2-Adic Valuations of Quadratic Sequences
Will Boultinghouse

Kentucky Wesleyan College

Division of Natural Sciences and Mathematics

3000 Frederica Street

Owensboro, KY 42301

USA

Jane Long

Stephen F. Austin State University

Department of Mathematics and Statistics

Nacogdoches, TX 75962-3040

USA

Olena Kozhushkina

Ursinus College

Department of Mathematics and Computer Science

Collegeville, PA 19426

USA

Justin Trulen

Kentucky Wesleyan College

Division of Natural Sciences and Mathematics

3000 Frederica Street

Owensboro, KY 42301

USA

###### Abstract

We determine properties of the 2-adic valuation sequences for general
quadratic polynomials with integer coefficients directly from the
coefficients. These properties include boundedness or unboundedness,
periodicity, and valuations at terminating nodes. We completely describe the
periodic sequences in the bounded case. Throughout, we frame results in terms
of trees and sequences.

##  1 Introduction

For $p$ prime and $n\in\mathbb{N}=\\{0,1,2,3,\ldots\\}$, the exponent of the
highest power of $p$ that divides $n$ is called the $p$-adic valuation of $n$,
which we denote $\nu_{p}(n)$. The valuation of $0$ is defined to be $+\infty$.
Formally, the valuation of a positive integer $n$ of the form $n=p^{k}d$,
where $k\in\mathbb{N}$ and $d$ is an integer not divisible by $p$, is
$\nu_{p}(n)=k$. We can find $p$-adic valuations of sequences by finding the
valuation of each successive term. The present work considers 2-adic
valuations of sequences generated from the natural numbers by evaluating
quadratic functions of the form $f(n)=an^{2}+bn+c$ where $a,b,c\in\mathbb{Z}$
and $a\neq 0$.

Information about sequences of valuations can be viewed in two different ways:
in terms of sequences and in terms of trees. We let $(\nu_{2}(f(n)))_{n\geq
0}$ denote the sequence of 2-adic valuations for the quadratic function
$f(n)$. Since every positive natural number $n$ can be written in the form
$n=2^{k}d$, where $d$ is not divisible by 2, in many cases, we can determine
the valuations of outputs of the quadratic function $f(n)$ using
characteristics of the coefficients $a$, $b$, and $c$. The main results are
given in Theorems 1 and 2; one would anticipate these results can be extended
to odd primes with some modifications, which will be addressed in future work.

######  Theorem 1.

Let $f(n)=an^{2}+bn+c$ where $a,b,c\in\mathbb{Z}$ with $a\neq 0$ and, without
loss of generality, $a,b,c$ are not all even. Then

  1. 1.

If $a$ and $b$ are even and $c$ is odd, then $\nu_{2}(f(n))=0$ for all
$n\in\mathbb{N}$.

  2. 2.

If $a$ is even and $b$ is odd, then $(\nu_{2}(f(n)))_{n\geq 0}$ is an
unbounded sequence.

  3. 3.

If $a$ is odd and $b$ is even, then

    1. (a)

if $b^{2}-4ac=0$, then $(\nu_{2}(f(n)))_{n\geq 0}$ is an unbounded sequence;

    2. (b)

if $b^{2}-4ac=4^{\ell}\Delta$ for $\ell\in\mathbb{Z}^{+}$ as large as possible
and $\Delta\equiv 1\ \mbox{\rm(mod}\ 8\mbox{\rm)}$, then
$(\nu_{2}(f(n)))_{n\geq 0}$ is an unbounded sequence;

    3. (c)

if $b^{2}-4ac=4^{\ell}\Delta$ for $\ell\in\mathbb{Z}^{+}$ as large as
possible, $\Delta\equiv m\ \mbox{\rm(mod}\ 8\mbox{\rm)}$, and
$m\in\left\\{2,3,5,6,7\right\\}$, then the sequence $(\nu_{2}(f(n)))_{n\geq
0}$ is bounded and its minimal period length equals $2^{\ell}$.

  4. 4.

If $a$ and $b$ are odd and $c$ is even, then $(\nu_{2}(f(n)))_{n\geq 0}$ is an
unbounded sequence.

  5. 5.

If $a$, $b$, and $c$ are odd, then $\nu_{2}(f(n))=0$ for all $n\in\mathbb{N}$.

Theorem 1 is proved in Section 3. Henceforward, we will refer to the minimal
period length simply as the period. In Case 3, we use the discriminant to
determine whether roots to $f(n)=0$ lie in the 2-adic numbers $\mathbb{Q}_{2}$
or the ring of 2-adic integers $\mathbb{Z}_{2}$. Corollary 10 takes care of
Case 3(a). Even though the statement of this theorem only classifies these
sequences as unbounded, the proofs of Cases 2 and 4 reveal more information
about the 2-adic valuations. Theorem 1 represents a complete answer to when
$\nu_{2}(f(n))_{n\geq 0}$ is bounded or unbounded using only the coefficients
of the quadratic polynomial. Furthermore, Theorem 1 gives an explicit period
length for the bounded sequences which can be determined by the coefficients
of the quadratic polynomial. In the unbounded cases we are able to describe
what possible valuations will be for certain subsequences. Such statements are
easier to frame in the sense of trees, which are discussed in Section 2.
Theorem 2, proved in Sections 4 and 5, completely determines all valuations in
the non-trivial bounded case (3(c) of Theorem 1).

######  Theorem 2.

Let $f(n)=an^{2}+bn+c$ where $a,b,c\in\mathbb{Z}$. If $a$ is odd and $b$ is
even and $b^{2}-4ac=4^{\ell}\Delta$ for $\ell\in\mathbb{Z}^{+}$ as large as
possible with $\ell\geq 2$, $\Delta\equiv m\ \mbox{\rm(mod}\ 8\mbox{\rm)}$,
and $m\in\left\\{2,3,5,6,7\right\\}$, then the sequence
$(\nu_{2}(f(n)))_{n\geq 0}$ is bounded with period equal to $2^{\ell}$.
Furthermore, we have the following 2-adic valuations:

$\nu_{2}(f(n))=\begin{cases}0,&\ \text{if}\ n\equiv
a^{-1}\left(1-\frac{b}{2}\right)\ \mbox{\rm(mod}\ 2\mbox{\rm)};\\\ 2(i-1),&\
\text{if}\ n\equiv a^{-1}\left(2^{i-1}-\frac{b}{2}\right)\ \mbox{\rm(mod}\
2^{i}\mbox{\rm)}\ \text{with}\ 2\leq i<\ell;\\\ 2(\ell-1),&\ \text{if}\
n\equiv a^{-1}\left(2^{\ell-1}-\frac{b}{2}\right)\ \mbox{\rm(mod}\
2^{\ell}\mbox{\rm)}\ \text{and}\ m=6,2;\\\ 2\ell-1,&\ \text{if}\ n\equiv
a^{-1}\left(2^{\ell-1}-\frac{b}{2}\right)\ \mbox{\rm(mod}\
2^{\ell}\mbox{\rm)}\ \text{and}\ m=7,3;\\\ 2\ell,&\ \text{if}\ n\equiv
a^{-1}\left(2^{\ell-1}-\frac{b}{2}\right)\ \mbox{\rm(mod}\
2^{\ell}\mbox{\rm)}\ \text{and}\ m=5;\\\ 2\ell-1,&\ \text{if}\ n\equiv
a^{-1}\left(2^{\ell}-\frac{b}{2}\right)\ \mbox{\rm(mod}\ 2^{\ell}\mbox{\rm)}\
\text{and}\ m=6,2;\\\ 2(\ell-1),&\ \text{if}\ n\equiv
a^{-1}\left(2^{\ell}-\frac{b}{2}\right)\ \mbox{\rm(mod}\ 2^{\ell}\mbox{\rm)}\
\text{and}\ m=7,5,3;\\\ \end{cases}$

where $a^{-1}$ is the inverse of $a\ \mbox{\rm(mod}\ 2^{\ell}\mbox{\rm)}$.

The case $\ell=1$ is covered by Lemma 13. In this case, the sequence is
periodic with period equal to 2. Theorem 2 is proved in Proposition 12 and
Corollary 24. Both of these results are an extension of the work by Byrnes et
al. [3], which only considered quadratics of the form $f(n)=an^{2}+c$. The
work of Medina et al. [6] details conditions under which these sequences are
bounded or unbounded for general primes but we extend these results for $p=2$
by providing the exact conditions on the coefficients of quadratic equations.
Furthermore, we provide a closed form giving the exact valuation for the
bounded sequences relying only on the coefficients of the quadratic function.
Boundedness of $p$-adic valuations of polynomial sequences is also discussed
in Bell’s work [2].

##  2 Parity and trees

Consider a quadratic function of the form $f(n)=an^{2}+bn+c,$ where $a,b$ and
$c$ are integers and $a$ is nonzero. To prove the results stated in Theorems 1
and 2, we consider the eight possible cases based on the parity of the
coefficients $a$, $b$, and $c$. In the case where $a,b,$ and $c$ are all even,
there exists an $i\in\mathbb{N}$ such that $2^{i}$ divides $a,b$, and $c$ but
$2^{i+1}$ does not. Then $f(n)=2^{i}(a_{0}n^{2}+b_{0}n+c_{0})$ and it follows
that $\nu_{2}(f(n))=i+\nu_{2}(a_{0}n^{2}+b_{0}n+c_{0})$. Hence, this case can
be reduced to one of the other seven cases. So we assume, unless stated
otherwise, that $a$, $b$, and $c$ are not all even.

Two more cases of Theorem 1 are trivial (Case 1 where $a,b$ are even, and Case
5, where $a,b,$ and $c$ are odd), since $\nu_{2}(f(n))=0$ for all
$n\in\mathbb{N}$. For the remaining five cases, we classify the behavior using
trees. In the case that $a$ is odd and $b$ is even we show, with the help of
the discriminant, that $f(n)=0$ has a root in $\mathbb{Q}_{2}$. We must take
some care since some quadratics may not have a zero in $\mathbb{Q}_{2}$.

As discussed in Section 1, we can present information about the sequence of
valuations using a tree. We begin the construction of the tree with the top
node representing the valuation of the quadratic $f(n)$ evaluated at any
natural number $n$. If the $2$-adic valuation is constant for every $n$ in
this node, then we stop the construction, as $\nu_{2}(n)$ is completely
determined for the sequence. If $\nu_{2}(n)$ is not constant, this node splits
into two branches, where one branch represents all numbers of the form $n=2q$
and the other branch represents all numbers of the form $n=2q+1$, where in
both cases $q\in\mathbb{N}$. We then repeat this step as necessary to create
the tree. The nodes correspond to the sets
$\\{2^{i}q+r_{i-1}|q\in\mathbb{N}\\}$ where

$r_{i-1}=\sum_{k=0}^{i-1}\alpha_{k}2^{k},$ (1)

for fixed coefficients $\alpha_{k}\in\\{0,1\\}$. This process does not always
terminate. If it terminates, we say that the tree is finite; otherwise, the
tree is infinite. We say a node is non-terminating if $(\nu_{2}(f(n)))_{n\geq
0}$ is non-constant for every $n$ in that equivalence class. We say a node is
terminating if $(\nu_{2}(f(n)))_{n\geq 0}$ is constant for every $n$ in that
equivalence class. In practice, we label the node with this constant
valuation.

$\nu_{2}(f(n))$$\nu_{2}(f(2q+1))$$\nu_{2}(f(4q+3))$$\nu_{2}(f(4q+1))$$\nu_{2}(f(2q))$$\nu_{2}(f(4q+2))$$\nu_{2}(f(4q))$$2q+1$$2q$$4q+3$$4q+1$$4q+2$$4q$
Figure 1: Levels 0, 1, and 2 of a tree.

For each of the remaining five nontrivial cases on the parity of the
coefficients $a$, $b$ and $c$, either $(\nu_{2}(f(n)))_{n\geq 0}$ produces a
finite tree or an infinite tree. We say a finite tree has $\ell$ levels if
there exists $\ell\in\mathbb{Z}^{+}$ such that for all
$r_{\ell-1}\in\\{0,1,2,\ldots,2^{\ell}-1\\}$ we have
$(\nu_{2}(f(2^{\ell}q+r_{\ell-1})))_{q\geq 0}$ constant for all
$q\in\mathbb{N}$, and $\ell$ is the smallest possible value. Every node at
level $\ell$ in a finite tree has a constant valuation, which depends on
$r_{\ell-1}$.

Each node of a tree represents a subsequence of the sequence of 2-adic
valuations. A finite tree of $\ell$ levels represents a sequence with period
equal to $2^{\ell}$.

In the literature, these finite trees are also called finite automata. The
sequences generated via the 2-adic valuation are called 2-automatic sequences
and, in particular, the sequences $f(2^{i}q+r)$ are known as the 2-kernel
sequences. See Allouche and Shallit’s book [1] and Bell’s paper [2] for more
details.

###  2.1 2-adic numbers and selected lemmas

First, we state several well-known lemmas. The first is a well-established
fact about the $p$-adic valuation, which can also be defined on the set
$\mathbb{Q}$ and extends to $\mathbb{Q}_{2}$; see Lemma 3.3.2 in [5].

######  Lemma 3.

Let $x,y\in\mathbb{Q}$, then $\nu_{p}(xy)=\nu_{p}(x)+\nu_{p}(y)$.

An element $n$ in $\mathbb{Q}_{2}$ can be represented in the form

$n=\sum_{i=k}^{\infty}\alpha_{i}2^{i}$ (2)

where $k=-\nu_{2}(n)$ and $\alpha_{i}\in\left\\{0,1\right\\}$ for all $i$; it
is well-known that this representation is unique.

Lemma 3 and the construction of $\mathbb{Q}_{2}$ are well-known [5]. Medina et
al. [6] provide a useful characterization of the sequence of 2-adic valuations
of a polynomial. Before we state the result, we recall the following
characterization of the ring of 2-adic integers

$\mathbb{Z}_{2}=\left\\{n\in\mathbb{Q}_{2}:n=\sum_{i=0}^{\infty}\alpha_{i}2^{i}\
\text{where}\ \alpha_{i}\in\left\\{0,1\right\\}\right\\}.$

######  Lemma 4.

([6], Theorem 2.1) Let $f(n)\in\mathbb{Z}[n]$ be a polynomial that is
irreducible over $\mathbb{Z}$. Then $(\nu_{2}(f(n)))_{n\geq 0}$ is either
periodic or unbounded. Moreover, $(\nu_{2}(f(n)))_{n\geq 0}$ is periodic if
and only if $f(n)$ has no zeros in $\mathbb{Z}_{2}$. In the periodic case, the
minimal period length is a power of $2$.

We assume that the quadratic $f(n)$ is irreducible because, if not, by Lemma
3,

$\nu_{p}(f(n))=\nu_{p}(g(n)\cdot h(n))=\nu_{p}(g(n))+\nu_{p}(h(n)),$

where $g(n)$ and $h(n)$ are irreducible.

