1 Introduction
The problem of representing integers as sums of squares has garnered extensive attention in mathematical research. Jacobi’s theorem on sums of two squares provides a fundamental result in this domain, stating that the number of ways
$r_2(n)$
to express a positive integer n as a sum of two squares is given by
$$ \begin{align*}r_2(n)=4\left(d_1(n)-d_3(n)\right),\end{align*} $$
where
$d_i(n)$
is the number of divisors of n congruent to i modulo
$4$
for
$i=1,3$
. Let the letters
$p,q$
represent primes congruent to
$1,-1$
modulo 4 respectively. An equivalent formulation of Jacobi’s theorem can be stated as follows. Let
$n\in \mathbb {N}=\{1,2,3\ldots \}$
have the factorization
$$ \begin{align*}n=2^a\prod_{i=1}^kp_i^{b_i}\prod_{j=1}^lq_j^{c_j}\end{align*} $$
for positive integers
$b_i,c_j$
for
$1\leq i\leq k,1\leq j\leq l$
. Then, as in [Reference Hardy and Wright4, Theorem 278],
$$ \begin{align*}r_2(n) = \left\{ \begin{array}{ll} 4\prod_{i=1}^k(b_i+1) & \mbox{if} \hspace{1mm} c_j \hspace{1mm} \mbox{is even for all} \hspace{1mm} j\leq l, \\ 0 & \mbox{otherwise}. \end{array} \right.\end{align*} $$
In [Reference Hardy and Wright4], Hardy and Littlewood present a formula for the Dirichlet series corresponding to
$r_2(n)$
in terms of
$\zeta (s)$
and
$L_{-4}(s)$
. Here
$\zeta (s)$
is the Riemann zeta function and
$L_{-4}(s)$
is the L-function corresponding to the Kronecker symbol
$\genfrac {(}{)}{}{}{-4}{n}$
. The formula states
$$ \begin{align*}\sum_{n=1}^\infty\frac{r_2(n)}{n^s}=4\zeta(s)L_{-4}(s).\end{align*} $$
The Gauss circle problem, which seeks to determine the number of integer lattice points inside a circle of radius r centered at the origin, is linked to the study of the first moment of
$r_2(n)$
. Sierpinski’s result [Reference Landau6, Satz 509] provides insight into the first moment of
$r_2(n)$
indicating that
$$ \begin{align*}\sum_{n\leq x}r_2(n)= \pi x+O(x^{1/3}).\end{align*} $$
While it is conjectured that the exponent in the error term could be
$1/4+\epsilon $
, this conjecture remains elusive. Borwein and Choi [Reference Borwein and Choi2] derived the following asymptotic expression for the second moment of
$r_2(n)$
.
$$ \begin{align*}\sum_{n \leq x} r_{2}^{2}(n) = 4x\log{x} +4\alpha x+O(x^{2/3}),\end{align*} $$
where
$\alpha =2.0166\ldots $
. Their approach hinges on a Dirichlet series representation for
$r_2^2(n)$
:
$$ \begin{align*}\sum_{n=1}^\infty\frac{r_2^2(n)}{n^s}=\frac{(4\zeta(s)L_{-4}(s))^2}{(1+2^{-s}\zeta(2s))}.\end{align*} $$
Connors and Keating [Reference Connors and Keating3] explore higher moments of
$r_2(n)$
to analyze accidental degeneracies in energy levels for square billiards, quantifying deviations from a Poisson distribution. In addition to its connections with the aforementioned issues, the exploration of moments in the representation theory of quadratic forms is inherently intriguing.
