spectrum to a lower mass. In the next section will see the same effect for
$\rho(770)$ resonance that re-scattering will shift its mass to lower values.
This shift will be caused by a direct production plus the re-scattering adding
together creating a shifted $\rho(770)$[6]. We see that the $f_{0}$ is a
narrow resonance. The $f_{0}$ resonates at the $K$ $\overline{K}$ threshold.
Direct production of the $f_{0}$ gives a bump at the $K$ $\overline{K}$
threshold and the re-scattering of $\pi\pi$ also gives such a bump at the $K$
$\overline{K}$ threshold(Figure 2). Therefore we will only consider the
$f_{0}$ as a resonance being directly produced and decaying into $\pi\pi$ near
the $K$ $\overline{K}$ threshold.

##  4 The balance function for dipion effective mass

Up to this point in the paper we did not specify the charge of the pions
considered. With the idea of the balance function we look at the creation of
pairs of opposite charge pions. The QGP fireball begins mostly neutral without
a large excess of charge. Pairs of quarks and anti-quarks are created finally
forming hadrons mainly pions. Thus for every $\pi^{+}$ there is a $\pi^{-}$
which balance out the charge. The balance function measures the kinematic
variable between the balancing charge[7]. In our case we want to measure
effective mass of the $\pi^{+}\pi^{-}$ pairs. The dipion effective mass
balance function is defined by

$B(M_{\pi\pi})=\frac{1}{N}\sum_{events}\sum_{pairs}\frac{1}{2}\biggl{\\{}\frac{(M_{\pi^{+}\pi^{-}}-M_{\pi^{+}\pi^{+}})}{N_{+}}+\frac{(M_{\pi^{+}\pi^{-}}-M_{\pi^{-}\pi^{-}})}{N_{-}}\biggr{\\}},$
(12)

where $N$ is the number of events, $N_{+}$ is the number of positive pions per
event, and $N_{-}$ is the number of negative pions per event.

Let us consider the effective mass dipion balance function for p-p collisions
$\sqrt{s_{NN}}$=200 GeV[8]. The effective mass shows a large bump at the
$K^{0}$ mass from $K^{0}_{S}\rightarrow\pi^{+}\pi^{-}$. Above this bump there
is another mass bump which could be the $\rho(770)$. At this mass Pwave should
be the most important partial wave so let us assume this is the case. The
balance function drops off with mass so we assume an exponential fall off with
$M_{\pi^{+}\pi^{-}}$. The $\rho(770)$ should be directly produced and decay
through the Pwave phase shift. We will use equation 6 to fit the p-p balance
function. Figure 3 shows the fit to the STAR data, where the solid line is the
fit and dashed line is the $\rho(770)$. The exponential is amplitude A and the
dotted line is $A$ plus $A$ times the re-scattering of pions through the Pwave
phase shift. The value of $\alpha$ is 0.37 which means the radius of re-
scattering of the pions is 0.9 fm using equation 4 with the value of
$\alpha_{0}$ equal to 2. The solid line which is a good fit to the data peaks
at .73 GeV/c. We see that the shift of this peak is caused by the dotted line
added to the dashed line.

Figure 2: We plot for the Swave ($J^{PC}$ = $0^{++}$) $\pi\pi$ re-scattering
term $\left|PS(1+iT_{11})\right|^{2}$. The $T_{11}$ amplitude comes from a fit
to of Ref[3] using three K-matrix poles for the $\sigma$, $f_{0}$ and some
background from higher mass poles.

Figure 3: The fit to the p-p balance function STAR data[8], where the solid
line is the fit and dashed line is the $\rho(770)$. The exponential is
amplitude A and the dotted line is $A$ plus $A$ times the re-scattering of
pions through the Pwave phase shift.

We turn to the central Au-Au balance function measured by STAR[8]. Again using
the same functions that we used above we fit the data. The solid line is the
fit and dashed line is the $\rho(770)$. The exponential function is amplitude
A and the dotted line is $A$ plus $A$ times the re-scattering of pions through
the Pwave phase shift. The value of $\alpha$ is 0.44 which means the radius of
re-scattering of the pions is 0.88 fm using equation 4 with the value of
$\alpha_{0}$ equal to 2. We see it is a little smaller volume where pions can
re-scatter.