Therefore, to determine whether $(\nu_{2}(f(n)))_{n\geq 0}$ is periodic or
unbounded, it suffices to determine if $f(n)$ has zeros in $\mathbb{Q}_{2}$
and then determine whether the zeros are also in $\mathbb{Z}_{2}$. The
following lemmas will be used in Section 3 to identify when the square root of
a number is in $\mathbb{Z}_{2}$. The version of Hensel’s lemma stated below
determines when a polynomial in $\mathbb{Z}_{2}[x]$ has zeros in
$\mathbb{Z}_{2}$. Lemma 6, which follows from Lemma 5, specifically determines
whether the polynomial $f(x)=x^{2}-a$ has zeros in $\mathbb{Z}_{2}$.

######  Lemma 5.

(Hensel’s lemma, [7, Sec. 6.4]) Assume that $P\in\mathbb{Z}_{2}[x]$ and
$x_{0}\in\mathbb{Z}_{2}$ satisfies

$P(x_{0})\equiv 0\ \mbox{\rm(mod}\ 2^{n}\mbox{\rm)}$

If $\phi=\nu_{2}(P^{\prime}(x_{0}))<n/2$, then there exists a unique zero
$\xi$ of $P$ in $\mathbb{Z}_{2}$ such that

$\xi\equiv x_{0}\ \mbox{\rm(mod}\ p^{n-\phi}\mbox{\rm)}\ \text{and}\
\nu_{2}(P^{\prime}(\xi))=\nu_{2}(P^{\prime}(x_{0}))$

######  Lemma 6.

([7, Sec. 6.6]) The function $f(x)=x^{2}-a$ has a zero in
$\mathbb{Z}_{2}^{\times}$, the set of invertible elements of $\mathbb{Z}_{2}$,
if and only if $a\equiv 1\ \mbox{\rm(mod}\ 8\mbox{\rm)}$.

##  3 Proof of Theorem 1: unbounded cases and infinite trees

We now prove Theorem 1. The main idea is to describe the roots to $f(n)=0$ in
$\mathbb{Q}_{2}$ simply using the quadratic formula, the parity of the
coefficients, and the lemmas presented in Section 2.1. Moreover, according to
Lemma 4, if a zero exists in $\mathbb{Z}_{2}$, it manifests as an infinite
branch in the tree. We begin with Cases 2 and 4.

To this end, note that in Case 2, we can write $a=2r$ and $b=2k+1$ for some
$r,k\in\mathbb{Z}$. Then $an^{2}+bn+c=0$ has roots of the form

$x=\frac{-2k-1\pm\sqrt{1-8(rc-\beta)}}{4r},$ (3)

where $\beta=(k^{2}+k)/2$. Set $j=rc-\beta$.

Also, in Case 4, we can write $a=2r+1$, $b=2k+1$, and $c=2p$. Then
$an^{2}+bn+c=0$ has roots of the form

$x=\frac{-2k-1\pm\sqrt{1-8((2r+1)p-\beta)}}{2(2r+1)},$ (4)

where $\beta=(k^{2}+k)/2$. Set $j=(2r+1)p-\beta$. Observe that in either case
the roots contain $\sqrt{1-8j}$ where $j\in\mathbb{Z}$. Since $\sqrt{1-8j}$ is
a zero of the function $g(x)=x^{2}-(1-8j)$, by Lemma 6 the zero is in
$\mathbb{Z}_{2}$.

Notice that both roots (3) and (4) have an even denominator. We still need to
check if these roots are in $\mathbb{Q}_{2}$ or $\mathbb{Z}_{2}$. Therefore,
in light of Lemma 4, Case 2 (Proposition 7) and Case 4 (Proposition 8) are
proved by an inductive argument on the behavior of the tree. It turns out
that, in Case 2, $f(n)$ has exactly one zero in $\mathbb{Z}_{2}$ and in Case
4, $f(n)$ has two zeros in $\mathbb{Z}_{2}$. See Figure 3 in the Appendix for
an example of a tree with one infinite branch and Figure 4 for an example of a
tree with two infinite branches.

######  Proposition 7.

If $a$ is even and $b$ is odd, then the 2-adic valuation tree of
$f(n)=an^{2}+bn+c$ has exactly one infinite branch. Furthermore, the valuation
of the terminating node at the $i^{th}$ level is $i-1$.

###### Proof.

Note that this Proposition corresponds to Case 2 of Theorem 1. Substituting
$a=2r$ and $b=2k+1$ for some $r,k\in\mathbb{Z}$, we get
$an^{2}+bn+c=2(rn^{2}+kn)+n+c$. Now suppose that $c$ is even. If $n$ is even,
then $2(rn^{2}+kn)+n+c$ is divisible by 2 and so $\nu_{2}(f(2n))\geq 1$. If
$n$ is odd, then $2(rn^{2}+kn)+n+c$ is not divisible by 2 and so
$\nu_{2}(f(2n+1))=0$. An analogous argument shows that, for $c$ odd,
$\nu_{2}(f(2n))=0$ and $\nu_{2}(f(2n+1))\geq 1$. Therefore, the conclusion of
the proposition is valid at the initial step.

Now, arguing inductively, suppose that $n=2^{i}q+r_{i-1}$ is the non-
terminating node, that is $\nu_{2}(f(n))\geq i$. So $f(n)\equiv 0\
\mbox{\rm(mod}\ 2^{i}\mbox{\rm)}$ or
$a(2^{i}q+r_{i-1})^{2}+b(2^{i}q+r_{i-1})+c=2^{i}\beta$ where
$\beta\in\mathbb{Z}$. Consider $f(n)$ evaluated at the next level:

$a(2^{i+1}q+r_{i-1})^{2}+b(2^{i+1}q+r_{i-1})+c\equiv
ar_{i-1}^{2}+br_{i-1}+c\equiv 2^{i}\beta\ \mbox{\rm(mod}\ 2^{i+1}\mbox{\rm)},$

and

$\displaystyle a(2^{i+1}q+2^{i}+r_{i-1})^{2}+b(2^{i+1}q+2^{i}+r_{i-1})+c$
$\displaystyle\equiv ar_{i-1}^{2}+2^{i}b+br_{i-1}+c$ $\displaystyle\equiv
2^{i}\beta+2^{i}b\equiv 2^{i}(\beta+b)\ \mbox{\rm(mod}\ 2^{i+1}\mbox{\rm)}.$

Since $b$ is odd it follows that the valuation of one node is $i$ and the
other is greater than $i$ depending on if $\beta$ is odd or even. Therefore
one node terminates and the other is non-terminating. ∎

######  Proposition 8.

If $a$ and $b$ are odd, and $c$ is even, then the 2-adic valuation tree of
$f(n)=an^{2}+bn+c$ has two infinite branches. Furthermore, the valuation of
the terminating node at the $i^{th}$ level is $i$.

###### Proof.

This proposition addresses Case 4 of Theorem 1. Write $a=2r+1$, $b=2k+1$, and
$c=2p$ for some integers $r,k,$ and $p$. First note that both
$a(2q)^{2}+b(2q)+c$ and $a(2q+1)^{2}+b(2q+1)+c$ are congruent to $0\
\mbox{\rm(mod}\ 2\mbox{\rm)}$. We now verify that the proposition holds at the
initial step.

In the $2q$ case, we check $4q$ and $4q+2$. Note that

$a(4q)^{2}+b(4q)+c\equiv c\ \mbox{\rm(mod}\ 4\mbox{\rm)}$

and

$a(4q+2)^{2}+b(4q+2)+c\equiv 2b+c\ \mbox{\rm(mod}\ 4\mbox{\rm)}.$

If $c\equiv 0\ \mbox{\rm(mod}\ 4\mbox{\rm)}$, then $2b+c\not\equiv 0\
\mbox{\rm(mod}\ 4\mbox{\rm)}$. If $c\not\equiv 0\ \mbox{\rm(mod}\
4\mbox{\rm)}$ then $c=2p$ with $p$ odd and $2b+c=2(b+p)\equiv 0\
\mbox{\rm(mod}\ 4\mbox{\rm)}$. That is, either

$\nu_{2}(f(4q))\geq 2\text{ and }\nu_{2}(f(4q+2))=1,\text{ or}$
$\nu_{2}(f(4q))=1\text{ and }\nu_{2}(f(4q+2))\geq 2.$

For the $2q+1$ case, we check $4q+1$ and $4q+3$. Note that

$a(4q+1)^{2}+b(4q+1)+c\equiv a+b+c\ \mbox{\rm(mod}\ 4\mbox{\rm)}$

and

$a(4q+3)^{2}+b(4q+3)+c\equiv a+3b+c\ \mbox{\rm(mod}\ 4\mbox{\rm)}.$

By hypothesis, $a+b+c=2(r+k+p)$ and $a+3b+c=2(r+3k+p+2)$. But note that
$r+3k+p+2=(r+k+p+1)+(2k+1)$. Now it is clear that $r+3k+p+2$ is even (odd) if
and only if $r+k+p+1$ is odd (even). Again, either

$\nu_{2}(f(4q+1))\geq 2\text{ and }\nu_{2}(f(4q+3))=1\text{, or}$
$\nu_{2}(f(4q+1))=1\text{ and }\nu_{2}(f(4q+3))\geq 2.$

For the inductive step, now suppose that $n=2^{i}q+r_{i-1}$ and
$n=2^{i}q+r_{i-1}^{*}$ are the non-terminating nodes where
$r_{i-1}=\sum_{k=1}^{i-1}\alpha_{k}2^{k}+1$ (the odd side branch) and
$r_{i-1}^{*}=\sum_{k=1}^{i-1}\alpha_{k}2^{k}$ (the even side branch) where
$\alpha_{k}\in\left\\{0,1\right\\}$. The fact that these branches are non-
terminating follows from the same argument as in the proof of Proposition 7. ∎

We now consider Case 3(b) of Theorem 1.

######  Proposition 9.

Let $a$ be odd, $b$ be even and $b^{2}-4ac=4^{\ell}\Delta$ for some
$\ell\in\mathbb{Z}^{+}$ as large as possible and $\Delta\equiv 1\
\mbox{\rm(mod}\ 8\mbox{\rm)}$, then the 2-adic valuation tree of
$f(n)=an^{2}+bn+c$ has two infinite branches.

###### Proof.

Let $a$ be odd and $b=2k$ for some $k\in\mathbb{Z}$. Fix
$\ell\in\mathbb{Z}^{+}$. Then $an^{2}+bn+c=0$ has roots of the form
$x=\frac{-k\pm\sqrt{k^{2}-ac}}{a}$. By the hypothesis
$4k^{2}-ac=2^{2\ell}\Delta$ where $\Delta\equiv 1\ \mbox{\rm(mod}\
8\mbox{\rm)}$.

If $\Delta<0$ then we can naturally write $\Delta=1-8j$ where
$j\in\left\\{1,2,3,\ldots\right\\}$.

If $\Delta>0$, then we can write $\Delta=1+8j=1-8(-j)$ where $j\in\mathbb{N}$.

Thus in either case $\Delta=1-8j$ where $j\in\mathbb{Z}$. Then it follows that
$\sqrt{4k^{2}-4ac}=2^{\ell}\sqrt{1-8j}$. By Lemma 6, $\sqrt{1-8j}$ is in
$\mathbb{Z}_{2}$. Furthermore, since the denominator of $x$ is odd this also
guarantees that $x\in\mathbb{Z}_{2}$. Therefore, there are two infinite
branches, one corresponding to each root. ∎

######  Corollary 10.

Under the conditions of Proposition 9, if $b^{2}-4ac=0$, the tree has one
infinite branch.

###### Proof.

In this case (3(a) of Theorem 1) roots take the form $x=-\frac{b}{2a}$. Since
$b=2k$, then $x=-\frac{k}{a}$ which has 2-adic form
$x=\sum_{i=0}^{\infty}\alpha_{i}2^{i}$ where $\alpha_{i}$ is either 0 or 1.
This guarantees that the one branch is infinite. ∎

######  Remark 11.

Note the connection between subsequences of $(\nu_{2}(f(n)))_{n\geq 0}$ and
the infinite branches of a tree. Proposition 7 asserts that for all
$i\in\mathbb{Z}^{+}$ there exists exactly one subsequence of the form
$n=2^{i}q+r_{i-1}$ such that $\nu_{2}(f(n))\geq i$ and exactly one subsequence
of the form $n=2^{i}q+r_{i-1}^{*}$ with $\nu_{2}(f(n))=i-1$. Similarly,
Proposition 8 asserts that for all $i\in\mathbb{Z}^{+}$ there are exactly two
subsequences corresponding to $n=2^{i}q+r_{i-1}$ such that $\nu_{2}(f(n))\geq
i+1$ and exactly two subsequences of the form $n=2^{i}q+r_{i-1}^{*}$ with
$\nu_{2}(f(n))=i$. For $r_{i-1}$ and $r_{i-1}^{*}$, the representations
presented in equation (1) of Section 2 equate the coefficients $\alpha_{k}$
and $\alpha_{k}^{*}$ for all $0\leq k\leq i-2$, and meanwhile
$\alpha_{i-1}^{*}\equiv\alpha_{i-1}+1\ \mbox{\rm(mod}\ 2\mbox{\rm)}$.

As for the cases of Proposition 9 and Corollary 10, we can apply Lemma 4 to
conclude that these sequences are unbounded. Much like Propositions 7 and 8,
we can say that the results of Proposition 9 yield that for all
$i\in\mathbb{N}$ there are exactly two subsequences of the form
$n=2^{i}q+r_{i-1}$, where $(\nu_{2}(f(n)))_{n\geq 0}$ is not constant, while
Corollary 10 asserts there is exactly one such subsequence.

##  4 Bounded cases and finite trees

In this section, we prove Case 3(c) of Theorem 1 and the first part of Theorem
2. The coefficients of these quadratics satisfy the following: $a$ is odd and
$b$ is even, and $b^{2}-4ac=4^{\ell}\Delta$, where $\ell\in\mathbb{Z}^{+}$ is
as large as possible, $\Delta\equiv m\ \mbox{\rm(mod}\ 8\mbox{\rm)}$, and
$m\in\\{2,3,5,6,7\\}$. Their trees are finite with $\ell$ levels. We can again
apply the reasoning of the proof of Proposition 9.

If $\Delta<0$ we can naturally write $\Delta=m-8j$ where $j\in\mathbb{N}$ and
if $\Delta>0$ then we write $\Delta=m+8j=m-8(-j)$ where $j\in\mathbb{N}$ or
$j=0$. Henceforth, we will write $\Delta=m-8j$ where $j\in\mathbb{Z}$. Again,
by Lemma 6 functions of the form $g(x)=x^{2}-(m-8j)$ do not have a zero in
$\mathbb{Z}_{2}$. By Lemma 4, the corresponding valuation sequences are
periodic. Figures 5 and 6 in the Appendix illustrate examples of finite trees
arising from functions $f_{3}(n)=15n^{2}+1142n+25559$ and
$f_{4}(n)=5n^{2}+106n+1125$.