An integer can be expressed as a sum of two squares if and only if it is the norm of a Gaussian integer, meaning it is the norm of an element in the ring
$\mathbb {Z}[i]$
. A natural analog over
$\mathbb {F}_q[T]$
is to consider norms of the extension ring
$\mathbb {F}_q[\sqrt {-T}]$
. Let
$\mathcal {M}_{n,q}$
denote the set of monic polynomials of degree n with coefficients in
$\mathbb {F}_q$
. For
$f\in \mathcal {M}_{n,q}$
, define
$$ \begin{align*} b_q(f)= \begin{cases} 1, & f=\alpha^2+T\beta^2 \hspace{2mm}\text{for} \hspace{2mm}\alpha,\beta\in\mathbb{F}_q[T],\\ 0, & \text{otherwise. } \end{cases} \end{align*} $$
The counting function
$B_q(n)=\sum _{f\in \mathcal {M}_{n,q}}b_q(f)$
was examined by Bary–Soroker, Smilansky and Wolf in [Reference Bary-Soroker, Smilansky and Wolf1], drawing parallels to Landau’s theorem on sums of squares. For
$f\in \mathcal {M}_{n,q}$
, let
$$ \begin{align*}r_2^*(f)=\# \{(\alpha,\beta), \alpha, \beta \in \mathbb{F}_q[T] : f=\alpha^2+T\beta^2 \}.\end{align*} $$
In this article, we investigate the function
$r_2^*(f)$
and its moments for polynomial rings
$\mathbb {F}_q[T]$
, where q is a prime power. A key focus will be comparing these results with the known results for
$r_2(n).$
When
$\mathbb {F}_q$
has characteristic 2, the Frobenius map
$\phi :\mathbb {F}_q\rightarrow \mathbb {F}_q$
defines an isomorphism, making the study trivial. When
$\mathbb {F}_q$
has odd characteristic, the formula for
$r_2^*(f)$
is analogous to the one we get in the classical case. We obtain
Theorem 1.1 Let
$f=T^aP_1^{c_1}P_2^{c_2}...P_k^{c_k}Q_1^{2d_1}...Q_l^{2d_l}$
be the prime factorization of
$f\in \mathcal {M}_{n,q}$
such that
$\genfrac {(}{)}{}{}{P_i}{T}=1$
and
$\genfrac {(}{)}{}{}{Q_j}{T}=-1, \forall i,j.$
Then
$$ \begin{align*}r_2^*(f)= 2\prod_{i=1}^k(c_i+1).\end{align*} $$
The main difference between the classical and the function field analog seems to be a factor of 2 reflecting the number of units in the ring of integers. The ring
$\mathbb {Z}[i]$
has four units of norm 1, whereas
$\mathbb {F}_q[\sqrt {-T}]$
only has two units of norm 1.
Let
$\mathcal {M}$
be the set of monic polynomials, and let
$\zeta _q(s)$
be the zeta function of
$\mathbb {F}_q[T]$
. Analogous to the Dirichlet series in the classical case, we find that the Dirichlet series for
$r_2^*$
has the form
$$ \begin{align*}\sum_{f \in\mathcal{M}}\frac{r_2^*(f)}{|f|^{s}}=2\zeta_q(s).\end{align*} $$
The Dirichlet series for the square of
$r_2^*$
is computed as
$$ \begin{align*}\sum_{f \in\mathcal{M}}\frac{(r_2^*)^2(f)}{|f|^{s}}=\frac{4\zeta_q^2(s)}{\zeta_q(2s)(1+q^{-s})}.\end{align*} $$
See Theorems 3.1 and 3.2 for more details. We note that the L-series
$L_\chi (s)$
corresponding to the quadratic character
$\chi (f)=\genfrac {(}{)}{}{}{f}{T}$
is trivially equal to 1 (see Section 4. This simplifies the above formulae even further in contrast.
Employing the Dirichlet series described above, we derive the following formulae.
Theorem 1.2 Let
$\mathbb {F}_q$
be a field of odd characteristic. The first and the second moments for
$r_2^*(f)$
are given by
$$ \begin{align*}\sum_{f\in\mathcal{M}_{n,q}}r_2^*(f)=2q^n,\end{align*} $$
and
$$ \begin{align*}\sum_{f\in\mathcal{M}_{n,q}}(r_2^*)^2(f)=4(n+1)q^n+4\sum_{k=1}^{n-1}(-1)^k(2n-2k+1)q^{n-k}+4(-1)^n.\end{align*} $$
The absence of error terms in the formulae can be attributed to the fact that the L-function
$L_\chi (s)$
is constant and equal to 1. While the Dirichlet series approach proves effective for the first and second moments, it encounters limitations with higher moments.
