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\begin{document}
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\begin{center}
\vskip 1cm{\LARGE\bf
Valuations of $v$-adic Power Sums and Zero \\
\vskip .06in
Distribution for the Goss $v$-adic Zeta Function \\
\vskip .13in
for $\FF_q[t]$
}
\vskip 1cm
\large
Dinesh S. Thakur\footnote{The author
was supported by NSA grant H98230-08-1-0049.}\\
Department of Mathematics\\
University of Arizona\\
Tucson, AZ 85721\\
USA\\
\href{mailto:thakur@math.arizona.edu}{\tt thakur@math.arizona.edu}
\end{center}
\vskip .2 in
\begin{abstract}
We study the valuation at an irreducible polynomial $v$ of the
$v$-adic power sum, for exponent $k$ (or $-k$), of polynomials of a
given degree $d$ in $\FF_q[t]$, as a sequence in $d$ (or $k$).
Understanding these sequences has immediate consequences, via standard
Newton polygon calculations, for the zero distribution of the
corresponding $v$-adic Goss zeta functions. We concentrate on $v$ of
degree one and two and give several results and conjectures describing
these sequences. As an application, we show, for example, that the
naive Riemann hypothesis statement which works in several cases, needs
modifications, even for a prime of degree two. In the last section, we
give an elementary proof of (and generalize) a product formula of Pink
for the leading term of the Goss zeta function.
\end{abstract}
\centerline{\it Dedicated to Jean-Paul Allouche on his 60th birthday}
\section{Introduction}
In a previous paper \cite{T09},
we investigated valuation sequences at the infinite place of the
power sums of polynomials in $\FF_q[t]$, with positive and negative
exponents. We gave
\begin{itemize}
\item[(i)] a simple recipe to find these;
\item[(ii)] a duality between valuations for positive and negative exponents;
\item[(iii)] a simple recursion in case $q$ was prime; and
\item[(iv)] applications to the zero distribution of the Goss
zeta function \cite{G, G2, T},
giving a cleaner approach to the Riemann hypothesis (due
to Wan and Sheats \cite{W, D, S}) and to the study of the multizeta
values in this case.
\end{itemize}
In this paper, we look at finite places $v$ of
$\FF_q[t]$ and study the $v$-adic valuations of power sums with the
$v$-factor removed. We give conjectural formulas for them describing
very interesting patterns, for $v$ of degree one and two, and
proving full results in some special cases.
We then give applications to the zero distribution
of the Goss $v$-adic zeta function, showing that
in the degree-two case, the Riemann hypothesis type
statement, holding at the infinite degree-one place, needs
modification. We also calculate a particular power
sum, which is a leading term of the Goss zeta, when
$q$ is a prime, and in particular, give an elementary
proof of (and generalize) Pink's nice product formula for the same when
$q=2$.
Patterns of these valuation sequences exhibit symmetries remarkably
similar to those occurring in several papers of Allouche, Shallit,
Mend\`es France, Lasjaunias, etc., as well as those found by the author
for continued fractions for analogs of $e$, Hurwitz numbers, and some
algebraic quantities.
\section{Power sums}
\subsection{Notation}
$$\begin{array}{rcl}
\ZZ&=&\mbox{\{integers\}}\\
\ZZ_+&=&\mbox{\{positive integers\}}\\
\ZZ_{\geq 0}&=&\mbox{\{nonnegative integers\}}\\
q&=&\mbox{a power of a prime} \ p, \ q=p^f\\
A&=&\FF_q[t]\\
A+&=&\mbox{\{monics in $A$\}}\\
A_{d}+&=&\mbox{\{monics in $A$ of degree $d$\}}\\
%A_{k_1>0$ would imply $\ell_d^{k_2-k_1} \equiv \modd{1} {t}$,
whereas it is divisible by $t$. On the other hand,
interestingly, the statement is true, when $q$ is prime, with $S_d$ replaced by $s_d$ by
\cite[Thm.\ 1]{T09}, and not in general as we saw above, for example.
(In my editing gaffe, these motivating remarks were inadvertently
left off in \cite{T09}).
\end{enumerate}
\subsection{Valuation sequence at $v$}
We have $v_d(k)\geq 0$ for all $k\in \ZZ$, with it being infinite
when the corresponding power sum is zero.
Note $v_d(pk)=pv_d(k)$, so without loss of generality, we can restrict to
$k$ prime to $p$.
Another simple remark is that two primes
related by automorphisms
$t\rightarrow t+\theta$, ($\theta\in \FF_q$) of $A$ (e.g., any two degree-one
primes) give the same
valuation sequence.
\subsection{Non-vanishing of power sums}
The power sums $S_{d, v}(k)$ are non-zero for $k>0$, as can be seen by
choosing a monic prime $P$, unequal to $v$ and of degree $d$ (which can be done unless
$q=2$, $v$ is of degree $2$ and $d=2$, which we can check separately),
and noticing that except for $a=P$ term, all other terms in
the power sum $S_{d, v}(k)$ have valuation $0$ at $P$.
For $k\leq 0$, these power sums can be zero. In fact, $S_{d,v}(k)$ is
zero if $d>\ell(-k)/(q-1)+\deg(v)$, by the Carlitz result
\cite[Cor. 5.6.2]{T} that
$S_d(k)=0$ if $d>\ell(-k)/(q-1)$ (with the converse of the latter
holding also for $q$ prime).