We should take a moment to note why we only need to consider these five values
of $m$. First note that in Cases 3(b) and 3(c) of Theorem 1, where $a$ is odd
and $b$ is even, we have the condition that $\ell$ is as large as possible.
This corresponds to factoring out as many powers of 4 as possible, ruling out
the possibilities $m\in\\{0,4\\}$. Now if $m=1$ (Case 3(b), covered in Section
3), an infinite tree is created. This leaves the cases $m\in\\{2,3,5,6,7\\}$.
As discussed above, the zeros of these quadratic functions are not elements of
$\mathbb{Q}_{2}$; therefore, their trees must be finite. The proofs of the
next two propositions follow the proofs of Propositions 7 and 8.

######  Proposition 12.

If $a$ is odd and $b$ is even, and $b^{2}-4ac=4^{\ell}\Delta$ where
$\ell\in\mathbb{Z}^{+}$ is as large as possible, $\Delta\equiv m\
\mbox{\rm(mod}\ 8\mbox{\rm)}$, and $m\in\\{2,3,5,6,7\\}$, then the 2-adic
valuation tree of $f(n)$ is finite with $\ell$ levels.

The proof of this proposition is broken down into Lemmas 13, 14, and 16.
Unless stated otherwise, let $b=2k$ for some $k\in\mathbb{Z}$. Lemma 13 covers
the case $\ell=1$, in which the 2-adic valuation tree has exactly one level.
Lemmas 14 and 16 describe valuations for finite trees with more than one
level; Lemma 16 describes the valuation at the final level and Lemma 14
describes the other levels. Under the assumptions of Proposition 12, with $a$
odd and $b$ even, we complete the square and use properties of the $p$-adic
valuation to obtain $\nu_{2}(an^{2}+bn+c)=\nu_{2}((an+k)^{2}-k^{2}+ac)$.

######  Lemma 13.

Let $\ell=1$, i.e., $b^{2}-4ac=4\Delta$, $\Delta\equiv m\ \mbox{\rm(mod}\
8\mbox{\rm)}$, and $m\in\\{2,3,5,6,7\\}$. If $m\in\\{2,7\\}$ and $b\equiv 0\
\mbox{\rm(mod}\ 4\mbox{\rm)}$ or if $m\in\\{3,6\\}$ and $b\equiv 2\
\mbox{\rm(mod}\ 4\mbox{\rm)}$, then

$\nu_{2}(an^{2}+bn+c)=\begin{cases}0,&\ \text{if}\ n\ \text{even};\\\ 1,&\
\text{if}\ n\ \text{odd}.\end{cases}$

If $m\in\left\\{3,6\right\\}$ and $b\equiv 0\ \mbox{\rm(mod}\ 4\mbox{\rm)}$ or
if $m\in\left\\{2,7\right\\}$ and $b\equiv 2\ \mbox{\rm(mod}\ 4\mbox{\rm)}$,
then

$\nu_{2}(an^{2}+bn+c)=\begin{cases}1,&\ \text{if}\ n\ \text{even};\\\ 0,&\
\text{if}\ n\ \text{odd}.\end{cases}$

If $m=5$ and $b\equiv 0\ \mbox{\rm(mod}\ 4\mbox{\rm)}$, then

$\nu_{2}(an^{2}+bn+c)=\begin{cases}0,&\ \text{if}\ n\ \text{even};\\\ 2,&\
\text{if}\ n\ \text{odd}.\end{cases}$

If $m=5$ and $b\equiv 2\ \mbox{\rm(mod}\ 4\mbox{\rm)}$, then

$\nu_{2}(an^{2}+bn+c)=\begin{cases}2,&\ \text{if}\ n\ \text{even};\\\ 0,&\
\text{if}\ n\ \text{odd}.\end{cases}$

###### Proof.

Using the convention that $\Delta=m-8j$ where $j\in\mathbb{Z}$ and
$m\in\left\\{2,3,5,6,7\right\\}$, consider the case where $m=7$ and $b\equiv
2\ \mbox{\rm(mod}\ 4\mbox{\rm)}$. Then, since $b=2k$, we have $k$ odd. If $n$
is even, then $(an+k)^{2}\equiv k^{2}\ \mbox{\rm(mod}\ 2\mbox{\rm)}$ and so it
follows that

$((an+k)^{2}-k^{2}+ac)\equiv k^{2}-7\equiv-6\equiv 0\ \mbox{\rm(mod}\
2\mbox{\rm)},$

but

$((an+k)^{2}-k^{2}+ac)\equiv k^{2}-7\equiv-6\equiv 2\ \mbox{\rm(mod}\
4\mbox{\rm)}.$

Therefore $\nu_{2}(an^{2}+bn+c)=1$ when $n$ is even. Similarly, when $m=7$ and
$b\equiv 2\ \mbox{\rm(mod}\ 4\mbox{\rm)}$ if $n$ is odd, then $(an+k)^{2}$ is
even. Therefore, $(an+k)^{2}-k^{2}+ac\equiv-7\equiv 1\ \mbox{\rm(mod}\
2\mbox{\rm)}$. Thus $\nu_{2}(an^{2}+bn+c)=0$ when $n$ is odd.

Now consider the case where $m=7$ and $b\equiv 0\ \mbox{\rm(mod}\
4\mbox{\rm)}$. We have $b=2k$ with $k$ even. Thus, if $n$ is odd we have

$(an+k)^{2}-k^{2}+ac\equiv-6\equiv 0\ \mbox{\rm(mod}\ 2\mbox{\rm)}$

and

$(an+k)^{2}-k^{2}+ac\equiv k^{2}-7\equiv-6\equiv 2\ \mbox{\rm(mod}\
4\mbox{\rm)}.$

Thus $\nu_{2}(an^{2}+bn+c)=1$ when $n$ is odd. When $n$ is even we have
$(an+k)^{2}-k^{2}+ac\equiv-7\equiv 1\ \mbox{\rm(mod}\ 2\mbox{\rm)}$. Thus
$\nu_{2}(an^{2}+bn+c)=0$ when $n$ is even.

The cases of $m\in\\{2,3,6\\}$ when $b\equiv 0\ \mbox{\rm(mod}\ 4\mbox{\rm)}$
or $b\equiv 2\ \mbox{\rm(mod}\ 4\mbox{\rm)}$ can be handled in the same
fashion. For $m=5$, the valuations are slightly different.

Consider the case where $m=5$. Recall that $b=2k$ for some $k\in\mathbb{Z}$.
Note that

$b^{2}-4ac=4(5-8j),$

and hence $k^{2}-ac=5-8j$. Thus,

$(an+k)^{2}-k^{2}+ac=(an+k)^{2}-5+8j.$

If $(an+k)$ is even, which is the case when both $n$ and $k$ are even or both
$n$ and $k$ are odd, then $(an+k)^{2}-5+8j$ is odd. Thus,
$\nu_{2}(an^{2}+bn+c)=0$.

Now suppose that $(an+k)$ is odd, which is true when $n$ and $k$ have
different parity. Then $(an+k)^{2}\equiv 1\ \mbox{\rm(mod}\ 4\mbox{\rm)}$, and
this implies

$(an+k)^{2}-5+8j\equiv 1-5+8j\equiv-4+8j\equiv 0\ \mbox{\rm(mod}\
4\mbox{\rm)}.$

Thus, $\nu_{2}(an^{2}+bn+c)\geq 2$.
Since $(an+k)$ is odd, let $an+k=2d+1$, for some $d\in\mathbb{Z}$. Then,

$\displaystyle(an+k)^{2}-5+8j$ $\displaystyle=(2d+1)^{2}-5+8j$
$\displaystyle\equiv 4(d^{2}+d-1)\ \mbox{\rm(mod}\ 8\mbox{\rm)}.$

Observe that $d^{2}+d-1$ is odd, regardless of whether $d$ is even or odd.
Thus, $\nu_{2}(an^{2}+bn+c)<3$. Therefore, $\nu_{2}(an^{2}+bn+c)=2$. ∎

######  Lemma 14.

Under the assumptions of Proposition 12 (Case 3(c) of Theorem 1) let $\ell\geq
2$ and suppose $0<i<\ell$. At the $i^{th}$ level there is one terminal and one
non-terminal node. Furthermore, the terminal node has valuation $2(i-1)$ and
the non-terminal node has valuation at least $2i$.

First we need:

######  Claim 15.

Let $a,k\in\mathbb{Z}$ with $a$ odd. Let $g(n)=an+k$, then
$(\nu_{2}(g(n)))_{n\geq 0}$ creates an unbounded sequence.

###### Proof.

First note that the root of $ax+k=0$ is $x=-\frac{k}{a}$. Also note that
$\nu_{2}(x)=\nu_{2}(-k)-\nu_{2}(a)$. Since $a$ is odd, $\nu_{2}(a)=0$.
Therefore $\nu_{2}(x)=\nu_{2}(-k)\geq 0$. By equation (2),
$x\in\mathbb{Z}_{2}$, so Lemma 4 implies that $(\nu_{2}(g(n)))_{n\geq 0}$ is
an unbounded sequence. ∎

###### Proof.

To prove Lemma 14, we proceed by an inductive argument on $i$. Again, using
the convention that $\Delta=m-8j$ where $j\in\mathbb{Z}$, for the base case
$i=1$, note that $b^{2}-4ac\equiv 4^{\ell}(m-8j)\equiv 0\ \mbox{\rm(mod}\
4\mbox{\rm)}$. Recall that $b=2k$. First, assume that $k$ is even. If $n$ is
even, then $an+k$ is even and so $(an+k)^{2}-k^{2}+ac\equiv 0\ \mbox{\rm(mod}\
4\mbox{\rm)}$. Thus $\nu_{2}(an^{2}+bn+c)=\nu_{2}((an+k)^{2}-k^{2}+ac)\geq 2$
by Claim 15. If $n$ is odd, then $(an+k)^{2}-k^{2}+ac\equiv 1\ \mbox{\rm(mod}\
2\mbox{\rm)}$, and again using the technique of completing the square,
$\nu_{2}(an^{2}+bn+c)=0$. If $k$ is odd, a similar argument shows that
$\nu_{2}(an^{2}+bn+c)\geq 2$ when $n$ is odd. Observe also that Claim 15 can
be used to show that $(\nu_{2}((an+k)^{2}))_{n\geq 0}$ forms an unbounded
sequence therefore $\nu_{2}((an+k)^{2}-k^{2}+ac)\geq 2$. Thus, the claim is
true for $i=1$.

For the inductive step, notice that since $i<\ell$, it follows that

$b^{2}-4ac\equiv 4^{\ell}(m-8j)\equiv 0\ \mbox{\rm(mod}\ 2^{2i}\mbox{\rm)}.$

Suppose there exists an $i-1\geq 0$ such that $n=2^{i-1}q+r_{i-2}$ splits into
two nodes: one node terminating with valuation $2(i-1)$ and the other node
having valuation of at least $2i$. We let $n=2^{i}q+r_{i-1}$ denote the non-
terminating node, where $r_{i-1}=\sum_{h=0}^{i-1}\alpha_{h}2^{h}$ with
$\alpha_{h}\in\left\\{0,1\right\\}$, for all $0\leq h\leq i-2$, and
$q\in\mathbb{Z}$. Then we have

$(an+k)^{2}-k^{2}+ac\equiv{(a(2^{i}q+r_{i-1})+k)^{2}}\equiv 0\ \mbox{\rm(mod}\
2^{2i}\mbox{\rm)},$

so $\nu_{2}(an^{2}+bn+c)\geq 2i$. This also implies that
$a(2^{i}q+r_{i-1})+k\equiv 0\ \mbox{\rm(mod}\ 2^{i}\mbox{\rm)}$. Thus
$ar_{i-1}+k=2^{i}\beta$ where $\beta\in\mathbb{Z}$. Now suppose that $k$ is
even. (The proof for $k$ odd can be handled in the same fashion, and thus is
omitted.) Since $k$ is even, then $r_{i-1}$ must be even.

Consider the $(i+1)^{st}$ level where $i+1<\ell$. Here again we have

$b^{2}-4ac=4^{\ell}(m-8j)\ \mbox{\rm(mod}\ 2^{2(i+1)}\mbox{\rm)}\equiv 0.$

Moving to the next level, in the case $n=2^{i+1}q+r_{i-1}$ we have

$\displaystyle\nu_{2}((an+k)^{2}-4^{\ell-1}(m-8j))$
$\displaystyle=\nu_{2}((a(2^{i+1}q+r_{i-1})+k)^{2}-4^{\ell-1}(m-8j))$
$\displaystyle=\nu_{2}((2^{i+1}aq+ar_{i-1}+k)^{2}-4^{\ell-1}(m-8j))$
$\displaystyle=\nu_{2}((2^{i+1}aq+2^{i}\beta)^{2}-4^{\ell-1}(m-8j))$
$\displaystyle=\nu_{2}(2^{2i}(2aq+\beta)^{2}-2^{2(\ell-1)}(m-8j)),$

and in the case $n=2^{i+1}q+2^{i}+r_{i-1}$ we have

$\displaystyle\nu_{2}((an+k)^{2}-4^{\ell-1}(m-8j))$ $\displaystyle\hskip
30.0pt=\nu_{2}((a(2^{i+1}q+2^{i}+r_{i-1})+k)^{2}-4^{\ell-1}(m-8j))$
$\displaystyle\hskip
30.0pt=\nu_{2}((2^{i+1}aq+2^{i}a+ar_{i-1}+k)^{2}-4^{\ell-1}(m-8j))$
$\displaystyle\hskip
30.0pt=\nu_{2}((2^{i+1}aq+2^{i}a+2^{i}\beta)^{2}-4^{\ell-1}(m-8j))$
$\displaystyle\hskip
30.0pt=\nu_{2}(2^{2i}(2aq+a+\beta)^{2}-2^{2(\ell-1)}(m-8j)).$

Since $\beta\in\mathbb{Z}$ either $2aq+\beta$ or $2aq+a+\beta$ is odd and the
other is even. As long as $i+1<\ell$ then in the odd case the valuation is
$2i$ and in the even case the valuation is at least $2(i+1)$. ∎

######  Lemma 16.

If $a$ is odd and $b$ is even with $b=2k$ for $k\in\mathbb{Z}$, and
$b^{2}-4ac=4^{\ell}\Delta$ where $\ell\in\mathbb{Z}^{+}$ is as large as
possible, $\Delta\equiv m\ \mbox{\rm(mod}\ 8\mbox{\rm)}$, and
$m\in\\{2,3,5,6,7\\}$, then at the $\ell^{th}$ level the nodes of the 2-adic
valuation tree terminate with valuations of $2(\ell-1)$, $2\ell-1$ or $2\ell$.