For higher moments, we split the study into two scenarios. First, in the large degree limit, we fix q and obtain asymptotic formulas as n gets large. In this scenario, asymptotic formulae are obtained rather than exact formulae, as the Dirichlet series cannot be represented solely in terms of the zeta function
$\zeta _q(s)$
and the L-function
$L_\chi (s)$
. Utilizing techniques similar to those in [Reference Connors and Keating3], we investigate the Dirichlet series for the ith power of
$r_2^*$
, leading to the following results.
Theorem 1.3 Let
$\mathbb {F}_q$
be a field of odd characteristic. For a fixed q and
$i\in \mathbb {N}$
with
$i\geq 3$
, the ith moment for
$r_2^*$
is given by
$$ \begin{align*}\sum_{f\in\mathcal{M}_{n,q}}(r_2^*)^i(f)=-B_i(1)2^i\binom{n+2^{i-1}-1}{n}q^n+O(q^{n-1}),\end{align*} $$
where
$B_i(1)$
is a fixed constant depending on i and q obtained from the Dirichlet series for
$(r_2^*)^i$
. The error term has a dependence on i and q.
Explicitly describing the constant
$B_i(1)$
using the above methods is a non-trivial problem.
In the second scenario, we obtain asymptotic formulas in the large finite field limit, i.e., when n is fixed and q gets large. Inspired by the techniques in [Reference Bary-Soroker, Smilansky and Wolf1], we obtain the main result of this article, which provides a second-order term for the higher moments. We make a note that in the integer setting, such formulae are not known.
Theorem 1.4 Let
$\mathbb {F}_q$
be a field of odd characteristic and
$f\in \mathcal {M}_{n,q}$
. For
$i\in \mathbb {N}$
, the ith moment for
$r_2^*$
for a fixed n is given by
$$ \begin{align*} \sum_{f\in\mathcal{M}_{n,q}}(r_2^*)^i(f)=2^i\gamma_{i,n}q^n- \alpha_{i,n} q^{n-1}+O(q^{n-2}), \end{align*} $$
where
$$ \begin{align*}\alpha_{i,n}=\left(2^{2i-1}-2^i\right)\gamma_{i,n-1}+\left(2^{2i-1}+2^{3i-2}-2^{i-1}-2^{i-1}3^i\right)\gamma_{i,n-2}\end{align*} $$
and
$\gamma _{i,n}$
is the binomial coefficient
$$ \begin{align*}\gamma_{i,n}=\binom{n+2^{(i-1)}-1}{n}.\end{align*} $$
The error term has a dependence on i and n.
Specifically, for
$i=1,2$
, the first two terms of the result align with the first and second moments outlined in Theorem 1.2.
2 Number of representations in the form
$A^2+TB^2$
Fix an odd prime power q. Define the embedding
$\mathbb {F}_q[T]\subset \mathbb {F}_q[S],$
where
$S=\sqrt {-T}.$
There are two automorphisms of
$\mathbb {F}_q[S]$
that fix
$\mathbb {F}_q[T]$
defined by the actions
$S\rightarrow \pm S.$
The norm map
$$ \begin{align*}N:\mathbb{F}_q[S]\rightarrow\mathbb{F}_q[T]\end{align*} $$
is given by
$$ \begin{align*}N(A+SB)=(A+SB)(A-SB)=A^2+TB^2, A,B\in\mathbb{F}_q[T].\end{align*} $$
For
$f\in \mathcal {M}_{n,q}$
, we define the Legendre symbol
$\genfrac {(}{)}{}{}{f}{T}$
as follows.