We have not investigated the exact
conditions corresponding to the non-vanishing. For a similar necessary
and sufficient condition for vanishing of $S_d(k)$, for general
$q$, due to Carlitz, Sheats and B\" ockle,
see \cite[A5]{T09} and \cite{B12}.
\section{Some evaluations and simple bounds for $k>0$}
\label{sec3}
We recall \cite[Sec.\ 5.6]{T} the formulas for power sums and valuations of $[n]$ and $\ell_n$.
$$S_d(r)=\frac{1}{\ell_d^r},\ \ 01$. We have
$v_1(3)=1$, $v_0(3)=0$.
\end{claim}
\begin{proof} We have
$$S_{d, v}(3)=\frac{[d+1]v^3+[d-1]^4[d]^3}{v^3[1]\ell_d^3}.$$
The valuation of the denominator is $3+3\lfloor d/2\rfloor$. The valuation
of the numerator is $2^d+2$ for $d$ even, and $2^d+4$ for $d$ odd, by
a straight calculation (we omit the details) as in the proof above.
\end{proof}
More generally, we have the following {\it conjectural recipe} guessed
from a small
numerical data computed:
If $q=2$, $v=t^2+t+1$, $d>1$,
$$v_d(2^n-1)=2^{n+d-2}-(2^n-1)\lfloor d/2\rfloor +(-1)^{d-1},
\ \mbox{if $n$ is
even}, $$
and
$$v_d(2^n-1)=2^{n+d-2}-(2^n-1)\lfloor d/2\rfloor-\lfloor (d+1)/2\rfloor+\lfloor
d/2\rfloor, \ \mbox{if $n$ is odd}.$$
\subsection{Trivial lower bound}
\begin{claim}
If $k>0$,
and $d>m\deg(v)$, then $v_d(k)\geq m$.
\end{claim}
\begin{proof} The terms in the sum
$S_{d, v}(k)$ can be grouped by orbits $1/(n+\theta v^m)^k$, as $\theta$
runs through elements of $\FF_q$. The terms of each orbit
add to zero (mod $v^m$).
\end{proof}
\subsection{Trivial upper bound}
\begin{claim}
For $k>0$, $d\geq \deg(v)$,
we have
$v_d(k)\leq dk(q^d-q^{d-\deg(v)}-1)/\deg(v)$.
\end{claim}
\begin{proof} By definition $S_{d, v}(k)$ is the sum of $q^d-q^{\deg(v)}$ terms of
the form $1/n^k$ with $n$ monic and prime to $v$, so with the common denominator the product of $n^k$'s, the numerator is sum of products of
$q^d-q^{d-\deg(v)}-1$ terms of degree $dk$.
\end{proof}
\subsection{Congruences and periodicity} In contrast to $(\ZZ/p^n)^*$
which is cyclic for odd prime $p$, the analog $(A/v^n)^*$ is far from cyclic in
general, when $n>1$. If $v$ has degree $D$, then $(A/v^n)^*$ has order
$(q^D-1)q^{D(n-1)}$, but it has exponent
$$e_n=(q^D-1)p^{\lceil \log_p(n)\rceil}.$$ So
$S_{d, v}(k)\equiv \modd{S_{d, v}(k+me_n)} {v^n}$. In particular,
$$\mbox{if } v_d(k)k>0$. Using the congruence idea at the
infinite prime together with a switching trick, let us now give
another proof of this equality claimed, but using a slightly stronger
hypothesis that $s_d(-k)\neq 0$ and $q^n>k\ell(k)/(q-1)$.
Let us temporarily denote by $w$ the prime $1/t$ of $\FF_q[1/t]$.
We have
$$S_d(-k)=t^{dk}\sum (1+\frac{f_{d-1}}{t}+\cdots +\frac{f_0}{t^d})^k, \ \
S_d(q^n-k)=t^{dk-dq^n}\sum (1+\cdots )^{k-q^n}.$$
Now the first sum is (term-by-term) congruent modulo
$w^{q^n}$ to the second sum. Since the first sum is non-zero, we get
a trivial upper bound $dk$ on its valuation at $w$. On the other hand,
we know by \cite[5.6.2]{T} that $d\leq \ell(k)/(q-1)$, because $s_d(-k)\neq 0$.
Hence the claim follows as before.
\section{Recipes and inter-relations for degree $1$ primes,
$k\in\ZZ$}
Let us write $a\oplus b\oplus c+\cdots$ to denote the sum
$a+b+c+\cdots$, where this sum has no carry over base $p$.
In Theorem~\ref{thm1}
below, by the ``greedy algorithm'' we mean first choosing among
valid decompositions with $m_d\leq \cdots \leq m_1$, the least $m_d$,
then among these, the least $m_{d-1}$, and so forth.
\begin{theorem}
\begin{enumerate}
\item[(i)] Let $k$ be negative and $m=-k$. Then either
$s_d(k)=-dm+\min(m_1+2m_2+\cdots +dm_d)$, where
$m=m_0\oplus \cdots \oplus m_d$, $m_i\geq 0$, and for $i\geq 1$,
$(q-1)$ divides $m_i>0$, or $s_d(k)$ is infinite, if there is no
such decomposition. When the decompositions exist, the minimum
is uniquely given by greedy algorithm.