##  5 The dipion effective mass cocktail over a $p_{t}$ range in Au-Au

In order to form this cocktail we need to consider three important ingredients
that come into play. First is the thermal production of resonances that decay
into $\pi^{+}\pi^{-}$ as a function of dipion $p_{t}$. Second is the minijet
production of $\pi^{+}\pi^{-}$ that is not through a resonance decay and
determine its dipion effective mass spectrum as a function of dipion $p_{t}$.
Also we need the break up of the spectrum into partial waves. Third is
rewriting equation 6 in a form that uses resonance or Breit-Wigner parameters
(mass, widths) instead of phase shifts plus modify the equation to use the
derived minijet amplitudes.

###  5.1 Thermal production of resonances

Thermal resonance production will have a Boltzmann weighting of the dipion
effective mass spectrum. Since we are projecting in $p_{t}$ this weighting
will be an exponential function of the transverse mass divided by the
temperature[6,9-13].

$Weight(M_{\pi\pi})=\frac{M_{\pi\pi}}{\sqrt{M_{\pi\pi}^{2}+p^{2}_{t}}}exp\frac{-\sqrt{M_{\pi\pi}^{2}+p^{2}_{t}}}{T}$
(13)

This weight times the Breit-Wigner line shape is the thermal production of the
resonance which decays into the dipion system. The Breit-Wigner line shape is
given by

$BW(M_{\pi\pi})=\frac{M_{\pi\pi}M_{0}\Gamma}{(M^{2}_{0}-M^{2}_{\pi\pi})^{2}+M^{2}_{0}\Gamma^{2}}.$
(14)

Where $\Gamma$ is the $M_{\pi\pi}$ dependent total width

$\Gamma=\Gamma_{0}{\frac{qB_{\ell}(q/q_{s})}{M_{\pi\pi}}\over{\frac{q_{0}B_{\ell}(q_{0}/q_{s})}{M_{0}}}}$
(15)

with $\Gamma_{0}$ being the total width at resonance, $B_{\ell}$ is the Blatt-
Weisskopf-barrier factor[5] for the $\ell$ of the resonance, $q$ is the
$\pi\pi$ center mass momentum, $q_{0}$ is $q$ at resonance, $M_{0}$ is the
mass of the resonance, and $q_{s}$ is center mass momentum related to the
size(1.0 fm is used $q_{s}$ = .200 GeV/c).

Figure 4: The fit to the Au-Au balance function STAR data[8], where the solid
line is the fit and dashed line is the $\rho(770)$. The exponential is
amplitude A and the dotted line is $A$ plus $A$ times the re-scattering of
pions through the Pwave phase shift.

###  5.2 Minijet production of dipions

Partons that under go a hard scattering fragment into hadrons[1, 14]. These
hadrons become part of the outward flow of hadrons with the thermal hadrons.
Hadrons that have long life times will decay outside the freeze-out volume.
For these long lived resonances the dipion spectrum will be given by the
Breit-Wigner line shape of the last subsection. Thus the source of these
resonances either thermal or minijet fragmentation will not be apparent.
Resonances like the $\eta$($c\tau$ = 154000 fm), the $\omega$($c\tau$ = 24
fm), the $\eta^{\prime}$($c\tau$ = 100 fm), the $K^{*}$($c\tau$ = 4 fm), and
the $\phi$($c\tau$ = 50 fm) are decaying outside the freeze-out volume. All
other resonances decay inside this volume like the $\sigma$($c\tau$ = 1/3 fm),
the $\rho(770)$($c\tau$ = 1.3 fm), and the $f_{2}(1270)$($c\tau$ = 1 fm).
Pions from these decays become a source for re-scattering with pion directly
produced or the ones that arise from decays.