$$ \begin{align*} \genfrac{(}{)}{}{}{f}{T} = \begin{cases} 1 & f \text{ is a square modulo } T,\hspace{1mm} T\nmid f, \\ -1 & f \text{ is a non-square modulo } T, \hspace{1mm}T\nmid f,\\ 0 & T\mid f. \end{cases} \end{align*} $$
Then, by [Reference Bary-Soroker, Smilansky and Wolf1, Theorem 2.5],
$b_q(f)=1$
if and only if every monic irreducible polynomial Q with
$\genfrac {(}{)}{}{}{Q}{T}=-1$
appears with even multiplicity in the prime factorization of f. Before we proceed any further, we reserve the letters P and Q for monic irreducibles in
$\mathbb {F}_q[T]$
with
$\genfrac {(}{)}{}{}{P}{T}=1$
and
$\genfrac {(}{)}{}{}{Q}{T}=-1$
respectively. It is clear that the irreducible element T is not in any of the categories we mentioned since
$\genfrac {(}{)}{}{}{T}{T}=0.$
We are now ready to prove Theorem 1.1.
Proof of Theorem 1.1
From the definitions, we see that
$b_q(f)=1$
if and only if
$f=N(h)$
for some
$h\in \mathbb {F}_q[S]$
. So the question is how many elements in
$\mathbb {F}_q[S]$
have norm f. Let us see what happens when f is a prime power.
We know from [Reference Bary-Soroker, Smilansky and Wolf1, Proposition 2.4] that
$\pm Q$
are the only primes of norm
$Q^2$
. We deduce that if
$f=Q^{2d}$
, where
$\genfrac {(}{)}{}{}{Q}{T}=-1$
, then
$\pm Q^d$
are the only elements in
$\mathbb {F}_q[S]$
with norm f.
If
$f=P^c$
, where
$\genfrac {(}{)}{}{}{P}{T}=1$
, then there are exactly
$2(c+1)$
elements with norm f. To see this, if
$N(q+rS)=P,$
then
$N(q-rS)=P$
as well and hence
$N(\pm (q+rS)^i(q-rS)^{c-i})=P^i$
for all
$0\leq i\leq c$
. The last thing we should check is the uniqueness of elements of norm P. If,
$$ \begin{align*}N(q+rS)=N(t+uS)=P\end{align*} $$
for
$q,r,t,u\in \mathbb {F}_q[T]$
. Then,
$$ \begin{align*}(q+rS)(q-rS)=(t+uS)(t-uS),\end{align*} $$
each of the factors being an irreducible polynomial in
$\mathbb {F}_q[S]$
(the irreducibilty comes from [Reference Bary-Soroker, Smilansky and Wolf1, Lemma 2.1]). Without loss of generality, assume that
$$ \begin{align*}q+rS=c(t+uS)\end{align*} $$
for some
$c\in \mathbb {F}_q.$
Taking the norms and observing that P is monic, we get that c has to be an element of order 2 and the only elements of order 2 in
$\mathbb {F}_q$
are
$\pm 1$
. Hence, the result is evident.