\item[(ii)] Let $k$ be positive. Then $s_d(k)=dk+\min(m_1+\cdots +dm_d)$,
with $(k-1)+m=(k-1)\oplus m$ and
$m=m_1\oplus\cdots\oplus m_d$, with $m_i$ positive and divisible by
$q-1$. The minimum is uniquely given by the greedy algorithm.
\item[(iii)] Let $k$ be negative and $m=-k$. Let $v$ be a prime of $A$ of
degree one. Then either $v_d(k)=\min(m_1+\cdots +dm_d)$, where
$m=m_0\oplus\cdots \oplus m_d$, where $(q-1)$ divides $m_i>0$ for
$0*0$ for $jk$).
\subsection{Remarks}
\begin{enumerate}
\item By part (iii) of Corollary~\ref{cor2}, the recursion in (iv) can
be continued, so that $v_2(k)$ determines $v_d(k)$, for $d\geq 2$,
for a prime $v$ of degree one.
\item All parts no longer hold if we drop the divisibility condition.
\item The second proof mentioned above
is achieved by developing the ideas
of Wan \cite{W} and Goss \cite[Prop.\ 9]{G2} further.
\end{enumerate}
\section{$v_d(k)$ when $q$ is a prime, $v$ is of degree $1$, and $k<0$}
\begin{theorem}
Let $q=p$ be a prime, $v$ a prime of $A$ of degree one and $-m=k<0$.
Write $m=\sum_{i=1}^{\ell}p^{e_i}$, with $e_i$ monotonically increasing and
with not more than $p-1$ of the consecutive values being the same
(i.e., consider the base $p$-digit expansion sequentially one digit at a time).
Also, let $r$ be the least non-negative residue of $m$ (mod $q-1$).
Then $v_d(k)$ is infinite if $\ell < (p-1)(d-1)$, and otherwise
$$v_d(k)=dr+\sum_{j=1}^{d-1}j\sum_{s=1}^{p-1}p^{e_{d-1-j(p-1)+s}}.$$
\label{thm3}
\end{theorem}
\begin{proof}
Note that $p^i \equiv \modd{1} {q-1}$ when $q$ is prime. Hence $p-1$ powers
together give divisibility by $q-1$. Hence the recipe in (iii) of
the Theorem~\ref{thm1} simplifies and
has for the minimum the choice
$m_d=r$, $m_{d-1}, \cdots, m_1$ obtained by picking $p-1$ digits
from the base $p$ expansion of $m$ starting from the lowest
digits (and dumping the rest of the expansion, if any, into $m_0$).
\end{proof}
Note that it looks even simpler for $q=2$, so that $r=0$ and inner
sums are singletons, and also that in this case $v_1(k)=0$, $v_2(k)$
is, in fact, the valuation of $k$ at $2$.
\section{$v_d(k)$ when $q=2$, $v$ is of degree $1$, and $k>0$}
First note that $v_d(2k)=2v_d(k)$, so we will focus on the case where $k$
is odd.
Next, observe that $v_1(k)=0$ for all $k$ and $v_2(k)=1$ for all odd $k$.
From \cite[Sec.\ A.2 (1)]{T09} and duality part (i) of Corollary~\ref{cor2}, we see that
$$w_n: =v_d(2^n-1)=2^{d+n-1}-2^nd+(d-1).$$
We define sequence $f_n$ (it is $v_{n-3}(5)-v_{n-3}(1)$) by
$$f_0=0 \mbox{ and } f_n=2f_{n-1}+4n.$$
Here is the recipe: For a given $d$, we describe the sequence of
$v_d(k)$ with $k$ odd, so that the $n$-th entry will correspond to
$k=2n-1$ and thus $w_n$ is $2^{n-1}$-th entry. Let $X_n$ be the vector
of the first $2^{n-1}-1$ entries, and write $X_{n+1}=X_n, w_n, X_n'$,
in two halves.
In other words, the whole sequence is of the form $X_1, w_1, X_1',
w_2, X_2', w_3, X_3',\cdots$ or equivalently,
$$X_n, w_n, X_n', w_{n+1}, \cdots, $$
\begin{theorem} The second half
$X_n'$ is obtained from the first half $X_n$ by adding $2^{n-2}f_m$ to the
entries with index (i.e., $(k+1)/2$) having the base-$2$ expansion
$$\sum_{i=w}^{n-2}b_i2^i, \ b_w\neq 0, \mbox{with exactly $d-3-m$ of
the $b_i$'s zero},$$
where $1\leq m\leq d-3$.
\label{thm4}
\end{theorem}
\begin{proof}
We replace $d$ by $d+1$ for convenience. We then need to prove
$v_{d+1}(k+2^n)-v_{d+1}(k)=2^{n-2}f_m$.
By using the duality (i) of Theorem~\ref{thm3}, we are reduced to
proving $s_d(k+2^n)-s_d(k)=d2^n+2^{n-2}f_m$, for $2^n-1>
k$ odd and $m$ as in Theorem~\ref{thm4}, but with $d$ replaced by $d+1$.