Suppose that $n=2^{\ell}q+r_{\ell-2}$. If $an+k\equiv 0\ \mbox{\rm(mod}\
2^{\ell}\mbox{\rm)}$, then

$\nu_{2}(f(n))=\begin{cases}2(\ell-1),&\ \text{if}\ m=7,5,3;\\\ 2\ell-1,&\
\text{if}\ m=6,2;\end{cases}$

and if $an+k\not\equiv 0\ \mbox{\rm(mod}\ 2^{\ell}\mbox{\rm)}$, then

$\nu_{2}(f(n))=\begin{cases}2(\ell-1),&\ \text{if}\ m=6,2;\\\ 2\ell-1,&\
\text{if}\ m=7,3;\\\ 2\ell,&\ \text{if}\ m=5.\end{cases}$

Suppose that $n=2^{\ell}q+2^{\ell-1}+r_{\ell-2}$. If $an+k\equiv 0\
\mbox{\rm(mod}\ 2^{\ell}\mbox{\rm)}$, then

$\nu_{2}(f(n))=\begin{cases}2(\ell-1),&\ \text{if}\ m=6,2;\\\ 2\ell-1,&\
\text{if}\ m=7,3;\\\ 2\ell,&\ \text{if}\ m=5;\end{cases}$

and if $an+k\not\equiv 0\ \mbox{\rm(mod}\ 2^{\ell}\mbox{\rm)}$, then

$\nu_{2}(f(n))=\begin{cases}2(\ell-1),&\ \text{if}\ m=7,5,3;\\\ 2\ell-1,&\
\text{if}\ m=6,2.\end{cases}$

###### Proof.

By Lemma 14 there exists a non-terminating node $n=2^{\ell-1}q+r_{\ell-2}$
with $q\in\mathbb{Z}$ and

$\nu_{2}((an+k)^{2}-k^{2}+ac)\geq 2(\ell-1).$

Consider $n=2^{\ell}q+r_{\ell-2}$ with $q\in\mathbb{Z}$. By the same argument
as in Lemma 14 and using the convention that $\Delta=m-8j$ where
$j\in\mathbb{Z}$, we have

$(an+k)^{2}-k^{2}+ac=(2^{\ell}aq+2^{\ell-1}\beta)^{2}-2^{2(\ell-1)}(m-8j)=2^{2(\ell-1)}((2aq+\beta)^{2}+8j-m),$

where $\beta\in\mathbb{Z}$. Recall that $a$ is odd. Then depending on whether
$\beta$ is even or odd, simple calculations show the first two results.

In the case when $n=2^{\ell}q+2^{\ell-1}+r_{\ell-2}$ with $q\in\mathbb{Z}$ we
have

$(an+k)^{2}-k^{2}+ac=2^{2(\ell-1)}((2aq+a+\beta)^{2}+8j-m),$

where $\beta\in\mathbb{Z}$. Then again depending on whether $\beta$ is odd or
even, it is straightforward to show the last two results. ∎

##  5 Structure of finite trees

The section describes the overall structure of finite trees, continuing the
discussion of Case 3(c) of Theorem 1, in which $a$ is odd, $b$ is even,
$b^{2}-4ac=4^{\ell}\Delta$ where $\Delta\equiv m\ \mbox{\rm(mod}\
8\mbox{\rm)}$, and $m\in\\{2,3,5,6,7\\}$. Throughout this section, we make use
of several operators. The operators allow us to track changes from very easily
described trees, which we call type $(\ell,1)$, to more complicated trees.

######  Definition 17 (Translation operator, [4]).

For quadratics of the form $f(n)=an^{2}+bn+c$ we define
$\tau^{s}(f)(n)=f(n-s)$ for $s\in\mathbb{R}$, namely
$\tau^{s}(f)(n)=a(n-s)^{2}+b(n-s)+c=an^{2}+(b-2as)n+(c+as^{2}-bs)$.

######  Proposition 18.

Let the assumptions of Proposition 12 hold for the function $f(n)=an^{2}+bn+c$
and suppose $s\in\mathbb{Z}$. Then we have the following relationship

$\nu_{2}(f(2^{i}q+r_{i-1}))=\nu_{2}(\tau^{s}f(2^{i}q+(r_{i-1}+s)\bmod{2^{i}})).$

That is the valuations $\nu_{2}(f(n))$ at the node of the form
$n=2^{i}q+r_{i-1}$ are moved to the node of the form $n=2^{i}q+(r_{i-1}+s)\
\mbox{\rm(mod}\ 2^{i}\mbox{\rm)}$ under the operation $\tau^{s}$.

###### Proof.

Note that finite trees with $\ell$ levels correspond to periodic sequences
with a period equal to $2^{\ell}$. Since $\tau^{s}$ is a translation operator,
every element in the sequence $(\nu_{2}(f(n)))_{n\geq 0}$ is moved over $s$
spaces. ∎

######  Definition 19 ($S$-operator).

Let $a$ be a positive, odd integer. For quadratics of the form
$f(n)=n^{2}+bn+ac$ we define $S^{a}(f)(n)=an^{2}+bn+c$. Likewise, for
quadratics of the form $f(n)=an^{2}+bn+c$ define
$S^{a^{-1}}(f)(n)=n^{2}+bn+ac$.

In general, the $S$-operator need not output a quadratic function with an
integer constant term. However, the present work only applies $S^{a}$ to
functions whose output has integer coefficients.

######  Definition 20 (Dilation operator, [4]).

For quadratics of the form $f(n)=an^{2}+bn+c$ we define
$\delta^{s}(f)(n)=f(sn)$ for $s\in\mathbb{R}$, namely
$\delta^{s}(f)(n)=a(sn)^{2}+b(sn)+c$.

######  Lemma 21.

Under the assumptions of Proposition 12 the trees created by
$f(n)=n^{2}+bn+ac$ and $S^{a}(f)(n)$ where $a\in\mathbb{Z}$ have the same
number of levels. Similarly, the trees created by $g(n)=an^{2}+bn+c$ and
$\tau^{s}(g)(n)$ where $s\in\mathbb{Z}$ have the same number of levels.

###### Proof.

The assumptions of Proposition 12 represent Cases 3(b) and 3(c) of Theorem 1.
Simple calculations show that the discriminants of $f(n)$ and $S^{a}(f)(n)$
are the same, and that the discriminants of $g(n)$ and $\tau^{s}(g)(n)$ are
the same. The conclusions then follow directly from Proposition 12. ∎

######  Proposition 22.

Let the assumptions of Proposition 12 hold and suppose $f(n)=n^{2}+bn+ac$.
Then we have the following relationship

$\nu_{2}(f(2^{i}q+r_{i-1}))=\nu_{2}(S^{a}(f(2^{i}q+a^{-1}\cdot r_{i-1}))).$

That is the valuation $\nu_{2}(f(n))$ at the node the form $n=2^{i}q+r_{i-1}$
is moved to the node of the form of $n=2^{i}q+(a^{-1}\cdot r_{i-1})\
\mbox{\rm(mod}\ 2^{i}\mbox{\rm)}$ under the operation $S^{a}$. In this context
$a^{-1}$ is the inverse of $a\ \mbox{\rm(mod}\ 2^{i}\mbox{\rm)}$.

###### Proof.

Since $a$ is odd, note that

$\nu_{2}(S^{a}(f)(n))=\nu_{2}((an^{2}+bn+c))=\nu_{2}((an)^{2}+b(an)+ac)=\nu_{2}(\delta^{a}(f)(n)),$

where $\delta^{a}(f)(n)=f(an)$ is the dilation operator. Thus, the valuation
of $f(n)$ for $n=2^{i}q+r_{i-1}$ is the same as the valuation of
$n^{\prime}=2^{i}(a^{-1}q)+a^{-1}\cdot r_{i-1}$ after the $S^{a}$-operator is
applied. ∎

Suppose that $f(n)=an^{2}+bn+c$ creates a finite tree. We say that this tree
is type $(\ell,1)$, for $\ell\geq 2$, if at every level the non-terminating
node is of the form $n=2q$ or $n=2^{i}q+2^{i-2}+\cdots+2^{1}+2^{0}$ for
$i<\ell$ and the tree has $\ell$ levels. We also say that a quadratic function
is type $(\ell,1)$ if it creates an $(\ell,1)$ tree. That is, $f(n)$ creates a
finite tree of the following form:

Figure 2: The form of trees of type $(\ell,1)$.

Here, we suppose that $\ell\geq 2$ because $\ell=1$ creates a tree with one
level, see Lemma 13, and the directional behavior we seek to classify is not
defined. The conditions $4a^{2}-4ac=4^{\ell}\Delta$ for
$\ell\in\mathbb{Z}^{+}$ as large as possible, $\Delta\equiv m\ \mbox{\rm(mod}\
8\mbox{\rm)}$, and $m\in\\{2,3,5,6,7\\}$ imply $c$ must be odd.

######  Proposition 23.

Under the assumptions of Proposition 12, if $c$ is odd and $\ell\geq 2$ is an
integer, then a quadratic of the form $f(n)=an^{2}+2an+c$ creates a tree that
is of type $(\ell,1)$. Furthermore, we have

$\nu_{2}(f(n))=\begin{cases}0,&\ \text{if}\ n\equiv 0\ \mbox{\rm(mod}\
2\mbox{\rm)};\\\ 2(i-1),&\ \text{if}\ n\equiv\sum_{k=0}^{i-2}2^{k}\
\mbox{\rm(mod}\ 2^{i}\mbox{\rm)}\ \text{with}\ 2\leq i<\ell;\\\ 2(\ell-1),&\
\text{if}\ n\equiv\sum_{k=0}^{\ell-2}2^{k}\ \mbox{\rm(mod}\
2^{\ell}\mbox{\rm)}\ \text{and}\ m=6,2;\\\ 2\ell-1,&\ \text{if}\
n\equiv\sum_{k=0}^{\ell-2}2^{k}\ \mbox{\rm(mod}\ 2^{\ell}\mbox{\rm)}\
\text{and}\ m=7,3;\\\ 2\ell,&\ \text{if}\ n\equiv\sum_{k=0}^{\ell-2}2^{k}\
\mbox{\rm(mod}\ 2^{\ell}\mbox{\rm)}\ \text{and}\ m=5;\\\ 2\ell-1,&\ \text{if}\
n\equiv\sum_{k=0}^{\ell-1}2^{k}\ \mbox{\rm(mod}\ 2^{\ell}\mbox{\rm)}\
\text{and}\ m=6,2;\\\ 2(\ell-1),&\ \text{if}\ n\equiv\sum_{k=0}^{\ell-1}2^{k}\
\mbox{\rm(mod}\ 2^{\ell}\mbox{\rm)}\ \text{and}\ m=7,5,3.\end{cases}$

###### Proof.

In light of Lemma 14, we know that if a node is non-terminating, then it
produces two nodes that either both terminate (i.e., these nodes are at the
$\ell^{th}$ level) or one node is non-terminating and the other is
terminating. So in order to show that the tree is of type $(\ell,1)$, we only
need to confirm that nodes corresponding to
$n=2^{i}q+2^{i-1}+\cdots+2^{1}+2^{0}$, where $1\leq i\leq\ell$, are always
non-terminating. Since $a$ is odd, completing the square and using the
convention that $4a^{2}-4ac=4^{\ell}\Delta$ where $\Delta=m-8j$ where
$j\in\mathbb{Z}$ gives

$\displaystyle\nu_{2}(f(n))$
$\displaystyle=\nu_{2}(an^{2}+2an+c)=\nu_{2}(a(n+1)^{2}-a+c)$
$\displaystyle=\nu_{2}(a^{2}(n+1)^{2}-a^{2}+ac)=\nu_{2}(a^{2}(n+1)^{2}-4^{\ell-1}(m-8j))$
$\displaystyle=\nu_{2}(a^{2}(2^{i}q+2^{i-1}+2^{i-2}+\cdots+2+1+1)^{2}-4^{\ell-1}(m-8j))$
$\displaystyle=\nu_{2}(a^{2}(2^{i}q+2^{i})^{2}-4^{\ell-1}(m-8j))$
$\displaystyle=\nu_{2}(a^{2}4^{i}(q+1)^{2}-4^{\ell-1}(m-8j)).$

If $q$ is odd, then $n=2^{i}q+2^{i-1}+\cdots+2^{1}+2^{0}$ is the non-
terminating node, provided $i<\ell$, and produces two nodes one of which does
not terminate. If $i=\ell$, then both nodes terminate by Proposition 12.

The nodes that terminate are of the form $n=2^{i}q+2^{i-2}+\cdots+2^{1}+2^{0}$
when $1\leq i<\ell$. The case when $n=2q$ is handled by the proof of Lemma 21.
For the case $1<i<\ell$, by the same calculation as above we have

$\nu_{2}(f(n))=\nu_{2}(a^{2}2^{2(i-1)}(2q+1)^{2}-4^{\ell-1}(m-8j))$

Since $2q+1$ is odd and $i<\ell$ the valuation must be $2(i-1)$.

In the case when $n=2^{\ell}q+2^{\ell-2}+\cdots+2^{1}+2^{0}$ we have

$\nu_{2}(f(n))=\nu_{2}(a^{2}2^{2(\ell-1)}(2q+1)^{2}-4^{\ell-1}(m-8j))$

Thus the valuation must be $2(\ell-1)$ if $m=6,2$, or $2\ell-1$ if $m=7,3$ or
$2\ell$ if $m=5$.

Finally if $n=2^{\ell}q+2^{\ell-2}+\cdots+2^{1}+2^{0}$ we have
$\nu_{2}(f(n))=\nu_{2}(a^{2}2^{2\ell}(q+1)^{2}-4^{\ell-1}(m-8j))$. Thus the
valuation must be $2(\ell-1)$ if $m=7,5,3$ or $2\ell-1$ if $m=6,2$. ∎

If the function $f(n)=an^{2}+bn+c$ meets the assumptions of Proposition 12
(Case 3(c) of Theorem 1) note if we define the function
$g(n)=n^{2}+2n-\left(1-\frac{b}{2}\right)^{2}+2\left(1-\frac{b}{2}\right)+ac$,
then it follows that $S^{a}(\tau^{1-\frac{b}{2}}(g))(n)=f(n)$. Therefore, by
Propositions 18, 22, and 23 we immediately have the following corollary.

######  Corollary 24.

If $f(n)=an^{2}+bn+c$ meets the assumptions of Proposition 12 (Case 3(c)) with
$\ell\geq 2$, then

$\nu_{2}(f(n))=\begin{cases}0,&\ \text{if}\ n\equiv
a^{-1}\left(1-\frac{b}{2}\right)\ \mbox{\rm(mod}\ 2\mbox{\rm)};\\\ 2(i-1),&\
\text{if}\ n\equiv a^{-1}\left(2^{i-1}-\frac{b}{2}\right)\ \mbox{\rm(mod}\
2^{i}\mbox{\rm)}\ \text{with}\ 2\leq i<\ell;\\\ 2(\ell-1),&\ \text{if}\
n\equiv a^{-1}\left(2^{\ell-1}-\frac{b}{2}\right)\ \mbox{\rm(mod}\
2^{\ell}\mbox{\rm)}\text{and}\ m=6,2;\\\ 2\ell-1,&\ \text{if}\ n\equiv
a^{-1}\left(2^{\ell-1}-\frac{b}{2}\right)\ \mbox{\rm(mod}\
2^{\ell}\mbox{\rm)}\ \text{and}\ m=7,3;\\\ 2\ell,&\ \text{if}\ n\equiv
a^{-1}\left(2^{\ell-1}-\frac{b}{2}\right)\ \mbox{\rm(mod}\
2^{\ell}\mbox{\rm)}\ \text{and}\ m=5;\\\ 2\ell-1,&\ \text{if}\ n\equiv
a^{-1}\left(2^{\ell}-\frac{b}{2}\right)\ \mbox{\rm(mod}\ 2^{\ell}\mbox{\rm)}\
\text{and}\ m=6,2;\\\ 2(\ell-1),&\ \text{if}\ n\equiv
a^{-1}\left(2^{\ell}-\frac{b}{2}\right)\ \mbox{\rm(mod}\ 2^{\ell}\mbox{\rm)}\
\text{and}\ m=7,5,3;\\\ \end{cases}$

where $a^{-1}$ is the inverse of $a\ \mbox{\rm(mod}\ 2^{\ell}\mbox{\rm)}$.