We will use the minijet fragmentation code of PYTHIA[14] in order to estimate
the dipion effective mass spectrum. We set up correct kinetics by using
minijets that are predicted by the program HIJING[15] for Au-Au collisions at
$\sqrt{s_{NN}}=$ 200 GeV. We cycle through the minijets produced by HIJING
forming dipion pair($\pi^{+}\pi^{-}$) combinations within a given minijet
fragmentation. We remove the long lived resonance listed above because they
can be handled through the direct thermal production. All pairs that come
directly from the second class short lived resonances are also not considered.
They also can be handled through the direct thermal production. However
combinations of their decay pions with other pions are considered as a source
of minijet dipions. We see from Figure 5 that the hadrons that fragment from
the minijets are moving in the same direction and will interact with each
other.

Figure 5: A minijet parton shower.

The dipions from this minijet source can be selected for its dipion $p_{t}$
range and decomposed into each partial wave $\ell$($\ell$wave) obtaining an
amplitude A for equation 6. We can separate out this dipion spectrum by using
statistical and kinematic weighting. At the highest dipion mass the S, P, D,
and F waves have a $2\ell+1$ statistical weighting. S is 1, P is 3, D is 5,
and F is 7 making 16 units of probability. The kinematic weighting is given by
the Blatt-Weisskopf-barrier factors. At each dipion mass we have a dipion
spectrum from the minijets with a selection on dipion $p_{t}$ which we call
$jet(M_{\pi^{+}\pi^{-}})$. Let us define Z as the ratio of the center mass
dipion momentum $q$ divided $q_{s}$ which is .200 GeV/c(size of 1.0 fm) all
squared

$Z=\frac{q^{2}}{q_{s}^{2}}.$ (16)

The Fwave minijet dipion weight is given by

$F=\frac{7}{16}\frac{Z^{3}}{(Z^{3}+6Z^{2}+45Z+225)}.$ (17)

The Dwave minijet dipion weight is given by

$D=\frac{5}{9}(1.0-F)\frac{Z^{2}}{(Z^{2}+3Z+9)}.$ (18)

The Pwave minijet dipion weight is given by

$P=\frac{3}{4}(1.0-D-F)\frac{Z}{(Z+1)}.$ (19)

The Swave minijet dipion weight is given by

$S=(1.0-P-D-F).$ (20)

Thus $F(M_{\pi^{+}\pi^{-}})=Fjet(M_{\pi^{+}\pi^{-}})$,
$D(M_{\pi^{+}\pi^{-}})=Djet(M_{\pi^{+}\pi^{-}})$,
$P(M_{\pi^{+}\pi^{-}})=Pjet(M_{\pi^{+}\pi^{-}})$,and
$S(M_{\pi^{+}\pi^{-}})=Sjet(M_{\pi^{+}\pi^{-}})$.

Let us choose a dipion $p_{t}$ range and plot the above minijet spectrum. We
choose 1.6 GeV/c $<$ $p_{t}$ $<$ 1.8 GeV/c and plot in Figure 6 the Swave,
Pwave, Dwave, and Fwave from minijets coming from HIJING for Au-Au collisions
at $\sqrt{s_{NN}}=$ 200 GeV.

Figure 6: The dipion $p_{t}$ range 1.6 GeV/c $<$ $p_{t}$ $<$ 1.8 GeV/c showing
the Swave, Pwave, Dwave, and Fwave from minijets coming from HIJING for Au-Au
collisions at $\sqrt{s_{NN}}=$ 200 GeV.

###  5.3 Equation 6 for Breit-Wigner parameters

In this subsection we will alter equation 6 so it can use Breit-Wigner
parameters (mass, width) instead of phase shifts. We will also need to modify
the re-scattering part of the equation in order to have the correct threshold
behavior we have just introduced into the minijet partial waves above. The
phase shift can be written for the $\ell^{th}$ wave as

$cot\delta_{\ell}=\frac{(M_{\ell}^{2}-M_{\pi\pi}^{2})}{M_{\ell}\Gamma_{\ell}},$
(21)

where $M_{\ell}$ is the mass of the resonance in the $\ell$wave and
$\Gamma_{\ell}$ is its total width.