By \cite[3.3]{T09}, we have
$$s_d(k)-dk=d \cdot 2^{e_0}+\cdots +1 \cdot 2^{e_{d-1}},$$
where we write the base $2$ expansion of $k$ as
$$k=\cdots 0_{e_{t+1}}0_{e_t}1\cdots 10_{e_{t-1}}1\cdots1\cdots 0_{e_2}
1\cdots 10_{e_1}1\cdots 1_{e_0}0\cdots 0.$$
Since, for us, $k$ is odd, less than $2^n$, (with $n=e_t+r$, $r\geq 0$)
$k+2^n$ has expansion of the form
$$\cdots 0_{e_{t+r}} 1 0_{e_{t+r-1}}\cdots 0_{e_t}1\cdots
10_{e_{t-1}}1\cdots1\cdots 0_{e_2}
1\cdots 10_{e_1}1\cdots 1_{e_0}.$$ Observe that
$(d+1)-3-m=t+r-2$, by counting the relevant
zeros in the expansion of $(k+1)/2$. If $m\leq 0$, the relevant
$e_i$'s are the same for $k$ and for $k+2^n$, and hence the left side
of the first formula is $d2^n$, which agrees with the right side
as $f_m=0$ then. Now we proceed by an induction on $m$. Now the
$e_i$'s which are different are $e_{d-m}, \cdots, e_{d-1}$ which
are $n, \cdots , n+(m-1)$ for $k$ whereas $n+1, \cdots , n+m$ for
$k+2^n$ resulting in the difference $X_m:=(2^{n+m}+2 \cdot 2^{n+m-1}+\cdots
+m2^{n+1})-(2^{n+m-1}+\cdots +m2^n)$. Hence
$X_{m+1}-2X_m=(m+1)\cdot 2^n=4 \cdot 2^{n-2}(m+1)$, matching the recursion for
$2^{n-2}f_m$'s.
\end{proof}
\begin{remark}
Using duality, we have converted
the recursion in $d$ for valuations
at infinity into proving nice fast `doubling' pattern for a fixed $d$.
\end{remark}
\subsection{Examples} For $d\leq 3$, $X_n'=X_n$, so there is block
repetition after new entries $w_n$. The case $d=0$ (resp., $d=1$)
corresponds to vectors with all entries zero (resp, one), as
mentioned above. On the other hand, for $d=3$, we get
$$\overline{4}, \overline{6}, 4, \overline{10}, 4, 6, 4, \overline{18},
4, 6, \cdots, 4, \overline{34}, 4, 6, 4, \cdots, $$
where the over-lined entries are the $w_n$'s.
For $d=4$, the quantity
$X_n'$ is obtained by keeping its first half the same and
adding $2^n=2^{n-2} \cdot 4$ to the half-way entry,
half of the next half-way, etc.,
entries (i.e., $\sum_{w}^{n-2}2^i$-th entries).
More generally, $X_n$ and $X_n'$ share their first $X_{n-(d-3)}$ part.
This also follows another way from 3.3.
\section{When $q=3$, $v$ is of degree $1$, and $k>0$}
We list valuation sequence $v_d(k)$ for $k$ not divisible by $3$.
For $d=1$, $v_d(k)$ is zero for even $k$ and $1$ for odd $k$.
So the sequence is periodic of period $1, 0, 0, 1$ of length $4$, when only $k$ not divisible by $3$ are used.
\subsection{Conjectural recipe for $d=2$}
The valuation sequence is of the form $X_1, a_1, X_2, a_2, X_3, a_3,
\cdots$ where
(i) $X_1=X_{3n}=X_{3n+1}=[6, 4, 2, 14, 12, 10, 2, 8, 6, 4, 2]$,
(ii) $X_{3n+2}$ is the same as $X_1$ except $a_{3n+1}-a_{3n+2}$ is added to
fourth, fifth and sixth terms of $X_1$ to get the corresponding entries,
(iii) The $a_n$'s (which correspond to $k=17+18(n-1)$) look like
$3^r+3^s+2$, with $r\geq s\geq 2$ (and $a_{3m+2}=20$, so that $r=s=2$
and for $n$ of the form $3m+1$, we have $s=2$). More precisely, we
describe the full sequence $[a_1, a_2, \cdots]$ as follows:
$a_{3^n}=3^{n+3}+3^{n+2}+2$, $a_{2 \cdot 3^n}=3^{n+2}+3^{n+2}+2$, the block
between $a_{2 \cdot 3^n+1}$ to $a_{3^{n+1}-1}$ is exactly the block between
$a_1$ to $a_{3^n-1}$, whereas you get the block between $a_{3^n+1}$ to
$a_{2 \cdot 3^n-1}$ by taking the block between $a_1$ to $a_{3^n-1}$ and
replacing the entries $3^{n+2}+3^k+2$ by the new entries
$3^{n+3}+3^k+2$.
So the sequence is $X_1$ followed by $38$ followed by 6, 4, 2, 32, 30, 28,
2, 8, 6, 4, 2, followed by 20, $X_1$, 110, $X_1$, 92, followed by
6, 4, 2, 86, 84, 82, 2, 8, 6, 4, 2, followed by 20, $X_1$, 56, $X_1$, 38,
....