###### Proof.

Simply note that $g$ is type $(\ell,1)$ and recall the ways in which the
operators affect the function $g$. Each terminating node, under the operators,
moves from $n=2^{i}q+r_{i-2}$ to
$n=2^{i}q+a^{-1}\left(r_{i-2}+1-\frac{b}{2}\right)\ \mbox{\rm(mod}\
2^{i}\mbox{\rm)}$. In the case of type $(\ell,1)$ we have
$r_{i-2}=\sum_{k=0}^{i-2}2^{k}$. Thus $r_{i-2}+1=2^{i-1}$ in each case. ∎

##  6 Acknowledgments

The authors would like to thank Dr. Victor Moll for suggesting this topic. We
would also like to thank following institutions for providing support to
collaborate: ICERM, AIM, Kentucky Wesleyan College, Ursinus College, and
Stephen F. Austin State University. We are grateful to our other colleagues
for their ongoing support: Dr. Maila Brucal-Hallare, Dr. Jean-Claude Pedjeu,
and Dr. Bianca Thompson. And, finally, we are very grateful to the reviewer
who provided many very helpful and insightful comments.

## Appendix: figures illustrating trees and tables of values for $2$-adic
valuation sequences of some quadratic functions

In the following tree representations, a closed circle indicates a terminating
node and an open circle indicates a non-terminating node.

$n$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11
---|---|---|---|---|---|---|---|---|---|---|---|---
$f_{1}(n)$ | $-25$ | $-8$ | 17 | 50 | 91 | 140 | 197 | 262 | 335 | 416 | 505 | 602
$\nu_{2}(f_{1}2(n))$ | 0 | 3 | 0 | 1 | 0 | 2 | 0 | 1 | 0 | 5 | 0 | 1

$n$ | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19 | 20 | 21
---|---|---|---|---|---|---|---|---|---|---
$f_{1}(n)$ | 707 | 820 | 941 | 1070 | 1207 | 1352 | 1505 | 1666 | 1835 | 2012
$\nu_{2}(f_{1}(n))$ | 0 | 2 | 0 | 1 | 0 | 3 | 0 | 1 | 0 | 2

Figure 3: The 2-adic valuation tree for $f_{1}(n)=4n^{2}+13n-25$. Theorem 1
predicts that $(\nu_{2}(f_{1}(n)))_{n\geq 0}$ is an unbounded sequence, as it
satisfies Case 2.

$n$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
---|---|---|---|---|---|---|---|---|---|---|---|---
$f_{2}(n)$ | $-28$ | $-3$ | 48 | 125 | 228 | 357 | 512 | 693 | 900 | 1133 | 1392 |
$\nu_{2}(f_{2}(n))$ | 2 | 0 | 4 | 0 | 2 | 0 | 9 | 0 | 2 | 0 | 4 |

$n$ | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19
---|---|---|---|---|---|---|---|---|---
$f_{2}(n)$ | 1677 | 1988 | 2325 | 2688 | 3077 | 3492 | 3933 | 4400 | 4893
$\nu_{2}(f_{2}(n))$ | 0 | 2 | 0 | 7 | 0 | 2 | 0 | 4 | 0

Figure 4: The 2-adic valuation tree and data for $f_{2}(n)=13n^{2}+12n-28$.
Notice that Theorem 1 predicts that $(\nu_{2}(f_{2}(n)))_{n\geq 0}$ is an
unbounded sequence, as it satisfies Case 3(a) since $12^{2}-4\cdot
13(-28)=4^{3}(1-8(-3))$.

$n$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7
---|---|---|---|---|---|---|---|---
$f_{3}(n)$ | 25559 | 26716 | 27903 | 29120 | 30367 | 31644 | 32951 | 34288
$\nu_{2}(f_{3}(n))$ | 0 | 2 | 0 | 6 | 0 | 2 | 0 | 4

$n$ | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15
---|---|---|---|---|---|---|---|---
$f_{3}(n)$ | 35655 | 37052 | 38479 | 39936 | 41423 | 42940 | 44487 | 46064
$\nu_{2}(f_{3}(n))$ | 0 | 2 | 0 | 10 | 0 | 2 | 0 | 4

Figure 5: The 2-adic valuation tree and data for
$f_{3}(n)=15n^{2}+1142n+25559$. Notice that Theorem 1 predicts that
$(\nu_{2}(f_{3}(n))_{n\geq 0}$ is a bounded sequence, as it satisfies Case
3(c) since $1142^{2}-4\cdot 15\cdot 25559=4^{7}(2-8\cdot 2)$.

$n$ | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9
---|---|---|---|---|---|---|---|---|---|---
$f_{4}(n)$ | 1125 | 1236 | 1357 | 1488 | 1629 | 1780 | 1941 | 2112 | 2293 | 2484
$\nu_{2}(f_{4}(n))$ | 0 | 2 | 0 | 4 | 0 | 2 | 0 | 6 | 0 | 2

$n$ | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 | 19
---|---|---|---|---|---|---|---|---|---|---
$f_{4}(n)$ | 2685 | 2896 | 3117 | 3348 | 3589 | 3840 | 4101 | 4372 | 4653 | 4944
$\nu_{2}(f_{4}(n))$ | 0 | 4 | 0 | 2 | 0 | 8 | 0 | 2 | 0 | 4

Figure 6: The 2-adic valuation tree and data for $f_{4}(n)=5n^{2}+106n+1125$.
Notice that Theorem 1 predicts that $(\nu_{2}(f_{4}(n))_{n\geq 0}$ is a
bounded sequence, as it satisfies Case 3(c) since $106^{2}-4\cdot 5\cdot
1125=4^{5}(5-8\cdot 2)$.

## References

  * [1] J.-P. Allouche and J. Shallit, Automatic Sequences, Theory, Applications, Generalizations, Cambridge University Press, 2003\.
  * [2] J. P. Bell, $p$-adic valuations and $k$-regular sequences, Discrete Math. 307 (2007), 3070–3075.
  * [3] A. N. Byrnes, J. Fink, G. Lavigne, I. Nogues, S. Rajasekaran, A. Yuan, L. Almodovar, X. Guan, A. Kesarwani, L. A. Medina, E. Rowland, and V. H. Moll, A closed-form solution might be given by a tree. Valuations of quadratic polynomials, Sci. Ser. A Math. Sci. 29 (2019), 11–28.
  * [4] L. Grafakos, Classical Fourier Analysis, Second Edition, Springer, 2008.
  * [5] F. Gouvêa, p-Adic Numbers: An Introduction, Springer-Verlag, 1997.
  * [6] L. A. Medina, V. H. Moll, and E. Rowland, Periodicity in the $p$-adic valuation of a polynomial, J. Number Theory 180 (2017), 139–153.
  * [7] A. M. Robert, A Course in p-adic Analysis, Springer, 2000.


# Proofs and Supplementary Material:
Unified Characterization and Precoding for Non-Stationary Channels

Zhibin Zou, Maqsood Careem, Aveek Dutta Ngwe Thawdar Department of Electrical
and Computer Engineering US Air Force Research Laboratory University at Albany
SUNY, Albany, NY 12222 USA Rome, NY, USA {zzou2, mabdulcareem,
<EMAIL_ADDRESS><EMAIL_ADDRESS>
Instructions: This document provides the supplementary material including a
comprehensive related work, the complete proofs and extended evaluation
results that support the manuscript, “Unified Characterization and Precoding
for Non-Stationary Channels”, that was accepted for publication at IEEE
International Conference on Communications (ICC) 2022. Equations (1)–(34)
refer to the equations from the main manuscript, and the Theorem, Lemma and
Corollaries correspond to those from the manuscript.

##  Appendix A Related work

We categorize the related work into three categories:

Characterization of Non-Stationary Channels: Wireless channel characterization
in the literature typically require several local and global (in space-time
dimensions) higher order statistics to characterize or model non-stationary
channels, due to their time-varying statistics. These statistics cannot
completely characterize the non-stationary channel, however are useful in
reporting certain properties that are required for the application of interest
such as channel modeling, assessing the degree of stationarity etc.
Contrarily, we leverage the 2-dimensional eigenfunctions that are decomposed
from the most generic representation of any wireless channel as a spatio-
temporal channel kernel. These spatio-temporal eigenfunctions can be used to
extract any higher order statistics of the channel as demonstrated in Section
iii@, and hence serves as a complete characterization of the channel.
Furthermore, since this characterization can also generalize to stationary
channels, it is a unified characterization for any wireless channel. Beyond
characterizing the channel, these eigenfunctions are the core of the precoding
algorithm.

Precoding Non-Stationary Channels: Although precoding non-stationary channels
is unprecedented in the literature [1], we list the most related literature
for completeness. The challenge in precoding non-stationary channels is the
lack of accurate models of the channel and the (occasional) CSI feedback does
not fully characterize the non-stationarities in its statistics. This leads to
suboptimal performance using state-of-the-art precoding techniques like Dirty
Paper Coding which assume that complete and accurate knowledge of the channel
is available, while the CSI is often outdated in non-stationary channels.
While recent literature present attempt to deal with imperfect CSI by modeling
the error in the CSI [2, 3, 4, 5, 6, 7, 8, 9], they are limited by the
assumption the channel or error statistics are stationary or WSSUS at best.
Another class of literature, attempt to deal with the impact of outdated CSI
[10, 11] in time-varying channels by quantifying this loss or relying
statistical CSI. These methods are not directly suitable for non-stationary
channels, as the time dependence of the statistics may render the CSI (or its
statistics) stale, consequently resulting in precoding error.

Space-Temporal Precoding: While, precoding has garnered significant research,
spatio-temporal interference is typically treated as two separate problems,
where spatial precoding at the transmitter aims to cancel inter-user and
inter-antenna interference, while equalization at the receiver mitigates
inter-carrier and inter-symbol interference. Alternately, [12] proposes to
modulate the symbols such that it reduces the cross-symbol interference in the
delay-Doppler domain, but requires equalization at the receiver to completely
cancel such interference in practical systems. Moreover, this approach cannot
completely minimize the joint spatio-temporal interference that occurs in non-
stationary channels since their statistics depend on the time-frequency domain
in addition to the delay-Dopper domain (explained in Section ii@). While
spatio-temporal block coding techniques are studied in the literature [13]
they add redundancy and hence incur a communication overhead to mitigate
interference, which we avoid by precoding. These techniques are capable of
independently canceling the interference in each domain, however are incapable
of mitigating interference that occurs in the joint spatio-temporal domain in
non-stationary channels. We design a joint spatio-temporal precoding that
leverages the extracted 2-D eigenfunctions from non-stationary channels to
mitigate interference that occurs on the joint space-time dimensions, which to
the best of our knowledge is unprecedented in the literature.

##  Appendix B Proofs on Unified Characterization

###  B-A Proof of Lemma 1: Generalized Mercer’s Theorem

###### Proof.

Consider a 2-D process $K(t,t^{\prime})\in L^{2}(Y\times X)$, where $Y(t)$ and
$X(t^{\prime})$ are square-integrable zero-mean random processes with
covariance function $K_{Y}$ and $K_{X}$, respecly. The projection of
$K(t,t^{\prime})$ onto $X(t^{\prime})$ is obtained as in (35),

$\displaystyle C(t)=\int K(t,t^{\prime})X(t^{\prime})\leavevmode\nobreak\
dt^{\prime}$ (35)

Using Karhunen–Loève Transform (KLT), $X(t^{\prime})$ and $C(t)$ are both
decomposed as in (36) and (37),

$\displaystyle X(t^{\prime})=\sum_{i=1}^{\infty}x_{i}\phi_{i}(t^{\prime})$
(36) $\displaystyle C(t)=\sum_{j=1}^{\infty}c_{j}\psi_{j}(t)$ (37)

where $x_{i}$ and $c_{j}$ are both random variables with
$\mathbb{E}\\{x_{i}x_{i^{\prime}}\\}{=}\lambda_{x_{i}}\delta_{ii^{\prime}}$
and
$\mathbb{E}\\{c_{j}c_{j^{\prime}}\\}{=}\lambda_{c_{j}}\delta_{jj^{\prime}}$.
$\\{\lambda_{x_{i}}\\}$, $\\{\lambda_{x_{j}}\\}$ $\\{\phi_{i}(t^{\prime})\\}$
and $\\{\psi_{j}(t)\\}$ are eigenvalues and eigenfuncions, respectively. Let
us denote $n{=}i{=}j$ and $\sigma_{n}{=}\frac{c_{n}}{x_{n}}$, and assume that
$K(t,t^{\prime})$ can be expressed as in (38),

$\displaystyle
K(t,t^{\prime})=\sum_{n}^{\infty}\sigma_{n}\psi_{n}(t)\phi_{n}(t^{\prime})$
(38)

We show that (38) is a correct representation of $K(t,t^{\prime})$ by proving
(35) holds under this definition. We observe that by substituting (36) and
(38) into the right hand side of (35) we have that,

$\displaystyle\int K(t,t^{\prime})X(t^{\prime})\leavevmode\nobreak\
dt^{\prime}$
$\displaystyle=\int\sum_{n}^{\infty}\sigma_{n}\psi_{n}(t)\phi_{n}(t^{\prime})\sum_{n}^{\infty}x_{n}\phi_{n}(t^{\prime})\leavevmode\nobreak\
dt^{\prime}$
$\displaystyle=\int\sum_{n}^{\infty}\sigma_{n}x_{n}\psi_{n}(t)|\phi_{n}(t^{\prime})|^{2}$
$\displaystyle+\sum_{n^{\prime}\neq
n}^{\infty}\sigma_{n}x_{n^{\prime}}\psi_{n}(t)\phi_{n}(t^{\prime})\phi_{n^{\prime}}^{*}(t^{\prime})\leavevmode\nobreak\
dt^{\prime}$ $\displaystyle=\sum_{n}^{\infty}c_{n}\psi_{n}(t)=C(t)$ (39)

which is equal to the left hand side of (35). Therefore, (38) is a correct
representation of $K(t,t^{\prime})$.