3 First and second moments of
$r_2^*$
using Dirichlet series
In this section, we will study the Dirichlet series to arrive at the first and second moments of
$r_2^*$
. For
$f=T^cP_1^{a_1}P_2^{a_2}...P_k^{a_k}Q_1^{2c_1}Q_2^{2c_2}...Q_r^{2c_r}$
, we have
$r_2^*(f)= 2\prod _{i=1}^k(a_i+1).$
Let
$\delta _2(f)=\frac {r_2^*(f)}{2}.$
Then
$\delta _{2}(f)$
is a multiplicative function. Let
$\mathcal {M}$
and
$\mathcal {P}$
be the set of monic polynomials and the set of monic irreducibles in
$\mathbb {F}_q[T]$
, respectively. We want to study the Dirichlet series
$$ \begin{align*}L_i(s)=\sum_{f \in\mathcal{M}}\delta_2^i(f)|f|^{-s}\end{align*} $$
for
$i=1,2,$
where
$|f|=q^{\mathrm {deg}(f)}.$
Since
$\delta _2(f)$
is multilplicative, we have a product of local factors
$$ \begin{align*}L_i(s)=\prod_{R \in\mathcal{P}}\left(\sum_{k=0}^\infty \delta_2^i(R^k)|R|^{-ks}\right),\end{align*} $$
where
$$ \begin{align*}\delta_2^i\left(R^k\right)= \begin{cases}(k+1)^i & \text{if}\hspace{2mm}\left(\frac{R}{T}\right)=1, \\ 1 & \text{if}\hspace{2mm}R=T, \\ 1 & \text{if}\hspace{2mm}\left(\frac{R}{T}\right)=-1 \text { and } 2 \mid k, \\ 0 & \text{if}\hspace{2mm}\left(\frac{R}{T}\right)=-1 \text { and } 2 \nmid k.\end{cases} \end{align*} $$
Let
$\chi $
be the quadratic character given by
$\chi (f)=\genfrac {(}{)}{}{}{f}{T}.$
The associated L series is
$$ \begin{align} L_{\chi}(s)=\sum_{f \in\mathcal{M}} \chi(f)|f|^{-s}=\prod_{\genfrac{(}{)}{}{}{P}{T}=1}\left(1-|P|^{-s}\right)^{-1} \prod_{\genfrac{(}{)}{}{}{Q}{T}=-1}\left(1+|Q|^{-s}\right)^{-1}. \end{align} $$
Since
$\chi $
is defined modulo a linear polynomial,
$L_\chi (s)=1.$
The zeta function in
$\mathbb {F}_{q}$
is given by
$$ \begin{align} \begin{aligned} \zeta_q(s) & =\sum_{f \in\mathcal{M}}|f|^{-s}=(1-q^{-s})^{-1}\prod_{\genfrac{(}{)}{}{}{P}{T}=1}\left(1-|P|^{-s}\right)^{-1} \prod_{\genfrac{(}{)}{}{}{Q}{T}=-1}\left(1-|Q|^{-s}\right)^{-1}\\ & = \frac{1}{1-q^{1-s}}. \end{aligned} \end{align} $$
See [Reference Rosen7, Chapter 2, Equation 1] for a proof.
Theorem 3.1 We have
$$ \begin{align*}L_1(s)=\zeta_q(s),\end{align*} $$
where
$\zeta _q(s)$
is the zeta function in
$\mathbb {F}_{q}[T]$
.
Proof From the Euler product and the description of
$\delta _2(P^k)$
provided above, we get
$$ \begin{align} \begin{aligned} L_1(s) & = \left(\sum_{k=0}^\infty|T|^{-ks}\right)\prod_{\genfrac{(}{)}{}{}{P}{T}=1}\left(\sum_{k=0}^\infty (k+1)|P|^{-ks}\right)\prod_{\genfrac{(}{)}{}{}{Q}{T}=-1}\left(\sum_{k=0}^\infty |Q|^{-2ks}\right)\\ & = (1-q^{-s})^{-1}\prod_{\genfrac{(}{)}{}{}{P}{T}=1}\left(1-|P|^{-s}\right)^{-2}\prod_{\genfrac{(}{)}{}{}{Q}{T}=-1}\left(1-|Q|^{-2s}\right)^{-1}. \end{aligned} \end{align} $$