\section{When $q=4$, $v$ is of degree $1$, and $k>0$}
Here is the {\it conjectural recipe} for $d=1$:
We describe the sequence $v_1(k)$ with $k$ odd,
so that the $n$-th entry will correspond to $k=2n-1$ and thus $w_n$ is
$2^{n-1}$-th entry:
The whole sequence is limit of vectors $X_n$ (of increasing sizes with
initial portion being $X_{n-1}$) with
(i) $X_1$ being $[2, 0, 1, 8, 0, 1]$ and
(ii) $X_n$ consisting of $X_{n-1}$ followed by $X_{n-1}'$, where the
entries of $X_{n-1}'$ are the same as that of $X_{n-1}$, except one entry is
changed as follows: If $n$ is odd, change the $k=2^n-1$-th entry (which is
$2^n$) to $2^{n+2}$ and if $n$ is even, change $k=2^{n+1}-1$-th entry
(which is $2^{n+1}$) to $2^n+1$.
So the sequence is $X_1$ followed by 2, 0, 1, 5, 0, 1, 2, 0, 1, 32, 0, 1,
2, 0, 1, 5, 0, 1, $X_1$, ....
\section{When $q=2$, $v$ is of degree $2$, and $k>0$}
For a given $d$, we describe the sequence of $v_d(k)$ with $k$ odd,
so that the $n$-th entry will correspond to $k=2n-1$.
By definition, it is easy to see that $v_1(k)=1$ or $0$, respectively,
and $v_2(k)=0$ or $1$, respectively, according to whether $k$ is divisible by
$3$ or not. Let us check this for $d=1$: the numerator of
$1/t^k+1/(t+1)^k$ is congruent to
$t^k+(t+1)^k\equiv t^k+(t^2)^k\equiv t^k(1+t^k)$
modulo $v$, but $t^{3m}\equiv 1$ and $(t^{3m}+1)/(v(t+1))\equiv
t^{3m-3}+t^{3m-6}+\cdots +1\equiv m\not\equiv 0$ (mod $v$) (as $m$ is
odd).
Now we fix $d\geq 3$, so it will be dropped from the notation sometimes.
We write the sequence in the form $X_1, w_1, X_1', w_2, X_2', w_3,
X_3',\cdots$ where $X_{n+1}=X_n, w_n, X_n'$. In other words,
For every $n$, we write the sequence in the form
$$X_n, w_n, X_n', w_{n+1}, \cdots, $$ where
$X_n$ and $X_n'$ are vectors of entries of length $3 \cdot 2^{n-1}-1$, with
$X_n$ containing entries (odd $k$'s) from $v_d(1)$ to $v_d(3 \cdot 2^n-3)$
and $w_n=v_d(3 \cdot 2^n-1)$. We can further subdivide $X_n$ in `thirds'.
More precisely, $X_n=(A_n, B_n, C_n)$ with $A_n$ consisting of first
$2^{n-1}$ entries, namely for (odd) $k=1$ to $k=2^n-1$, i.e. those
with at most $n$ (base $2$) digits, with $B_n$ consisting of entries
with $k$ from $2^n+1$ to $2^{n+1}-1$, i.e., with $n+1$ digits and with
$C_n$ consisting of entries with $k$ from $2^{n+1}+1$ to $3 \cdot 2^n-3$,
i.e. with those entries in $X_n$ with $k$ of $n+2$ digits.
We write $X_n'=(A_n', B_n', C_n')$ in the obvious fashion.
Then as we have already proved in Section~\ref{sec3}, we have
\medskip
\noindent (i) {\bf Initial value} $X_1=[2^{d-1}-\lfloor (d+1)/2\rfloor, 2^d+(-1)^{d-1}-3\lfloor d/2\rfloor]$,
\medskip
Here is the {\it conjectural recipe} for the rest:
\medskip
\noindent (ii) {\bf The sequence $w_n=w_{n, d}$}
For $d$ odd, put
$$w_1=2^d-(5d-1)/2, \ \ w_{n+1}=2w_n-(d-1)/2,$$
so that $w_{n, 1}=0$, put $w_{n, 2}=1$, and for $d>2$ even, put
$$w_1=2^d+6-5d/2, \ \ w_{2n}=2w_{2n-1}-d/2+5, \ \ \
w_{2n+1}=2w_{2n}-4-d/2. $$
We now give conjectural description of the value of
$$t_k:=
t_{k, n, d}:=v_d(k+3 \cdot 2^n)-v_d(k)$$ depending on whether
$k$ belongs to $A_n, B_n, C_n$, respectively:
\medskip
\noindent (iii) {\bf Description of $A_n$ to $A_n'$ transition}:
$t_k=v_{d-i}(2^{n+2}-1)-v_{d-i}(2^n-1)$ when $k$ has
$n-i$ ones, with $0\leq i\leq d-3$, otherwise $t_k=0$.
In other words, add $v_{d-i}(2^{n+2}-1)-v_{d-i}(2^n-1)$ (which is
$3 \cdot 2^n(2^{d-i-2}-\lfloor (d-i)/2\rfloor)$ by conjecture above) to the
entry in $A_n$ with $k$ having $n-i$ ones, with $0\leq i\leq d-3$, to
get the corresponding entry in $A_n$. All other entries remain
unchanged.
\medskip
\noindent (iv) {\bf Description of $B_n$ to $B_n'$ transition}:
Let us temporarily write $f(n, d):=v_d(5 \cdot 2^n-1)-v_d(2^{n+1}-1)$.
Then $t_{k, n, d}=-f(n, d-i-1)$ and $0$, respectively,
if the base $2$ expansion of $k$ has $n-i$ number of ones (there
is exactly one term with $i=-1$, otherwise $i\geq 0$),
with $i< d-4$ and $i>d-4$, respectively. The special case $i=d-4$
has $t_k<0$ and is described later.