$\Gamma_{\ell}=\Gamma_{0\ell}{\frac{qB_{\ell}(q/q_{s})}{M_{\pi\pi}}\over{\frac{q_{\ell}B_{\ell}(q_{\ell}/q_{s})}{M_{\ell}}}}$
(22)

with $\Gamma_{0\ell}$ the total width at resonance, $B_{\ell}$ is the Blatt-
Weisskopf-barrier factor for the $\ell$ of the resonance, $q$ is the $\pi\pi$
center mass momentum, $q_{\ell}$ is $q$ at resonance, $M_{\ell}$ is the mass
of the resonance, and $q_{s}$ is center mass momentum related to the size(1.0
fm is used $q_{s}$ = .200 GeV/c).

Using equation 21 we rewrite equation 6 as

$|T_{\ell}|^{2}=|D_{\ell}|^{2}\frac{sin^{2}\delta_{\ell}}{PS_{\ell}}+\frac{|A_{\ell}|^{2}sin^{2}\delta_{\ell}}{PS_{\ell}}\left|\alpha+PS_{\ell}cot\delta_{\ell}\right|^{2}$
(23)

The $D_{\ell}$ is the thermal production term and is constant except for the
Boltzmann weight. The expected threshold behavior $q^{2\ell+1}$ comes from the
$sin\delta_{\ell}$ term. Since there is $sin^{2}\delta_{\ell}$ one of the
$q^{2\ell+1}$ is killed off by dividing by $PS_{\ell}$. In Figure 6 we have
put into our minijet $A_{\ell}$ the correct threshold $q^{2\ell+1}$ so we need
kill off the $q^{2\ell+1}$ of the other $sin\delta_{\ell}$ term. Therefore
equation 6 for our minijet $A_{\ell}$ we will use

$|T_{\ell}|^{2}=|D_{\ell}|^{2}\frac{sin^{2}\delta_{\ell}}{PS_{\ell}}+\frac{|A_{\ell}|^{2}sin^{2}\delta_{\ell}}{PS_{\ell}^{2}}\left|\alpha+PS_{\ell}cot\delta_{\ell}\right|^{2}$
(24)

Rewriting equation 6 for each partial wave with Breit-Wigner parameters the
first term becomes

$|T_{\ell}|_{1}^{2}=|D_{\ell}|^{2}\frac{M_{\pi\pi}^{2}}{\sqrt{M_{\pi\pi}^{2}+p^{2}_{t}}}exp\frac{-\sqrt{M_{\pi\pi}^{2}+p^{2}_{t}}}{T}\frac{M_{\ell}\Gamma_{\ell}}{(M_{\ell}^{2}-M_{\pi\pi}^{2})^{2}+M_{\ell}^{2}\Gamma_{\ell}^{2}},$
(25)

while the second term

$|T_{\ell}|_{2}^{2}=|A_{\ell}|^{2}\frac{M_{\ell}^{2}\Gamma_{\ell}^{2}}{(M_{\ell}^{2}-M_{\pi\pi}^{2})^{2}+M_{\ell}^{2}\Gamma_{\ell}^{2}}\left|\alpha+\frac{2qB_{\ell}(\frac{q}{q_{s}})(M_{\ell}^{2}-M_{\pi\pi}^{2})}{M_{\pi\pi}M_{\ell}\Gamma_{\ell}}\right|^{2}\left(\frac{M_{\pi\pi}^{2}}{4q^{2}B_{\ell}^{2}(\frac{q}{q_{s}})}\right).$
(26)

$|T|^{2}=\sum_{\ell}|T_{\ell}|^{2}$ (27)

where

$|T_{\ell}|^{2}=|T_{\ell}|_{1}^{2}+|T_{\ell}|_{2}^{2}$ (28)

and
$|A_{0}|^{2}=S(M_{\pi^{+}\pi^{-}})$,$|A_{1}|^{2}=P(M_{\pi^{+}\pi^{-}})$,$|A_{2}|^{2}=D(M_{\pi^{+}\pi^{-}})$,
and $|A_{3}|^{2}=F(M_{\pi^{+}\pi^{-}})$.