Comparing (3.3), (3.1), and (3.2), we get
$$ \begin{align*}L_1(s)=L_\chi(s)\zeta_q(s)=\zeta_q(s).\end{align*} $$
Theorem 3.2 We have
$$ \begin{align*}L_2(s)=\frac{\zeta_q^2(s)}{\zeta_q(2s)(1+q^{-s})}.\end{align*} $$
Proof From the Euler product and the description of
$\delta _2(R^k)$
provided at the beginning of the section, we get
$$ \begin{align*} \begin{aligned} L_2(s) & = \left(\sum_{k=0}^\infty|T|^{-ks}\right)\prod_{\genfrac{(}{)}{}{}{P}{T}=1}\left(\sum_{k=0}^\infty (k+1)^2|P|^{-ks}\right)\prod_{\genfrac{(}{)}{}{}{Q}{T}=-1}\left(\sum_{k=0}^\infty |Q|^{-2ks}\right)\\ & = (1-q^{-s})^{-1}\prod_{\genfrac{(}{)}{}{}{P}{T}=1}\left(2\left(1-|P|^{-s}\right)^{-3}-\left(1-|P|^{-s}\right)^{-2}\right)\prod_{\genfrac{(}{)}{}{}{Q}{T}=-1}\left(1-|Q|^{-2s}\right)^{-1}\\ &= \zeta_q(s)L_{\chi}(s)\left[\prod_{\genfrac{(}{)}{}{}{P}{T}=1}\left(2\left(1-|P|^{-s}\right)^{-1}-1\right)\right]\\ &=\zeta_q(s)L_{\chi}(s)\prod_{\genfrac{(}{)}{}{}{P}{T}=1}\frac{(1+|P|^{-s})}{(1-|P|^{-s})}\\ &=\zeta_q(s)L_{\chi}(s)\prod_{\genfrac{(}{)}{}{}{P}{T}=1}\frac{(1-|P|^{-2s})}{(1-|P|^{-s})^2}. \end{aligned} \end{align*} $$
Here we used the fact that
$$ \begin{align*}\frac{2}{(1-x)^3}-\frac{1}{(1-x)^2}=\sum_{k=0}^\infty (k+1)^2x^{k}.\end{align*} $$
We now multiply and divide the equation by
$\prod _{\genfrac {(}{)}{}{}{Q}{T}=-1}(1-|Q|^{-2s})$
. We get
$$ \begin{align*} \begin{aligned} L_2(s) &=\zeta_q(s)L_{\chi}(s)\prod_{\genfrac{(}{)}{}{}{P}{T}=1}\frac{(1-|P|^{-2s})}{(1-|P|^{-s})^2}\prod_{\genfrac{(}{)}{}{}{Q}{T}=-1}\frac{(1-|Q|^{-2s})}{(1+|Q|^{-s})(1-|Q|^{-s})}. \end{aligned} \end{align*} $$
Using the fact that
$$ \begin{align*}\zeta_q(2s)=(1-q^{-2s})^{-1}\prod_{\genfrac{(}{)}{}{}{P}{T}=1}\left(1-|P|^{-2s}\right)^{-1} \prod_{\genfrac{(}{)}{}{}{Q}{T}=-1}\left(1-|Q|^{-2s}\right)^{-1},\end{align*} $$
we see that
$$ \begin{align*}L_2(s) =\frac{\zeta_q(s)L_{\chi}(s)}{\zeta_{q}(2s)(1-q^{-2s})}\prod_{\genfrac{(}{)}{}{}{P}{T}=1}(1-|P|^{-s})^{-2}\prod_{\genfrac{(}{)}{}{}{Q}{T}=-1}(1+|Q|^{-s})^{-1}(1-|Q|^{-s})^{-1}.\end{align*} $$
Further, using (3.1) and (3.2), we get
$$ \begin{align*}L_2(s)=\frac{\zeta_q^2(s)L^2_\chi(s)(1-q^{-s})}{\zeta_q(2s)(1-q^{-2s})}.\end{align*} $$
Making use of the fact that
$L_\chi (s)=1$
, we have the result.
Using the Dirichlet series
$L_i(s)$
for
$i=1,2$
, we prove Theorem 1.2.