∎

###  B-B Proof of Theorem 1: High Order Generalized Mercer’s Theorem (HOGMT

###### Proof.

Given a 2-D process $X(\gamma_{1},\gamma_{2})$, the eigen-decomposition using
Lemma 1 is given by,

$X(\gamma_{1},\gamma_{2})=\sum_{n}^{\infty}x_{n}e_{n}(\gamma_{1})s_{n}(\gamma_{2})$
(40)

Letting
$\psi_{n}(\gamma_{1},\gamma_{2}){=}e_{n}(\gamma_{1})s_{n}(\gamma_{2})$, and
substituting it in (40) we have that,

$X(\gamma_{1},\gamma_{2})=\sum_{n}^{\infty}x_{n}\phi_{n}(\gamma_{1},\gamma_{2})$
(41)

where $\phi_{n}(\gamma_{1},\gamma_{2})$ are 2-D eigenfunctions with the
property (42).

$\iint\phi_{n}(\gamma_{1},\gamma_{2})\phi_{n^{\prime}}(\gamma_{1},\gamma_{2})\leavevmode\nobreak\
d\gamma_{1}\leavevmode\nobreak\ d\gamma_{2}=\delta_{nn^{\prime}}$ (42)

We observe that (41) is the 2-D form of KLT. With iterations of the above
steps, we obtain Higher-Order KLT for $X(\gamma_{1},\cdots,\gamma_{Q})$ and
$C(\zeta_{1},\cdots,\zeta_{P})$ as given by,

$\displaystyle
X(\gamma_{1},\cdots,\gamma_{Q})=\sum_{n}^{\infty}x_{n}\phi_{n}(\gamma_{1},\cdots,\gamma_{Q})$
(43) $\displaystyle
C(\zeta_{1},\cdots,\zeta_{P})=\sum_{n}^{\infty}c_{n}\psi_{n}(\zeta_{1},\cdots,\zeta_{P})$
(44)

where $C(\zeta_{1},\cdots,\zeta_{P})$ is the projection of
$X(\gamma_{1},\cdots,\gamma_{Q})$ onto
$K(\zeta_{1},\cdots,\zeta_{P};\gamma_{1},\cdots,\gamma_{Q})$.

Then following similar steps as in Appendix B-A we get (B-B).

$\displaystyle K(\zeta_{1},\cdots,\zeta_{P};\gamma_{1},\cdots,\gamma_{Q})$
$\displaystyle=\sum_{n}^{\infty}\sigma_{n}\psi_{n}(\zeta_{1},\cdots,\zeta_{P})\phi_{n}(\gamma_{1},\cdots,\gamma_{Q})$
(45)

∎

##  Appendix C Proofs on Eigenfunction based Precoding

###  C-A Proof of Lemma 2

###### Proof.

Using 2-D KLT as in (13), $x(u,t)$ is expressed as,

$x(u,t)=\sum_{n}^{\infty}x_{n}\phi_{n}(u,t)$ (46)

where $x_{n}$ is a random variable with
$E\\{x_{n}x_{n^{\prime}}\\}{=}\lambda_{n}\sigma_{nn^{\prime}}$ and
$\phi_{n}(u,t)$ is a 2-D eigenfunction.

Then the projection of $k_{H}(u,t;u^{\prime},t^{\prime})$ onto
$\phi_{n}(u^{\prime},t^{\prime})$ is denoted by $c_{n}(u,t)$ and is given by,

$c_{n}(u,t)=\iint
k_{H}(u,t;u^{\prime},t^{\prime})\phi_{n}(u^{\prime},t^{\prime})\leavevmode\nobreak\
du^{\prime}\leavevmode\nobreak\ dt^{\prime}$ (47)

Using the above, (28) is expressed as,

$\displaystyle||s(u,t)-Hx(u,t)||^{2}=||s(u,t)-\sum_{n}^{\infty}x_{n}c_{n}(u,t)||^{2}$
(48)

Let $\epsilon(x){=}||s(u,t)-\sum_{n}^{\infty}x_{n}\phi_{n}(u,t)||^{2}$. Then
its expansion is given by,

$\displaystyle\epsilon(x)=\langle
s(u,t),s(u,t)\rangle-2\sum_{n}^{\infty}x_{n}\langle c_{n}(u,t),s(u,t)\rangle$
(49) $\displaystyle+\sum_{n}^{\infty}x_{n}^{2}\langle
c_{n}(u,t),c_{n}(u,t)\rangle+\sum_{n}^{\infty}\sum_{n^{\prime}\neq
n}^{\infty}x_{n}x_{n^{\prime}}\langle c_{n}(u,t),c_{n^{\prime}}(u,t)\rangle$

Then the solution to achieve minimal $\epsilon(x)$ is obtained by solving for
$\partialderivative{\epsilon(x)}{x_{n}}=0$ as in (50).

$\displaystyle x_{n}^{opt}$ $\displaystyle{=}\frac{\langle
s(u,t),c_{n}(u,t)\rangle+\sum_{n^{\prime}\neq n}^{\infty}x_{n^{\prime}}\langle
c_{n^{\prime}}(u,t),c_{n}(u,t)\rangle}{\langle c_{n}(u,t),c_{n}(u,t)\rangle}$
(50)

where $\langle a(u,t),b(u,t)\rangle{=}\iint
a(u,t)b^{*}(u,t)\leavevmode\nobreak\ du\leavevmode\nobreak\ dt$ denotes the
inner product. Let $\langle c_{n^{\prime}}(u,t),c_{n}(u,t)\rangle=0$, i.e.,
the projections $\\{c_{n}(u,t)\\}_{n}$ are orthogonal basis. Then we have a
closed form expression for $x^{opt}$ as in (51).

$\displaystyle x_{n}^{opt}$ $\displaystyle{=}\frac{\langle
s(u,t),c_{n}(u,t)\rangle}{\langle c_{n}(u,t),c_{n}(u,t)\rangle}$ (51)

Substitute (51) in (49), it is straightforward to show that $\epsilon(x){=}0$.
∎

###  C-B Proof of Theorem 2: Eigenfunction Precoding

###### Proof.

The 4-D kernel $k_{H}(u,t;u^{\prime},t^{\prime})$ is decomposed into two
separate sets of eigenfunction $\\{\phi_{n}(u^{\prime},t^{\prime})\\}$ and
$\\{\psi_{n}(u,t)\\}$ using Theorem 1 as in (30). By transmitting the
conjugate of the eigenfunctions, $\phi_{n}(u^{\prime},t^{\prime})$ through the
channel $H$, we have that,

$\displaystyle H\phi_{n}^{*}(u^{\prime},t^{\prime})=\iint
k_{H}(u,t;u^{\prime},t^{\prime})\phi_{n}^{*}(u^{\prime},t^{\prime})\leavevmode\nobreak\
du^{\prime}\leavevmode\nobreak\ dt^{\prime}$
$\displaystyle{=}\iint\sum_{n}^{\infty}\\{\sigma_{n}\psi_{n}(u,t)\phi_{n}(u^{\prime},t^{\prime})\\}\phi_{n}^{*}(u^{\prime},t^{\prime})\leavevmode\nobreak\
dt^{\prime}\leavevmode\nobreak\ df^{\prime}$
$\displaystyle{=}\iint\sigma_{n}\psi_{n}(u,t)|\phi_{n}(u^{\prime},t^{\prime})|^{2}$
$\displaystyle+\sum_{n^{\prime}\neq
n}^{\infty}\sigma_{n^{\prime}}\psi_{n^{\prime}}(u,t)\phi_{n^{\prime}}(u^{\prime},t^{\prime})\
\phi_{n}^{*}(u^{\prime},t^{\prime})\leavevmode\nobreak\
du^{\prime}\leavevmode\nobreak\ dt^{\prime}$
$\displaystyle{=}\sigma_{n}\psi_{n}(u,t)$ (52)

where $\psi_{n}(u,t)$ is also a 2-D eigenfunction with the orthogonal property
as in (31).

From Lemma 2, if the set of projections, $\\{c_{n}(u,t)\\}$ is the set of
eigenfunctions, $\\{\psi_{n}(u,t)\\}$, which has the above orthogonal
property, we achieve the optimal solution as in (51). Therefore, let $x(u,t)$
be the linear combination of $\\{\phi_{n}^{*}(u,t)\\}$ with coefficients
$\\{x_{n}\\}$ as in (53),

$x(u,t)=\sum_{n}^{\infty}x_{n}\phi_{n}^{*}(u,t)$ (53)

Then (48) is rewritten as in (54),

$\displaystyle||s(u,t)-Hx(u,t)||^{2}=||s(u,t)-\sum_{n}^{\infty}x_{n}\sigma_{n}\psi(u,t)||^{2}$
(54)

Therefore, optimal $x_{n}$ in (51) is obtained as in (55),

$x_{n}^{opt}=\frac{\langle s(u,t),\psi_{n}(u,t)\rangle}{\sigma_{n}}$ (55)

Substituting (55) in (53), the transmit signal is given by (56),

$x(u,t)=\sum_{n}^{\infty}\frac{\langle
s(u,t),\psi_{n}(u,t)\rangle}{\sigma_{n}}\phi_{n}^{*}(u,t).$ (56)

∎

###  C-C Proof of Corollary 1

###### Proof.

First we substitute the 4-D kernel $k_{H}(u,t;u^{\prime},t^{\prime})$ with the
2-D kernel $k_{H}(u,u^{\prime})$ in Theorem 2 which is then decomposed by the
2-D HOGMT. Then following similar steps as in Appendix C-B it is
straightforward to show (34). ∎

##  Appendix D Results on Interference

Figure 1: Kernel $k_{H}(u,t;u^{\prime},t^{\prime})$ for $u{=}1$ at a) $t{=}1$,
b) $t{=}10$, c) $t{=}50$ and d) $t{=}100$.

Figure 1 shows the channel response for user $u{=}1$ at $t{=}1$, $t{=}10$,
$t{=}50$ and $t{=}100$, where at each instance, the response for user $u{=}1$
is not only affected by its own delay and other users’ spatial interference,
but also affected by other users’ delayed symbols. This is the cause of joint
space-time interference which necessitates joint precoding in the
2-dimensional space using eigenfunctions that are jointly orthogonal.

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# The indication for 40K geo-antineutrino flux with Borexino phase-III data

L. B. Bezrukov, I. S. Karpikov, V. V. Sinev
Institute for Nuclear Research, Moscow, 117312 Russia

###### Abstract

Abstract $-$ We provide the indication of high flux of 40K geo-antineutrino
and geo-neutrino (40K-geo-($\bar{\nu}+\nu$)) with Borexino Phase III data.
This result was obtained by introducing a new source of single events, namely
40K-geo-($\bar{\nu}+\nu$) scattering on electrons, in multivariate fit
analysis of Borexino Phase III data. Simultaneously we obtained the count
rates of events from 7Be, $pep$ and CNO solar neutrinos. These count rates are
consistent with the prediction of the Low metallicity Sun model SSM
B16-AGSS09. MC pseudo-experiments showed that the case of High metallicity Sun
and absence of 40K-geo-($\bar{\nu}+\nu$) can not imitate the result of
multivariate fit analysis of Borexino Phase III data with introducing
40K-geo-($\bar{\nu}+\nu$) events. We also provide arguments for the high
abundance of potassium in the Earth.

Keywords: Borexino detector, potassium geo-antineutrino, Low metallicity Sun,
Earth intrinsic heat.

##  1 Introduction

The Borexino collaboration reported in [1] an improved measurement of the
Carbon-Nitrogen-Oxygen (CNO) solar neutrino interaction rate at Earth obtained
with the complete Borexino Phase-III dataset. The measured rate, $R_{\rm
CNO}=6.7^{+2.0}_{-0.8}$ cpd/100t (counts/(day $\times$ 100 tonnes)).

The main idea of this article is to see what could happen if we introduce in
analysis the new source of single events from 40K geo-antineutrino and
neutrino (40K-geo-($\bar{\nu}+\nu$)) scattering on electrons. Such source was
proposed in [2, 3] and such analysis were done in [4] by using Borexino phase
II dataset from [5] and in [6] by using Borexino phase III dataset.

We will use in this paper the Phase-III dataset which is different from Phase-
II dataset [5]. The Phase-III dataset was taken when the radiopurity and
thermal stability of the detector was maximal, i. e. between January 2017 and
October 2021. The Phase-II dataset was taken from July 2016 until February
2020. The Phase-III statistics is higher than the Phase-II statistics on about
30%.

We hope that the obtained result will attract the attention of the scientific
community to an alternative interpretation of the Borexino data and encourage
further research of solar neutrino and potassium geo-antineutrino fluxes.

##  2 Energy spectrum of recoil electrons from 40K-geo-($\bar{\nu}+\nu$)

40K abundance in natural mixture of K isotopes is 0.0117%. It decays with
$T_{1/2}=1.248\cdot 10^{9}$ y through two modes: to 40Ca with efficiency
89.28% emitting $\bar{\nu_{e}}$ with maximal energy 1.311 keV and to 40Ar
emitting one of two possible mono-energetic $\nu_{e}$-s with energies 43.5 keV
(10.67%) and 1.504 keV ($\sim$0.05%). Decay scheme of 40K can be found in [7].
Low energy neutrinos cannot be detected but high energy ones can be seen in
large volume detector as a small addition to antineutrino’s effect.

Figure 1:  40K antineutrinos and neutrinos spectra $-$ solid line. The spectra
are normalized to the probablity of emission in decay: 0.8928 for
$\bar{\nu_{e}}$-s and 0.0005 for $\nu_{e}$-s. Energy spectrum of electrons
emitted in 40K decay $-$ dashed line.

Figure 1 demonstrates the total beta, antineutrino and neutrino energy spectra
for 40K decay. The antineutrino spectrum was calculated using the beta one
taken in [7]. The beta spectrum is the same as was measured in [8] and
calculated in [9].

Figure 2:  The energy spectrum of recoil electrons from
40K-geo-($\bar{\nu}+\nu$). The pedestal at higher energies corresponds to
neutrinos. The total counting rate of recoil electrons for this spectrum is
2.18 cpd/100t.

In Figure 2 one can see the calculated energy spectrum of recoil electrons for
40K-geo-($\bar{\nu}+\nu$) that follows from the spectrum shape shown in Figure
1. The pedestal at higher energies corresponds to 1.5-MeV neutrinos. The total
counting rate of recoil electrons for the spectrum shown in Figure 2 is equal
to 2.18 cpd/100t that corresponds to potassium abundance 1.0% of the Earth
mass in the case of uniform distribution of potassium in the Earth.

The energy spectrum normalized to unit is the probability density function and
we will use below the notation PDF for such normalized spectrum.

Figure 3:  Recoil electron spectrum from 40K-geo-($\bar{\nu}$ \+ $\nu$)
transferred to $N_{h}$ (numder of hit PMTs) scale according to the algorithm
from [10]$-$ solid line. Energy spectrum of electrons emitted in 40K decay $-$
dashed line.

We will use PDFs in multivariate fit analysis (MF) as a function of number of
hit PMTs $N_{h}$ following the Borexino Collaboration. The calculated recoil
electron energy spectrum from 40K-geo-($\bar{\nu}+\nu$) was transferred from
the energy scale to $N_{h}$ scale using the algorithm described in [10]. In
Figure 3 one can find 40K-geo-($\bar{\nu}+\nu$) PDF used in the MF analysis.
40K $\beta$ spectrum PDF is also shown here for comparison.