###  5.4 STAR data dipion $p_{t}$ range (1.6 GeV/c $<$ $p_{t}$ $<$ 1.8 GeV/c)

We have fitted a dipion $p_{t}$ range using equation 27 above for the STAR
data Au-Au collisions at $\sqrt{s_{NN}}=$ 200 GeV 40% to 80% centrality. We
included minijets up to $\ell$ = 3 and resonances $\sigma$ $\ell$ = 0,
$\rho(770)$ $\ell$ = 1, and $f_{2}(1270)$ $\ell$ = 2. Using the arguments of
Sec. 3 we added the $f_{0}$ as a direct thermal term ($|T_{0}|_{1}^{2}$) and
only the $\sigma$ interfered with $\ell$ = 0 minijet background. Two other
thermal terms are present in the cocktail, the $K^{0}_{S}$ and the
$\omega_{0}$.

Finally the threshold effective mass region .280 GeV to .430 GeV is dominated
by the Swave and receives contributions from minijet fragmentation, $\pi\pi$
Swave phase shift, $\eta$ decay, HBT adding to the like sign $\pi\pi$
distribution that has been subtracted away from the unlike sign $\pi\pi$ and
the coulomb correction between the charged pions. The minijet fragmentation is
the least known of the effects since we relied on PYTHIA, however there are
large uncertainty in all the other effects. So for this fit we let the minijet
fragmentation be free to fit the data and let the Breit-Wigner parameters for
the $\sigma$ determine the Swave phase shifts plus leaving out all other
effects. The results of this fit is shown in Figure 7. Table I shows the
Breit-Wigner parameters used in the fit.

Table I. The Bret-Wigner Parameters of the fit.

Table I
---
resonance | mass(GeV) | width(GeV)
$\sigma$ | 1.011 | 1.015
$f_{0}$ | 0.973 | 0.041
$\rho$ | 0.748 | 0.147
$f_{2}$ | 1.275 | 0.185

Figure 7: Fit to STAR dipion effective mass distribution (1.6 GeV/c $<$
$p_{t}$ $<$ 1.8 GeV/c) for Au-Au collisions at $\sqrt{s_{NN}}=$ 200 GeV 40% to
80% centrality using equation 27. See text for complete information.

The value of $\alpha$ for this $p_{t}$ range is .806. For coherent photo-
production of the $\rho(770)$ from STAR data[2] the value of $\alpha$ is 2.0
or $\alpha_{0}$ because the two pions emerge from a point source(see Appendix
B). Thus the radius of the size associated with re-scattering is .773 fm for
this $p_{t}$ range from equation 4. Since the thermal term for resonances
could also come from minijet fragmentation, we need additional studies to
determine how much production comes from minijets and how much comes from the
fireball directly[16].

##  6 Summary and Discussion

In this article we start with the basic definition of elastic $\pi\pi$
scattering. Next we show how re-scattering of pions depends on the unitary
condition that interactions present in the phase shift of an orbital state
must interact all the time. The process of parton fragmentation into dipion
states through unitarity leads to a equation of production and re-scattering
in a given orbital quantum number. This equation (equation 6) has two
components in each orbital state: one being the thermal production of
resonances in a dipion orbital state, the other is the re-scattering of
dipions coming from parton or minijet fragmentation into the dipion orbital
state which do not come directly from the resonance. Unitarity requires that
there most be re-scatter through the resonance of the phase shift. This
equation is used over and over again with the details presented in the
appendix.

Equation 6 considers only elastic scattering of the $\pi\pi$ system. We
considered the Dwave which couples to the $f_{2}(1270)$ ($J^{PC}$ = $2^{++}$)
with 85% of the cross section in the $\pi\pi$ channel. The Pwave which coupled
to the $\rho(770)$ ($J^{PC}$ = $1^{--}$) where 100% is in the $\pi\pi$
channel. The Swave $\pi\pi$ ($J^{PC}$ = $0^{++}$) couples to two resonances
the $\sigma$ and the $f_{0}(980)$. The $\sigma$ is purely elastic while the
$f_{0}(980)$ is split between the $\pi$ $\pi$ and $K$ $\overline{K}$ channels.
We saw that the $f_{0}$ was a narrow resonance. The $f_{0}$ resonates at the
$K$ $\overline{K}$ threshold. Direct production of the $f_{0}$ gives a bump at
the $K$ $\overline{K}$ threshold and the re-scattering of $\pi\pi$ also gives
such the same bump at the $K$ $\overline{K}$ threshold Sec. 3 (Figure 2).
Therefore we considered the $f_{0}$ as a resonance being directly produced and
decaying into $\pi\pi$ near the $K$ $\overline{K}$ threshold.