Proof of Theorem 1.2
For the first moment, we have from Theorem 3.1 that
$$ \begin{align*}L_1(s)=\zeta_q(s).\end{align*} $$
Expanding to retrieve the coefficient of
$q^{-ns}$
, we get
$$ \begin{align*}L_1(s)=\sum_{n=1}^\infty\frac{q^n}{q^{-ns}}.\end{align*} $$
Hence we get that
$$ \begin{align*}\sum_{f\in\mathcal{M}_{n,q}}\delta_2(f)=q^n.\end{align*} $$
Now, since
$r_2^*(f)=2\delta _2(f)$
, we have the first moment. For the second moment, we have from Theorem 3.2 that
$$ \begin{align*}L_2(s)=\frac{\zeta_q^2(s)}{\zeta_q(2s)(1+q^{-s})}.\end{align*} $$
We know that
$$ \begin{align*}\zeta_q(s) = \frac{1}{1-q^{1-s}}= \sum_{n=0}^\infty \frac{q^n}{q^{ns}}.\end{align*} $$
Raising to the power of 2, we get
$$ \begin{align*}\zeta_q^2(s) = \left(\frac{1}{1-q^{1-s}}\right)^2= \sum_{n=0}^\infty \frac{(n+1)q^n}{q^{ns}}.\end{align*} $$
Lastly, we note that
$$ \begin{align*}\frac{1}{(1+q^{-s})}=\sum_{n=0}^\infty\frac{(-1)^n}{q^{ns}}.\end{align*} $$
Combining all of the above, we get
$$ \begin{align*}L_2(s)=\frac{\zeta_q^2(s)}{\zeta_q(2s)(1+q^{-s})}=\left(\sum_{n=0}^\infty\frac{(n+1)q^n}{q^{ns}}\right)\left(\sum_{n=0}^\infty\frac{(-1)^n}{q^{ns}}\right)\left(1-q^{1-2s}\right).\end{align*} $$
Extracting the coefficient of
$q^{-ns}$
from the right hand side and using the fact that
$r_2^*(f)=2\delta _2(f)$
, we have the second moment.
We make a note that the formulae obtained for the first and second moments in this section have no associated error and hence work for all values of q and n.
4 Higher moments of
$r_2^*$
in the large degree limit
In this section, we will first compute the L-series corresponding to
$\delta _2^i$
for
$i\geq 3$
. We then make some remarks about the higher moments of
$r_2^*$
from the aforementioned L functions. We use the approach taken by Connors and Keating in [Reference Connors and Keating3]. The L-series
$L_i(s)$
for
$\delta _2^i(f)$
is given by
$$ \begin{align*} \begin{aligned} L_i(s) & = \left(\sum_{k=0}^\infty|T|^{-ks}\right)\prod_{\genfrac{(}{)}{}{}{P}{T}=1}\left(\sum_{k=0}^\infty (k+1)^i|P|^{-ks}\right)\prod_{\genfrac{(}{)}{}{}{Q}{T}=-1}\left(\sum_{k=0}^\infty |Q|^{-2ks}\right)\\ & = (1-q^{-s})^{-1}\prod_{\genfrac{(}{)}{}{}{P}{T}=1}\left(\sum_{k=0}^\infty (k+1)^i|P|^{-ks}\right)\prod_{\genfrac{(}{)}{}{}{Q}{T}=-1}\left(1-|Q|^{-2s}\right)^{-1}. \end{aligned} \end{align*} $$
To further simplify this, we want to simplify the expression
$$ \begin{align*}\sum_{k=0}^\infty (k+1)^i|P|^{-ks}.\end{align*} $$
To this aim, we define
$$ \begin{align*}\gamma_i(x):=\sum_{k=0}^\infty (k+1)^ix^k.\end{align*} $$
Then,
$\gamma _i(x)$
satisfies the recurrence