To prove that we correctly perform the calculation of 40K-geo-($\bar{\nu}$ \+
$\nu$) PDF, one can compare our 7Be PDF and the one taken from Borexino plot
[5] shown in Figure 4.

Figure 4: PDF of recoil electron spectrum from 7Be neutrinos transferred to
$N_{h}$ (numder of hit PMTs) scale according to the algorithm from [10]. Solid
line $-$ our calculation, dashed line $-$ [5].

##  3 Energy spectrum of recoil electrons from CNO-$\nu$

The CNO-$\nu$ spectrum consists of three components: the 13N, 15O and 17F
neutrino spectra. The shape of these spectra is well known because the
transitions are allowed and can be easily calculated. However, the fluxes
depend on the solar model, and many flux predictions exist. In Table 1 we can
see a number of flux predictions for 13N, 15O and 17F according to different
models.

Table 1: Neutrino fluxes produced in the Sun by nuclei 13N, 15O and 17F in units cm-2 s-1. Model |  13N$\times 10^{8}$ |  15O$\times 10^{8}$ |  17F$\times 10^{6}$
---|---|---|---
B16 [11] | 5.03 | 1.34 |  $<$8.5
GS98 [12] | 2.78 | 2.05 | 5.29
AGSS09 [12] | 2.04 | 1.44 | 3.26
Caffau11 [13] | 2.801 | 2.123 | 4.648
used in Borexino [14] | 2.78 | 2.05 | 5.29

We have calculated neutrino spectra from 13N, 15O and 17F ourselves in the
moment of production inside the Sun. These calculated spectra undergo
oscillations according to the MSW-mechanism [15]. As a result each spectrum
split in two: the electron neutrino spectrum and $\nu_{\mu}$ \+ $\nu_{\tau}$
spectrum. For each one we calculated a recoil electron spectrum and composed
them in one. The sum of three CNO components appear similar to the CNO
spectrum used in the Borexino analysis but not exactly. So, we decided to use
in our analysis the CNO spectrum digitized from the plot of the Borexino
publication. We can see in Table 1 that spectrum used in the Borexino analysis
corresponds to High metallicity Sun model GS98. Later we are going to make the
analysis using CNO spectrum corresponding to Low metallicity Sun model AGSS09,
see Table 1.

##  4 Borexino experimental data multivariate fit analysis

We will use here the dataset digitized from the figure from [1] with
suppressed contribution of the cosmogenic 11C background. The histogram of the
number of events of Phase III $N$ depending on the number of hit PMTs $N_{h}$
is shown in Table 2.

Table 2: Experimental histogram of the number of events $N$ of Phase III depending on the number of hit PMTs $N_{h}$ from [1]. Bin centers are shown in columns labeled as $N_{h}$. Events were collected over 871.31 days in 71.3 t. $N_{h}$ | $N$ | $N_{h}$ | $N$ | $N_{h}$ | $N$ | $N_{h}$ | $N$ | $N_{h}$ | $N$ | $N_{h}$ | $N$ | $N_{h}$ | $N$ | $N_{h}$ | $N$ | $N_{h}$ | $N$
---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---
140 | 1021 | 230 | 931 | 320 | 167 | 410 | 55 | 500 | 54 | 590 | 34 | 680 | 25 | 770 | 23 | 860 | 40
145 | 1098 | 235 | 834 | 325 | 123 | 415 | 68 | 505 | 46 | 595 | 34 | 685 | 22 | 775 | 35 | 865 | 36
150 | 1251 | 240 | 725 | 330 | 110 | 420 | 43 | 510 | 53 | 600 | 31 | 690 | 22 | 780 | 26 | 870 | 37
155 | 1573 | 245 | 663 | 335 | 105 | 425 | 56 | 515 | 50 | 605 | 31 | 695 | 20 | 785 | 34 | 875 | 39
160 | 1801 | 250 | 677 | 340 | 76 | 430 | 71 | 520 | 50 | 610 | 26 | 700 | 15 | 790 | 31 | 880 | 35
165 | 2097 | 255 | 602 | 345 | 712 | 435 | 62 | 525 | 39 | 615 | 25 | 705 | 20 | 795 | 33 | 885 | 28
170 | 2614 | 260 | 797 | 350 | 71 | 440 | 43 | 530 | 44 | 620 | 31 | 710 | 20 | 800 | 34 | 890 | 35
175 | 3044 | 265 | 555 | 355 | 60 | 445 | 46 | 535 | 47 | 625 | 21 | 715 | 14 | 805 | 37 | 895 | 32
180 | 3341 | 270 | 562 | 360 | 55 | 450 | 65 | 540 | 36 | 630 | 21 | 720 | 19 | 810 | 42 | 900 | 26
185 | 3471 | 275 | 479 | 365 | 59 | 455 | 57 | 545 | 36 | 635 | 30 | 725 | 24 | 815 | 26 | 905 | 33
190 | 3341 | 280 | 425 | 370 | 56 | 460 | 45 | 550 | 26 | 640 | 28 | 730 | 24 | 820 | 36 | 910 | 25
195 | 3285 | 285 | 406 | 375 | 51 | 465 | 62 | 555 | 35 | 645 | 25 | 735 | 24 | 825 | 38 | 915 | 22
200 | 2869 | 290 | 365 | 380 | 44 | 470 | 71 | 560 | 46 | 650 | 16 | 740 | 19 | 830 | 38 | 920 | 27
205 | 2443 | 295 | 316 | 385 | 59 | 475 | 70 | 565 | 35 | 655 | 25 | 745 | 24 | 835 | 29 | 925 | 22
210 | 2151 | 300 | 265 | 390 | 49 | 480 | 58 | 570 | 39 | 660 | 17 | 750 | 29 | 840 | 39 | 930 | 20
215 | 1741 | 305 | 233 | 395 | 46 | 485 | 52 | 575 | 35 | 665 | 25 | 755 | 22 | 845 | 33 | 935 | 13
220 | 1368 | 310 | 191 | 400 | 51 | 490 | 40 | 580 | 24 | 670 | 25 | 760 | 24 | 850 | 44 | 940 | 19
225 | 1160 | 315 | 198 | 405 | 56 | 495 | 57 | 585 | 37 | 675 | 23 | 765 | 32 | 855 | 52 | 945 | 11

We have calculated PDFs for most of the components used in the Borexino
analysis and used them in our multivariate fit (MF) analysis (7Be, $pep$, 11C,
210Bi, 85Kr and CNO). Four components were digitized from [5], they are: 210Po
alpha-peak, 8B and two backgrounds caused by external gammas from 208Tl and
the summed spectrum of 214Bi and 40K gammas.

The $\chi^{2}$ function was used to estimate the goodness of our fit:

$\chi^{2}=\sum_{i=1}^{162}\frac{(N_{i}-\sum_{k=1}^{10}w_{k}\cdot
f_{k,i})^{2}}{\sigma^{2}_{i}},$ (1)

where $N_{i}$ is the experimental value of counting rate of the Borexino
detector events in the $i$-th bin of the number of hitted PMTs,
$w_{k}=R_{k}\cdot 0.713\cdot t_{meas}-$ the weight for the PDF of the $k$-th
component with $R_{k}-$ the total counting rate of the $k$-th component in
cpd/100t, 0.713 $-$ the ratio of 71.3 t and 100 t and $t_{meas}-$ the
measurement time in days, $f_{k,i}-$ the PDF of the $k$-th component and
$\sigma_{i}$ is experimental uncertainty. We used a standard tool for
analysis, ROOT 6.22/08, to find the set of parameters $R_{k}\pm\Delta R_{k}$
which minimizes $\chi^{2}$. The $\Delta R_{k}$ corresponds to the confidence
level of $68\%$ in ROOT 6.22/08. To check the results of ROOT we used also
another minimization code.

Table 3: Total counting rates and $\chi^{2}$ obtained without (column 1 and 1a) and with (column 2 and 2a) the 40K-geo-($\bar{\nu}+\nu$) component of the fit, cpd/100t. Comparison of the experimental results of solar neurtino counting rates from MFs and theoretical predictions for High metallicity (column 3) and Low metallicity (column 4) [10], cpd/100t. | Mod.1,exp. | Mod.1,cpd range | Mod.2,exp. | Mod.2,cpd range | B16CS98 | B16AGSS09
---|---|---|---|---|---|---
|  |  |  |  | HZ | LZ
| 1 | 1 a | 2 | 2 a | 3 | 4
7Be |  48.4 $\pm$ 0.9 |  46$\div$49 |  46.0 $\pm$ 0.84 |  42$\div$49 |  47.9 $\pm$ 2.8 |  43.7 $\pm$ 2.5
$pep$ | 2.7 | 2.7 |  2.78 $\pm$ 0.07 |  2.7$\div$2.78 |  2.74 $\pm$ 0.04 |  2.78 $\pm$ 0.04
CNO |  6.5 $\pm$ 0.7 |  4.5$\div$8 |  3.67 $\pm$ 0.73 |  3.5$\div$20 |  4.92 $\pm$ 0.55 |  3.52 $\pm$ 0.37
11C |  1.7$\pm$ 0.1 |  1.35$\div$2 |  1.83 $\pm$ 0.07 |  1.4$\div$2 |  |
210Po |  41.5$\pm$0.4 |  40$\div$46 |  41.1$\pm$ 0.08 |  35$\div$45 |  |
210Bi | 11.8 |  9.8$\div$11.8 |  11.0 $\pm$ 0.23 |  10.7$\div$11 |  |
85Kr | 8.5 |  3.5$\div$8.5 |  8.5 $\pm$ 0.24 |  3.5$\div$8.5 |  |
208Tl |  4.76$\pm$ 0.2 |  4.3$\div$5 | 4.76 | 4.7/6 |  |
214Bi |  1.8$\pm$ 0.3 |  1.7$\div$2 | 1.79 | 1.79 |  |
40Kgeo | 0.0 | 0.0 |  11 $\pm$ 0.6 |  0$\div$11 |  |
$\chi^{2}_{1},\chi^{2}_{2}$ | 199.5 |  | 173.5 |  |  |
Figure 5:  Counting rate of recoil electrons and alphas in Borexino detector
in 5 hits bin versus number of hit PMTs. The values of points with errors from
[1] are given in Table 2. The curves are the contributions of different
components of the fit and the sum of all components obtained by the MF
analysis. Dark black dashed curve is contribution of 40K-geo-($\bar{\nu}+\nu$)
events. The obtained total counting rates for components of the fit are
presented in the second column of the Table 3.

The PDF for 40K-geo-($\bar{\nu}$ \+ $\nu$) was added to the analysis. So, in
total we have 10 PDFs to fit: 7Be, $pep$, 11C, 210Po, 210Bi, 85Kr, external
backgrounds 1 and 2, CNO and 40K-geo-($\bar{\nu}$ \+ $\nu$). 8B was fixed.

At the first stage we made the fit of Borexino dataset by minimization of
$\chi^{2}$ from (1) using our calculated PDFs for the components but
40K-geo-($\bar{\nu}$ \+ $\nu$) was set to zero. We used the energy range from
140 to 945 hits following the Borexino analysis. The experimental dataset was
described with a sum of 8 variable components. The $pep$ was fix and other
parameters were constrained as column 1a of Table 3 shows. The upper limits of
constraints for 210Bi and 85Kr we took from Borexino measurements of the upper
limits of 210Bi and 85Kr concentrations in scintillator. As a result we
obtained total counting rates that are on the first column of Table 3. We can
observe the similarity of the obtained values of the parameters to the
parameters presented by Borexino Collaboration in [1, 10] and references
inside them. The obtained values of $R$(7Be), $R(pep)$, $R$(CNO) support the
Sun model of High metallicity. Theoretical predictions for High metallicity
Sun are shown in column 3 of Table 3.

The PDF of 40K-geo-($\bar{\nu}+\nu$) was added at the second stage of the
analysis. We made 8 parameters free (constrained) and external backgrounds 1
and 2 were fixed as column 2a of Table 3 shows.

The result of this fit is shown in Figure 5 and obtained total counting rates
are in the second column of Table 3. We obtained the value of
$R(^{40}$K-geo-$(\bar{\nu}+\nu))=11\pm 0.6$ cpd/100t. The fit found also the
reasonable values of total counting rates for all components. The obtained
values in second stage of ananlysis of $R$(7Be), $R(pep)$, $R$(CNO) support
the Sun model of Low metallicity in contrary to the first stage analysis.
Theoretical predictions for Low metallicity Sun are shown in column 4 of Table
3. We note here that $\chi^{2}_{2}$ values obtained at the second stage is
lower (better) than obtained at the first stage $\chi^{2}_{1}$. The
$\chi^{2}_{1}$ and $\chi^{2}_{2}$ values are shown in last row of Table 3.
This means that Low metallicity Sun and high abundance of potassiun in the
Earth is more realistic than High metallicity Sun and low abundace of
potassium in the Earth.

The inclusion of the scattering of 40K-geo-($\bar{\nu}+\nu$) on electrons in
the analysis mainly reduces the number of events from 7Be and CNO. But the
number of events from $pep$ becomes a little higher.

Note that the error for the counting rate of 40K-geo-$(\bar{\nu}+\nu)$ events
given in column 2 of the Table 3 characterizes the minimum of the function
$\chi^{2}$ and is not an experimental error in measuring the value of
$R(^{40}$K-geo-$(\bar{\nu}+\nu))$.

##  5 Simulation of the experiment

We performed MC simulated experiments to understand the influence of the
finite sample of data, the parameter constraints and the other conditions of
the fit on reconstructed results. The MC simulated experiments help also to
understand the possibilities of a next generation Borexino-type detector to
measure the 40K-geo-$(\bar{\nu}+\nu)$ flux accurately.

We took a set of PDFs with certain fixed counting rates and simulated the
Borexino-like spectrum of single events many times via the MC method. Each
time we had a new finite sample of Borexino-like data. We reconstructed the
values of counting rates using the procedure (1) described above for each
sample and, as a result, obtained probability distributions of the
reconstructed counting rates and $\chi^{2}_{1}$ and $\chi^{2}_{2}$. We
obtained wide non-symmetric distributions with their mean values different
from the values fixed in MC mainly for components with low counting rates
$R(^{40}$K-geo) and $R$(CNO). The main reason for this is the effect of
statistical fluctuations of components with high counting rates. These
fluctuations affect the reconstructed counting rates of components with low
counting rates. The value of this effect depends on statistics. Therefore, we
found that the reconstructed counting rates in the Borexino-like data samples
obtained using this procedure can have systematic biases relative to their
true values.

Below we will give an examples of our pseudo-experiments. We have two sets of
components after our MFs of the experimental Borexino data sample. One set is
in the first column of Table 3 and another set is in the second column of
Table 3. The main difference is the reconstructed
$R(^{40}$K-geo-($\bar{\nu}+\nu)$), $R(^{7}$Be) and $R$(CNO) rates.