We used equation 6 to fit the dipion effective mass balance function. We
assumed the Pwave was the most important partial wave in the dipion effective
mass range of the fit. The balance function drops off with mass so we assumed
an exponential fall off with $M_{\pi^{+}\pi^{-}}$. The two parts of equation 6
where the first part being the $\rho(770)$ directly produced and decaying
through the Pwave phase shift with the second part being the exponential
function plus re-scattering through the $\rho(770)$. With this simple model we
were able to fit the p-p balance (Figure 3) and the central Au-Au
balance(Figure 4). The observed mass shifts were due to the re-scattering
effect, being $\sim$40 MeV.

We chose a dipion $p_{t}$ range to do a cocktail fit to the effective dipion
mass spectrum. We needed three important ingredients in order to do this fit.
First is the thermal production of resonances that decay into $\pi^{+}\pi^{-}$
as a function of dipion $p_{t}$. Second is the dipion effective mass spectrum
as a function of dipion $p_{t}$ coming from minijet production not through
resonance decay. Third we needed to rewrite equation 6 in a form that uses
resonance or Breit-Wigner parameters (mass, widths) instead of phase shifts.
Once these three ingredients were developed we were successful in doing a
cocktail fit (Figure 7). Since the thermal term for resonances could also come
from minijet fragmentation, we need additional studies to determine how much
production comes from minijets and how much comes from the fireball
directly[16]. Other fits to more $p_{t}$ ranges are considered in Ref.[17].

##  7 Acknowledgments

This research was supported by the U.S. Department of Energy under Contract
No. DE-AC02-98CH10886. The author thanks William Love for the STAR analysis of
the angular correlation data from Run 4. Also for his assistance in the
production of figures. It is sad that he is gone.

##  Appendix A Appendix

Starting with equation (4.5) from Ref[4]

$T=\frac{V_{1}U^{\prime}_{1}}{D_{1}}+\frac{\left(V_{2}+\frac{D_{12}V_{1}}{D_{1}}\right)\left(U^{\prime}_{2}+\frac{D_{12}U^{\prime}_{1}}{D_{1}}\right)}{D_{2}-\frac{D_{12}^{2}}{D_{1}}}$
(29)

In order to have the correct threshold kinematics, we define

$U^{\prime}_{1}=U_{1}\sqrt{q^{2\ell+1}}$ (30)
$U^{\prime}_{2}=U_{2}\sqrt{q^{2\ell+1}}$ (31)

where $q$ is the $\pi\pi$ center of mass momentum and $\ell$ is the value of
the angular momentum. The amplitude $A$ of the text is given by

$\frac{V_{1}U_{1}}{D_{1}}=A$ (32)

Thus we have

$\frac{V_{1}U^{\prime}_{1}}{D_{1}}=A\sqrt{q^{2\ell+1}}$ (33)

The phase shift for the $\ell^{th}$ partial wave will be given by
$\delta_{\ell}$, where

$\frac{U^{\prime}_{2}U^{\prime}_{2}}{D_{2}}=e^{i\delta_{\ell}}sin\delta_{\ell}$
(34)

The above equality is true if the $D_{1}$ mode plays no role in the $\pi\pi$
scattering in the $\ell^{th}$ partial wave. But in the initial state there is
a large production of $D_{1}$. The $U^{\prime}$s are the basic coupling of the
$D^{\prime}$s to the $\pi\pi$ system. In order to decouple $D_{1}$ from the
$\pi\pi$ system $U_{1}$ must go to zero. We can maintain a finite production
of $D_{1}$ if we define