\medskip
\noindent (v) {\bf Description of $C_n$ to $C_n'$ transition}:
Put $g(n, i):=w_{n+1, i}-w_{n, i}$,
when $i\geq 0$ and $0$ otherwise, and
$r(n, i):=(-1)^{n+i-1} \cdot 3 \cdot 2^{n-i}$, when $1\leq i\leq n-1$.
Note $g(n, i)=0$ for $i\leq 3$.
Let $k$ have $n-m$ number of ones in its base-$2$ expansion, so that
$0\leq m\leq n-2$. If $m\geq d-4$, then $t_k=0$. So fix $m>d-4$.
If $d$ and $m$ have the same parity, then $t_k=g(n, d-(m+1))$.
If $d$ and $m$ have opposite parity, list such $k$'s
(there are ${n-1\choose n-m-2}={n-1\choose m+1}$ of these,
as out of the $n+2$ digits,
first two are $1, 0$ and the last is $1$) in increasing
order and add to corresponding entries the amounts
$$g(n, d-(m+1))+r(n, i_1)+r(n, i_2)+\cdots +r(n, i_{m+1}),$$
with $1\leq i_1<\cdots 3$ odd, while for $d=3$, it is $2^n$ for
$n$ odd and $2^n-1$ for $n$ even. Note $f(n, 5)=0$, so that
by (iv) more entries are unchanged than listed in the part 2 of these Remarks.
\end{enumerate}
\noindent {\bf Special case (iv), $i=d-4$}:
Since $k$ is odd with $n+1$ digits, we have $3\leq d\leq n+2$ and
there are ${n-1\choose d-3}$ such entries in $B_n$. Let
$a_n$ denote $2^n-1$, if $n$ is even and $2^n$, if $n$ is odd.
For $d=3$ (resp., $d=n+2$), we have a unique such $k$ and we
have $t_k=-a_n$ (resp., $t_k=-2^n$).
For $d=4$, the $n-1$
differences $-t_k-a_n$ are $(-1)^{i-1}(2^{n-i}-1)$ with increasing $1\leq i\leq
n-1$, if $n$ is even and they are $2^{n-1}+1, -2^{n-2}, \cdots, 2^2-1,
-2$ if $n$ is odd.
For $d=n+1$, the differences $-t_k-a_n$ are $0, 2^3, -(2^2-1), 2^5,
\cdots, 2^{n-1}, -(2^{n-2}-1)$ if $n$ is even and $2^2+1, -2, \cdots,
2^{n-1}+1, -2^{n-2}$ if $n$ is odd.
For $d=5$, $-t_k-a_n$ are given as follows. Write these differences in
$d=4$ case described above as $c_{n-1}, \cdots, c_1$, then in $d=5$
case, the first $n-2$ differences are (for $k$'s in the special case
written in increasing order) $0, c_{n-1}+c_{n-3},
c_{n-1}+c_{n-2}+c_{n-3}+c_1, c_{n-1}+c_{n-5},
c_{n-1}+c_{n-4}+c_{n-5}+c_1, \cdots$, (thus ending with
$c_{n-1}+c_4+c_3+c_1, c_{n-1}+c_1$ when $n$ is even and with
$c_{n-1}+c_2, c_{n-1}+c_3+c_2+c_1$ when $n$ is odd. These are followed
by $n-3$ copies of $-a_{n-2}$, followed by the differences at
$(n-2, d)$ level (there are ${n-3\choose 2}$ of them.)
For $d\geq 3$, for $k$ written in increasing order, for the
${n-1\choose d-3}$ $k$'s that we consider at the $(n, d)$-level,
the differences $t_{k, n, d}-a_n$ are given by repeating these differences
(${n-3\choose d-5}$ of them) at the $(n-2, d-2)$ level, followed by a
portion denoted by $Q$, as we have not been able to guess it yet. It consists
of ${n-3\choose d-4}$ entries), followed by $-a_{n-2}$ repeated
${n-3\choose d-4}$ times, followed by the differences at $(n-2, d)$
level (${n-3\choose d-3}$ of them.
The portion $Q$ is
$2^{n-1}+(-1)^{n-1}$ for $d=4$. For $d=5$, it is sum of first and third
differences at $(n, 4)$ level, followed by $3 \cdot 2^{n-3}$+ top
entries ($n-4$ of them) from $(n-2, 5)$ level.
The first term of the portion $Q$ in the
$d=6$ case is $2^{n-1}+(-1)^{n-1}$. For $d=n+1$, the portion $Q$ is
$2^{n-1}+1$ and $2^{n-1}$, according to whether
$n$ is odd or even, respectively.
More generally, the first (resp., last) entry of the portion
$Q$ is $2^{n-1}+(-1)^{n-1}$ (resp., $2^{n-1}$) according to whether
$d$ is even (resp., $d$ is odd and $n$ is even).
\medskip
\noindent {\bf Special cases of low $d$}:
We consider the $v_d(k)$ sequence for $k$ odd.
($d=1$) The pattern is 0, 1, 0, 0, 1, 0, ..., periodic with period $3$.
($d=2$) The pattern is 1, 0, 1, 1, 0, 1, ... , periodic with period $3$.