Figure 6:  Reconstructed counting rate
$R_{\rm{res}}(^{40}$K-geo-$(\bar{\nu}+\nu))$ distributions for $10^{5}$
pseudo-experiments. The curve labled $R_{\rm{in}}(^{40}$K) = 0 cpd/100t is the
result of pseudo-experiments with parameters from column 1 of the Table 3. The
mean value of this distribution is
$R_{\rm{res}}(^{40}$K-geo-$(\bar{\nu}+\nu))$ = 1.67 cpd/100t and $\sigma=1.53$
cpd/100t. The curve labled $R_{\rm{in}}(^{40}$K) = 7 cpd/100t is the result of
pseudo-experiments with parameters from column 2 of the Table 3 but with
$R(^{40}$K-geo-$(\bar{\nu}+\nu))$ = 7 cpd/100t.

At first we chose for our example of pseudo-experiments the parameters from
first column of Table 3. But when we reconstructed the parameters, we included
the PDF of $R(^{40}$K-geo-($\bar{\nu}+\nu)$) in procedure (1) and applied the
constraints as column 2a of Table 3 shows. The idea was to test whether
reality without potassium could simulate our result with
$R(^{40}$K-geo-$(\bar{\nu}+\nu))=11$ cpd/100t. The curve labled a
$R_{\rm{in}}(^{40}$K) = 0 cpd/100t on Figure 6 shows the distribution of
reconstructed $R_{\rm{res}}(^{40}$K-geo-$(\bar{\nu}+\nu))$ for $10^{5}$
pseudo-experiments. We see that probability to observe
$R(^{40}$K-geo-$(\bar{\nu}+\nu))>7$ cpd/100t is less than $10^{-5}$.

Then we chose for our another example of pseudo-experiments the parameters
from second column of Table 3. Knowing about the existence of systematic
biases, we took for MC simulation of data samples the counting rate
$R(^{40}$K-geo-($\bar{\nu}+\nu)$) = 7 cpd/100t instead of the value in the
second column of Table 3. We applied the constraints as column 2a of Table 3
shows but $R(^{40}$K-geo-$(\bar{\nu}+\nu))$ was free. The curve labled
$R_{\rm{in}}(^{40}$K) = 7 cpd/100t on Figure 6 shows the distribution of
reconstructed $R_{\rm{res}}(^{40}$K-geo-$(\bar{\nu}+\nu))$ for $10^{5}$
pseudo-experiments. We see that our result
$R(^{40}$K-geo-$(\bar{\nu}+\nu))=11$ cpd/100t has the high probability in this
case. We see also a systematic bias and the need a big statistic to measure
the potassium abundance more accurately. We constructed the $\chi^{2}_{1}$ and
$\chi^{2}_{2}$ distributions obtained as a result of approximation these MC
data samples by the sets of single event sources without of
$R(^{40}$K-geo-$(\bar{\nu}+\nu))$ and with of
$R(^{40}$K-geo-$(\bar{\nu}+\nu))$ correspondently . The distributions has the
gaussian form with $\bar{\chi}^{2}_{1}=184.4\pm 21.17.$ and
$\bar{\chi}^{2}_{2}=167.4\pm 19.79.$ The obtained values $\chi^{2}_{1}=199.5$
and $\chi^{2}_{2}=173.5$ for experimental Borexino phase III dataset (see
Table 3) is not contradict to these MC values. At first we chose for our
example of pseudo-experiments the parameters from first column of Table 3. But
when we reconstructed the parameters, we included the PDF of
$R(^{40}$K-geo-($\bar{\nu}+\nu)$) in procedure (1) and applied the constraints
as column 2a of Table 3 shows. The idea was to test whether reality without
potassium could simulate our result with $R(^{40}$K-geo-$(\bar{\nu}+\nu))=11$
cpd/100t. The curve labled a $R_{\rm{in}}(^{40}$K) = 0 cpd/100t on Figure 6
shows the distribution of reconstructed
$R_{\rm{res}}(^{40}$K-geo-$(\bar{\nu}+\nu))$ for $10^{5}$ pseudo-experiments.
We see that probability to observe $R(^{40}$K-geo-$(\bar{\nu}+\nu))>7$
cpd/100t is less than $10^{-5}$.

Then we chose for our another example of pseudo-experiments the parameters
from second column of Table 3. Knowing about the existence of systematic
biases, we took for MC simulation of data samples the counting rate
$R(^{40}$K-geo-($\bar{\nu}+\nu)$) = 7 cpd/100t instead of the value in the
second column of Table 3. We applied the constraints as column 2a of Table 3
shows but $R(^{40}$K-geo-$(\bar{\nu}+\nu))$ was free. The curve labled
$R_{\rm{in}}(^{40}$K) = 7 cpd/100t on Figure 6 shows the distribution of
reconstructed $R_{\rm{res}}(^{40}$K-geo-$(\bar{\nu}+\nu))$ for $10^{5}$
pseudo-experiments. We see that our result
$R(^{40}$K-geo-$(\bar{\nu}+\nu))=11$ cpd/100t has the high probability in this
case. We see also a systematic bias and the need a big statistic to measure
the potassium abundance more accurately. We constructed the $\chi^{2}_{1}$ and
$\chi^{2}_{2}$ distributions obtained as a result of approximation these MC
data samples by the sets of single event sources without of
$R(^{40}$K-geo-$(\bar{\nu}+\nu))$ and with of
$R(^{40}$K-geo-$(\bar{\nu}+\nu))$ correspondently . The distributions has the
gaussian form with $\bar{\chi}^{2}_{1}=184.4\pm 21.17.$ and
$\bar{\chi}^{2}_{2}=167.4\pm 19.79.$ The obtained values $\chi^{2}_{1}=199.5$
and $\chi^{2}_{2}=173.5$ for experimental Borexino phase III dataset (see
Table 3) is not contradict to these MC values.

Figure 7:  Distribution of $\Delta\chi^{2}=\chi^{2}_{1}-\chi^{2}_{2}$ for
$10^{4}$ pseudo-experiments for which we chose the parameters from the second
column of Table 3 but $R(^{40}$K-geo-$(\bar{\nu}+\nu))$ = 7 cpd/100t.
$\chi^{2}_{1}-$ result of fit of pseudo-experiment dataset with the sum of
event souces without 40K-geo-$(\bar{\nu}+\nu)$. $\chi^{2}_{2}-$ result of fit
of the same dataset but with 40K-geo-$(\bar{\nu}+\nu)$.

It is interesting also to compare the experimental
$\Delta\chi^{2}=\chi^{2}_{1}-\chi^{2}_{2}=26$ from Table 3 with MC
distribution of this value under assumption that Low metallicty Sun and the
high 40K-geo-$(\bar{\nu}+\nu)$ flux exists in nature. Figure 7 showes the MC
distribution of $\Delta\chi^{2}$ for this case. The value $\Delta\chi^{2}=26$
is not contradict to this MC distributuon.

##  6 Discussion

The Phase III dataset is different from Phase II dataset. The main difference
is in the energy dependence at low energies. This difference leads to
different MF results in the case of inclusion in the analysis of
40K-geo-$(\bar{\nu}+\nu)$ [4].

We put the constraint for $R(^{40}$K-geo-$(\bar{\nu}+\nu))\leq 11$ cpd/100t
(see column 2a of Table 3) and obtained that $R(^{40}$K-geo-$(\bar{\nu}+\nu))$
is equal to upper limit of constraint and $R$(CNO) = 3.67 cpd/100t. This value
is close to the prediction of LZ solar model. This is why we use here the
constraint $R(^{40}$K-geo-$(\bar{\nu}+\nu))\leq 11$ cpd/100t. If we put
$R(^{40}$K-geo-$(\bar{\nu}+\nu))$ free we can observe a further improvement of
$\chi^{2}$, an increase in the reconstructed value
$R(^{40}$K-geo-$(\bar{\nu}+\nu))$ and a decrease in $R$(CNO).

Such behavior of the MF result for Phase III dataset can be explained by the
existence in the Earth of a huge amount of potassium.

Taking into account the discovered bias of the reconstructed value, we propose
for further discussion to adopt the following value of the counting rate:

$R(^{40}$K-geo-$(\bar{\nu}+\nu))=7^{+2}_{-3}$ cpd/100t.

This mean that abundunce of potassium in the Earth is in the range from 1.8%
to 4.1% from the Earth mass.

##  7 Radiogenic terrestrial heat production

Suppose that the counting rate we have obtained
$R(^{40}$K-geo-$(\bar{\nu}+\nu))=7$ cpd/100t is given by the actual 40K in the
Earth. This means that about 3% of the Earth’s mass is potassium.

Let calculate the intrinsic Earth heat flux for this potassium abundance. The
mass of 40K in the Earth in this case is: $m(^{40}\rm{K})=2.1\cdot 10^{22}$ g.

The equation relating masses and heat production (power) is

$H=m\cdot\frac{N_{Avog}}{A}\cdot\frac{E_{release}}{\tau}\cdot\alpha,$ (2)

where $N_{Avog}$ \- Avogadro number, $A$ \- atomic number, $E_{release}=0.6$
MeV - average energy release in 40K decay, $\tau=t_{1/2}/ln2$ \- mean lifetime
of isotope, $\alpha$ \- the conversion factor $1\ \rm{MeV}=1.6\cdot 10^{-13}$
J

$H(^{40}\rm{K})=2.1\cdot 10^{22}\rm{g}\frac{6\cdot
10^{23}}{40}\rm{g^{-1}}\cdot\frac{0.89\cdot 0.6+0.1\cdot 1.5}{1.8\cdot
10^{9}\cdot 3.15\cdot 10^{7}s}\cdot 1.6\cdot 10^{-13}J=616\ TW.$ (3)

Add the heat production from uranium and thorium following the work [16]:

$H(\rm{U})=16.1\ TW$ and $H(\rm{Th})=18\ TW$.

$H=H({\rm{U}})+H({\rm{Th}})+H(^{40}\rm{K})=650\ \rm{TW}.$ (4)

This estimated value is rather high, but it is consistent with predictions of
Earth model ”Hydridic Earth model (HE model)” [18, 17]. HE model predicts the
high potassium abundance (up to several persent of Earth mass).

The obtained value of heat production 4 is enough to explain the observed by
ARGO project the increase of the ocean temperature [19]. Enegy release in 300
TW during 30 years is enough to heat the ocean as ARGO observed. In the frame
of HE model extra heat produced inside can be absorbed by the Earth itself
through its expansion. Latter leads to the Earth cooling. The cause of the
Earth expansion is the following. The radiogenic heat of the Earth interior
leads to hydride decomposition. Protons and metal appear as a result. The
metal volume is larger than volume of initial hydrides. This is the reason of
Earth expansion.

How much energy is needed to enhance the Earth radius by $h=1$ cm?

$M_{m}\cdot g\cdot h=4.91\cdot 10^{24}\rm{kg}\cdot 6\
\rm{\frac{m}{s^{2}}}\cdot 10^{-2}m=3\cdot 10^{23}J,$ (5)

here $M_{m}$ is the Earth mantle mass.

What work can be done by the energy release 350 TW during 30 years?

$3.5\cdot 10^{14}\rm W\cdot 3.15\cdot 10^{7}s\cdot 30y=3.3\cdot 10^{23}J.$ (6)

One can see that the expressions 5 and 6 are comparable. So, if the Earth
radius would be increaesd by more than 1 cm, then the decomposition of
hydrides be stopped and the cooling wave starts to spread through the Earth’s
body up to surface. This can cause a new ice age on the surface or make
decreasing of mean temperature at least.

##  8  Arguments in favor of the high potassium abundace in the Earth

The widespread belief in the fairness of Silicate Earth model and belief in
the validity of the results of work [20] that the heat production in the Earth
interior is equal to 47$\pm$2 TW do not allow to consider our result as the
evidence of high flux of 40K-geo-($\bar{\nu}+\nu$).

Often the HE model is critisized by the following points. The entire Earth
have to be melted due to radiogenic heat and spent the most part of its life
in this state if the Earth contains potassium more than 1% of the Earth mass.
Also the HE model predicts that the current heat production in the Earth’s
interior can be 200 TW and more which contradicts to the result of the work
[20].

However, these arguments are not fully reliable. In particular, the entire
Earth could not be melted because the HE model contains a subsurface cooling
mechanism. This mechanism is activated when the subsurface is heated enough to
decompose the metal hydrides. Therefore, in the HE model, the subsurface
temperature oscillates [18] and does not grow up till hydrides exist in the
Earth. This argument was used in [21, 22, 23, 24]. It is noted in these works
that thermal conductivity is not the main mechanism of heat transfer in the
Earth, but hot protons and hydrogen-containing gases carry out the heat away.
In these works the experimental evidences are provided that the heat
production in Earth interior can reach the several hundreds TW. These are the
heating of the oceans [19], the temperature profile of ultra-deep wells and
non-direct evidence – the heat production in Moon interior.

##  9 Next importat stepts in the Earth studies

Our analysis showed that the potential capabilities of the Borexino-type
detector are not exhausted.

Let consider a new detector of the Borexino-type with lower backgrounds and
higher statistics. This require the low background nylon for detector inner
vessel, better energy resolution, and detector must be plasced deeper
underground.

As a result of pure nylon inner vessel, a larger fiducial volume could be
achieved due to smaller amount of 210Po emanating into the scintillator. The
latter will make it possible to measure the concentration of 210Bi in the
scintillator by studying the Low Polonium Field in a stationary scintillator.
Exact knowledge of bismuth content in the scintillator will reduce the
uncertainty of the reconstructed values of the 40K-geo-($\bar{\nu}+\nu)$
counting rate.

##  10 Conclusion

We carried out our own analysis of the Borexino experimental data published in
Ref. [1].

  * •

We calculated the recoil electron energy spectrum from
40K-geo-($\bar{\nu}+\nu$) scattering on electrons and transferred it to the
numder of hit PMTs scale.

  * •

We intodused a new source of single events, namely 40K-geo-($\bar{\nu}+\nu$)
scattering on electrons, in multivariate fit analysis.

  * •

We provide the indication of high flux of 40K-geo-($\bar{\nu}+\nu$) with
Borexino phase III data. The abundance of potassium in the range $2\div 4\%$
from the Earth mass can give such flux.

  * •

We obtained the count rates of 7Be, $pep$ and CNO solar neutrinos. These count
rates are consistent with the prediction of the Low metallicity Sun model SSM
B16-AGS09.

  * •

MC pseudo-experiments showed that the High metallicity Sun and absence of
40K-geo-($\bar{\nu}+\nu$) can not simulate the result of multivariate fit
analysis with inrtodusing of 40K-geo-($\bar{\nu}+\nu$) events in analysis.

  * •

We consider our results as a support of Hydridic Earth model. We discussed the
problem of high thermal radiogenic flux from the Earth interior in the case of
the observed potassium abundance.

  * •

We proposed to bild the next generation Borexino-type detector to measure
40K-geo-($\bar{\nu}+\nu$) flux with better accuracy .

## Acknowledgments

We are grateful to G. V. Sinev for valuable discussions on transferring energy
loss to observable number of photoelectrons and idea to generate pseudo-
experiments of the Borexino data, to F. L. Bezrukov for discussions and
fruitful remarks and to I. I. Tkachev for common support and the possibility
to discuss the results in his seminar.

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See pages 1-last of RedditPerformanceStudy.pdf