$V_{1}=\frac{1}{U_{1}}$ (35)

Thus the first term in the equation becomes

$\frac{V_{1}U^{\prime}_{1}}{D_{1}}=\frac{\frac{1}{U_{1}}U_{1}\sqrt{q^{2\ell+1}}}{D_{1}}=\frac{\sqrt{q^{2\ell+1}}}{D_{1}}=A\sqrt{q^{2\ell+1}}$
(36)

The form of $D_{12}$ is given by

$D_{12}=\alpha U_{1}U_{2}+iq^{2\ell+1}U_{1}U_{2}$ (37)

$D_{12}$ is the real and imaginary part of the two pion loop from state 1 to
state 2. The $U^{\prime}$s are the $\pi\pi$ couplings and the imaginary part
goes to zero at the $\pi\pi$ threshold. The $\alpha$ factor is $\alpha_{0}$
for re-scattering coming from a point source, but goes to zero when re-
scattering is diffractive. A simple form for $\alpha$ is given by

$\alpha=(1.0-\frac{r^{2}}{r_{0}^{2}})\alpha_{0}$ (38)

where $r$ is the radius of re-scattering in fm’s and $r_{0}$ is 1.0 fm or the
limiting range of the strong interaction. The second term of the first
equation is

$\frac{\left(V_{2}+\frac{D_{12}V_{1}}{D_{1}}\right)\left(U^{\prime}_{2}+\frac{D_{12}U^{\prime}_{1}}{D_{1}}\right)}{D_{2}-\frac{D_{12}^{2}}{D_{1}}}$
(39)

Rewriting

$\frac{\left(V_{2}+\frac{\alpha
U_{1}U_{2}V_{1}}{D_{1}}+iq^{2\ell+1}\frac{U_{1}U_{2}V_{1}}{D_{1}}\right)\left(U^{\prime}_{2}+\frac{D_{12}U^{\prime}_{1}}{D_{1}}\right)}{D_{2}-\frac{D_{12}^{2}}{D_{1}}}$
(40)

Let us make substitutions

$V_{1}=\frac{1}{U_{1}},U_{2}=\frac{U^{\prime}_{2}}{\sqrt{q^{2\ell+1}}},\frac{1}{D_{1}}=A,D_{12}=0$
(41)

The second term becomes

$\frac{\left(V_{2}+\frac{A\alpha
U^{\prime}_{2}}{\sqrt{q^{2\ell+1}}}+i\sqrt{q^{2\ell+1}}AU^{\prime}_{2}\right)U^{\prime}_{2}}{D_{2}}$
(42)

The first term is

$\frac{V_{1}U^{\prime}_{1}}{D_{1}}=A\sqrt{q^{2\ell+1}}$ (43)

Adding the first and the second terms and substituting the phase shift,

$T=\frac{V_{2}}{U_{2}}\frac{e^{i\delta_{\ell}}sin\delta_{\ell}}{\sqrt{q^{2\ell+1}}}+A\left(\frac{e^{i\delta_{\ell}}\alpha
sin\delta_{\ell}}{\sqrt{q^{2\ell+1}}}+\sqrt{q^{2\ell+1}}e^{i\delta_{\ell}}cos\delta_{\ell}\right)$
(44)

The term with the factor $\frac{V_{2}}{U_{2}}$ is the direct production of the
dipion system. We shall call this amplitude $D$. The re-scattered amplitude is
$A$ and is modified by the dipion phase shift. These two amplitudes have some
random phase and are not coherent. Thus the cross section is

$|T|^{2}=|D|^{2}\frac{sin^{2}\delta_{\ell}}{PS}+\frac{|A|^{2}}{PS}\left|\alpha
sin\delta_{\ell}+PScos\delta_{\ell}\right|^{2}$ (45)

##  Appendix B Appendix

In this appendix we determine the value $\alpha$ coming from a point
source(thus $\alpha_{0}$ see equation 4 and 38) using photo-production data
reported in Ref.[2]. When a photon interacts with a strong field a $\rho$ or a
$\pi^{+}$ $\pi^{-}$ pair is formed in a Pwave, both states are coherent with
each other and must be added together(see equation 3 of Ref.[2]). We write
equation 3 using the phase shift of Pwave $\pi^{+}$ $\pi^{-}$ scattering and
the width($\Gamma_{0}$) of the $\rho$ resonance. Threshold behavior is added
using the phase space factor (PS) for Pwave production (see equation 5 of Sec.
2.1).