($d=3$) $w_n=1$ for $d=3$,
$X_n'$ is obtained from $X_n$ by adding $3 \cdot 2^n$ to the
$2^{n-1}$-th entry (i.e., $k=2^n-1$) and subtracting from $2^n$-th
entry (i.e., $k=2^{n+1}-1$) either $2^n$ or $2^n-1$, respectively, as
$n$ is odd or even.
So the pattern is 2, 6, 1, 8, 4, 1, 2, 18, 1, 5, 4, 1, 2,
6, 1, 32, 4, 1, 2, 10, $\ldots$ with $k$ being 5, 11, 1, 9 modulo 12 giving
the entries 1, 1, 2, 4, respectively.
\section{Zero distribution of Goss zeta functions}
For a given $y$, $\zeta(x, y)$ is a power series in $x$ with coefficients
in $K_{\infty}$. Wan \cite{W} noticed and proved using a Newton polygon
calculation, using estimates of $s_d(-y)$, that the zeros of $\zeta(x,y)$
are simple and
always lie in $K_{\infty}$,
when $q$ is a prime. This was later generalized by Sheats to any $q$. See
\cite[Sec.\ 5.8]{T}
and \cite{G, G2}
for the references for this development and discussion of higher genus case.
Noting that
$K_{\infty}$ and $C_{\infty}$ are, respectively, the
analogues of the real and complex number fields,
this restrictive behavior reminds one of the Riemann hypothesis
situation \cite{G, G2, T}. This looks even more remarkable, if one notices
further that algebraic degree of $C$ over $R$ is just 2, whereas
in our case, it is infinity.
We now turn to the $v$-adic case and the zero distribution for
the power series $\zeta_v(x, y)$ in $x$ with coefficients
in $K_v$, for a given $y$.
We can ask whether the zeros of $\zeta_v(x, y)$ are
in $K_v$ and whether they are simple.
Our results can be used to calculate Newton polygons
and the zero distribution. For this we can approach the $p$-adic integers $y$
through the sequence of positive $k$'s,
or through negative $k$, or indeed through
any dense subsequence (See \cite[5.8]{T} and \cite{T09}).
We leave this for a future paper and note here only two simple
applications requiring only a few things we have proved.
\medskip
\noindent (I) When degree of $v$ is one, and $q=2$, Wan \cite{W}
(see also \cite[Prop.\ 9]{G2}) already showed
that the zeros are simple and in $K_v$. For general
$q$, one has the same results for $y\in (q-1)S_v$.
(Note that when $q=2$, non-zero is the same as monic, otherwise we
restrict to `even' $k$ to kill the signs.)
This can be also derived immediately from Corollary~\ref{cor2}, which in fact
provides much more precise information to calculate the Newton polygon.
We remark that, when $q=2$, the one unit part of $S_d(k)$ when you substitute $1/t$ for $t$
is $S_{\leq d}(k)$ with $t$ factor removed. So $s_d(k)-k\geq \mbox{ min }
v_i(k)$ ($i\leq d$) with equality if there are no clashes. (But there can be
clashes, even for $d=1$, $k$ odd.)
To do the general $q$ case, without restriction of `evenness', for degree
one primes, we need to use
information provided by Theorems~\ref{thm1} and \ref{thm3}.
We leave this for a future paper.
\noindent (II) We now show that when $q=2$, $v=t^2+t+1$, the
zeros need not be in $K_v$.
For the first example, let $k=5$.
In fact, we saw that
$v_d(5)=0, 0, 1, 1, 12, 20, \ldots$. We do not need the full pattern.
We can say that the first
two slopes are $0$ and $1/2$ because degree of $v$ being
$2$, we have easy estimate $v_d(k)\geq m$, if $d>2m$, by the trivial
lower bound described above. (In fact, since the valuations are
not zero infinitely often, the slope would be at most $1/2$ in any case.)
Here is a second example: $q=2$, $v=t^2+t+1$, and $k=3$.
Now $v_d(3)$ for $d=0, 1, 2, \cdots$ is $0, 1, 0, 6, 9, 27, \ldots$,
so that the first two
slopes are $0$ and $9/2$. To see this, trivial lower bounds
above are not sufficient, but in this case, we know $v_d(3)$, by
Section~\ref{sec3},
so a straightforward calculation justifies this.
\section{Leading term formulas}
It follows \cite[Cor. 5.6.2]{T} from Carlitz' work that for $k>0$,
$S_d(-k)=0$ if
and only if $d>\ell(k)/(q-1)$, for $q$ prime.
(See \cite{B12} and \cite[A.5]{T09} for the
general situation). Hence, for $k>0$, $S_{\lfloor \ell(k)/(q-1)\rfloor}(-k)$
is the leading term of the Goss zeta series, at least when $q$
is prime, and also for general $q$, when $\ell(k)$ is the minimum of
$\ell(p^ik)$.
\begin{theorem}
Let $q$ be any prime power. Let $k>0$
and $\ell(k) =(q-1)d+r$, with $0\leq r<(q-1)$, so that
$d=\lfloor \ell(k)/(q-1)\rfloor$. Write
the base $q$-expansion $k=\sum_1^{d(q-1)+r} q^{k_i}$. Then
$$S_d(-k)=(-1)^d\sum t^{\sum_{i=1}^{d-1} i\sum_1^{q-1} q^{k_j}+d\sum_1^r q^{k_m}},$$
where the sum is over all assignments to $i$'s of groups of $q-1$
of the powers $q^{k_j}$'s corresponding to indices in
partitions of $d(q-1)+r$ indices
into $d$ groups of $q-1$ each and one group of $r$ powers.
\label{thm5}
\end{theorem}
\begin{proof}
Let us first see the simplest $q=2$ case. Then $r=0$ and we have
\begin{eqnarray*}S_d(-k)& = & \sum_{f_0, \ldots, f_{d-1}\in \FF_q}(t^d+f_{d-1}t^{d-1}
+\cdots +f_0)^k\\
& = & \sum \prod_{i=1}^{d(q-1)}(t^{dq^{k_i}}+f_{d-1}t^{(d-1)q^{k_i}}+\cdots +f_0)
\end{eqnarray*}
Keep the sum and consider the terms obtained by expanding the product.
Any term not containing all $f_i$'s will vanish after summing over
that missing $f_i$ (compare proof of \cite[Thm. 5.1.2]{T}). So terms
that matter are of the form $f_0\cdots f_{d-1}t^{\sum_1^{d-1}
iq^{k_{j_i}}}$, where there is really only one non-zero term corresponding
to $f_i=1$, so that $S_d(-k)$ is exactly the sum of these $t$ powers,
over $j_i$'s which are permutations of $1$ to $d$.
Now consider the general $q$ case.
Then, as before,
we have exactly the same displayed expression, and as before, when we
expand the product, all $f_i$'s need to be there and each $f_i$ with minimal
power $q-1$ to get the non-zero sum, so only terms that matter
have coefficient $(f_0\cdots f_{d-1})^{q-1}$ (as the relation between
$d$ and $\ell(k)$ shows) so that we get the expression as claimed.
\end{proof}
\subsection{Example} Let $q=3, k=38=27+9+1+1$, so that $d=2$ and our
our formula gives $S_d(-k)=t^{27+9}+2 \cdot t^{27+1}+2 \cdot t^{9+1}+t^2$.
\begin{corollary}
With the notation as in Theorem~\ref{thm5},
when $q=2$, we have a product formula
$$S_d(-k)=\prod_{d\geq n>m}(t^{2^{k_n}}+t^{2^{k_m}}).$$
More generally, for any $q$, but for the special family
$k=(q-1)\sum_1^d q^{k_i}>0$ (with $k_i$ distinct) we have the leading term
$$S_d(-k)=(-1)^d\prod_{d\geq n>m} (t^{q^{k_n}}-t^{q^{k_m}})^{q-1}.$$
\label{cor6}
\end{corollary}
\begin{proof}
Put $T_n=t^{q^{k_n}}$. The product formula follows immediately,
when $q=2$, by simple counting of monomials in
$\prod (T_n+T_m)$. For general $q$, one has to only note, in addition, that
${q-1 \choose i} = (-1)^i$, so that $(T_n-T_m)^{q-1}=\sum T_n^aT_m^b$, where
the sum is over $a, b$ with $a+b=q-1$.
\end{proof}
\subsection{Remarks}
\begin{enumerate}
\item The product formula
in the $q=2$ case was obtained earlier by
Pink using a cohomological formula for the leading power sum.
See \cite[7.1]{B12} for this, as well as the
proof of the Corollary using the Vandermonde determinantal formula
combined with cohomological machinery.
\item When $q>2$, we do not have a product formula involving
only monomials in $[n]$'s, in the general
case, for the leading term, even if $q$ is a prime. For example,
when $q=3$, $k=13$, $S_1(-13)=-(t^3-t)(t^3-t+1)(t^3-t-1)$. On the
other hand, for many families of $q, k$,
we can prove the product expression (for the leading term $S_d(-k)$
as above)
$$c\prod (t^{q^j}-t^{q^i})^{r_{i, j}},$$
where $c\in \FF_q$ expressed in terms of multinomial coefficient,
product being over $i0$, where $k=\sum k_jq^j$ is the base $q$ expansion of $k$.
We leave it to a future paper to investigate
the exact scope of when it works, and the cohomological explanation of
the prime factors which enter in terms of the $p$-ranks of the
Jacobians (components) of the corresponding cyclotomic extensions.
\item When $q=2$, the product has $d(d-1)/2$ terms of two terms. When
expanded, it has $2^{(d-1)d/2}$ terms, and the sum has $d!$ terms
(some can cancel), whereas if we just use the definition there are
$2^d$ terms to be added each consisting of $(d+1)^k$ (or rather
$(d+1)^{\ell(k)}$ using $p-$th powers) terms.
\end{enumerate}
\section{Acknowledgments} I thank Alejandro Lara Rodriguez for
creating data (using SAGE) on which many of the guesses of this paper
are based. I also thank Gebhard B\" ockle and David Goss for
discussions on these issues and their encouragement.
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L. Carlitz, On certain functions connected with
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J. Diaz-Vargas, Riemann hypothesis for $F_q[t]$,
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D. Goss,
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J. Sheats,
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\end{thebibliography}
\bigskip
\hrule
\bigskip
\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11M38; Secondary 11M26, 11R58.
\noindent \emph{Keywords: }
zeros of the zeta function, valuation, power sum.
\bigskip
\hrule
\bigskip
\vspace*{+.1in}
\noindent
Received August 1 2012;
revised version received January 7 2013.
Published in {\it Journal of Integer Sequences}, March 2 2013.
\bigskip
\hrule
\bigskip
\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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