$\frac{dN}{dM_{\pi\pi}}=\left|A_{\rho}\frac{sin\delta
e^{i\delta}}{\Gamma_{0}^{1/2}(\frac{PS}{PS_{0}})^{1/2}}+(\frac{PS}{PS_{0}})^{1/2}B_{\pi\pi}\right|^{2}$
(46)

The Pwave phase factor is

$PS=\frac{\frac{2q(q/q_{s})^{2}}{(1+(q/q_{s})^{2})}}{M_{\pi\pi}},$ (47)

where $q$ is the center mass momentum of the $\pi$ $\pi$ and $q_{s}$ is
related to the range of interaction of the $\pi\pi$ scattering. 1 fm is the
usual interaction distance which implies that $q_{s}$ is .200 GeV/c.
$M_{\pi\pi}$ is the effective mass of the system. The Pwave phase factor at
the $\rho$ resonance is

$PS_{0}=\frac{\frac{2q(q_{0}/q_{s})^{2}}{(1+(q_{0}/q_{s})^{2})}}{M_{\rho}},$
(48)

where $q_{0}$ is the center mass momentum of the $\pi$ $\pi$ at the $\rho$
mass.

From equation 46 we pull out a common phase space ratio from both terms,

$\frac{dN}{dM_{\pi\pi}}=\frac{1}{(\frac{PS}{PS_{0}})}\left|A_{\rho}\frac{sin\delta
e^{i\delta}}{\Gamma_{0}^{1/2}}+(\frac{PS}{PS_{0}})B_{\pi\pi}\right|^{2}.$ (49)

The ratio of the $\pi$ $\pi$ production amplitude over the $\rho$ production
amplitude as averaged over world data is

$\frac{B_{\pi\pi}}{A_{\rho}}=0.85.$ (50)

Using equation 50 and substituting this relation for $A_{\rho}$ into equation
49 we obtain

$\frac{dN}{dM_{\pi\pi}}=\frac{B_{\pi\pi}^{2}}{(\frac{PS}{PS_{0}})}\left|\frac{sin\delta
e^{i\delta}}{0.85\Gamma_{0}^{1/2}}+(\frac{PS}{PS_{0}})\right|^{2}.$ (51)

The second term of equation 51 can be expanded because 1 =
$e^{-i\delta}e^{i\delta}$ becoming

$\frac{dN}{dM_{\pi\pi}}=\frac{B_{\pi\pi}^{2}}{(\frac{PS}{PS_{0}})}\left|\frac{sin\delta
e^{i\delta}}{0.85\Gamma_{0}^{1/2}}+(\frac{PS}{PS_{0}})cos\delta
e^{i\delta}-i(\frac{PS}{PS_{0}})sin\delta e^{i\delta}\right|^{2}.$ (52)

There is a common phase factor($e^{i\delta}$) on all three terms which has a
magnitude of 1. The magnitude for all three terms can be written as

$\frac{dN}{dM_{\pi\pi}}=(\frac{PS}{PS_{0}})B_{\pi\pi}^{2}sin^{2}\delta+\frac{B_{\pi\pi}^{2}}{PS(PS_{0})}\left|\frac{PS_{0}sin\delta}{0.85\Gamma_{0}^{1/2}}+PScos\delta\right|^{2}.$
(53)

The second term has an amplitude of the form of equation 6 with

$\alpha sin\delta+PScos\delta.$ (54)

This implies that the maximum value of $\alpha$ is $\alpha_{0}$ and given by

$\alpha_{0}=\frac{PS_{0}}{0.85\Gamma_{0}^{1/2}}\simeq 2.0.$ (55)

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