$$ \begin{align} \frac{d}{dx}\left(x\gamma_{i-1}(x)\right)=\gamma_i(x). \end{align} $$
Hence, we can write
where
$a_i$
are integers and 
is a polynomial in x with least exponent term
$x^2$
. Before we move to estimating
$L_i(s)$
, we compute the leading coefficient
$a_i$
as it will prove to be useful. Substituting the form of
$\gamma _i(x)$
given in (4.2) in the recurrence relation (4.1), we get
The coefficient of x in the numerator is
$(2a_i+i)$
. Hence,
$a_i$
satisfies the recurrence
$a_{i+1}=2a_i+i.$
Using induction, we see that
$$ \begin{align*}a_i=2^i-i-1.\end{align*} $$
Coming back to our estimation of
$L_i(s)$
, we have
$$ \begin{align*} \begin{aligned} L_i(s) & = (1-q^{-s})^{-1}\prod_{\genfrac{(}{)}{}{}{P}{T}=1}\left(\sum_{k=0}^\infty (k+1)^i|P|^{-ks}\right)\prod_{\genfrac{(}{)}{}{}{Q}{T}=-1}\left(1-|Q|^{-2s}\right)^{-1}\\ & = (1-q^{-s})^{-1}\prod_{\genfrac{(}{)}{}{}{P}{T}=1}\left(1+a_i|P|^{-s}+b_i|P|^{-2s}+\ldots\right) \\ &\quad \times\left(1-|P|^{-s}\right)^{-(i+1)}\prod_{\genfrac{(}{)}{}{}{Q}{T}=-1}\left(1-|Q|^{-2s}\right)^{-1}. \end{aligned} \end{align*} $$
The product
$$ \begin{align*}\prod_{\genfrac{(}{)}{}{}{P}{T}=1}\left(1+a_i|P|^{-s}+b_i|P|^{-2s}+\ldots\right)\end{align*} $$
does not converge near
$s=1$
when
$a_i\neq 0$
. Hence, we define
$$ \begin{align*}A_i(s)=\prod_{\genfrac{(}{)}{}{}{P}{T}=1}\left(1+a_i|P|^{-s}+b_i|P|^{-2s}+\ldots\right)\left(1-|P|^{-s}\right)^{a_i}.\end{align*} $$
Note that by construction,
$$ \begin{align*}A_i(s)=\prod_{\genfrac{(}{)}{}{}{P}{T}=1}\left(1+c_i|P|^{-2s}+\ldots\right)\end{align*} $$
for some
$c_i$
and hence
$A_i(s)$
converges at
$s=1$
since it does not have any linear term. Therefore, we have
$$ \begin{align} \begin{aligned} L_i(s) & = (1-q^{-s})^{-1}\prod_{\genfrac{(}{)}{}{}{P}{T}=1}(1-|P|^{-s})^{-(i+1+a_i)}\prod_{\genfrac{(}{)}{}{}{Q}{T}=-1}\left(1-|Q|^{-2s}\right)^{-1}A_i(s)\\ & = (1-q^{-s})^{-1}\prod_{\genfrac{(}{)}{}{}{P}{T}=1}(1-|P|^{-s})^{-2^i}\prod_{\genfrac{(}{)}{}{}{Q}{T}=-1}\left(1-|Q|^{-2s}\right)^{-1}A_i(s). \end{aligned} \end{align} $$
Now,
$$ \begin{align*} \zeta_q(s)=\zeta_q(s)L_\chi(s)=(1-q^{-s})^{-1}\prod_{\genfrac{(}{)}{}{}{P}{T}=1}(1-|P|^{-s})^{-2}\prod_{\genfrac{(}{)}{}{}{Q}{T}=-1}\left(1-|Q|^{-2s}\right)^{-1}. \end{align*} $$
Hence
$$ \begin{align*} \zeta_q^{2^{i-1}}(s)=(1-q^{-s})^{-2^{i-1}}\prod_{\genfrac{(}{)}{}{}{P}{T}=1}(1-|P|^{-s})^{-2^i}\prod_{\genfrac{(}{)}{}{}{Q}{T}=-1}\left(1-|Q|^{-2s}\right)^{-2^{i-1}}. \end{align*} $$
Using this in (4.3), we get
$$ \begin{align} \begin{aligned} L_i(s) & = \zeta_q^{2^{i-1}}(s)(1-q^{-s})^{2^{i-1}-1}\prod_{\genfrac{(}{)}{}{}{Q}{T}=-1}\left(1-|Q|^{-2s}\right)^{2^{i-1}-1}A_i(s)\\ & = \zeta_q^{2^{i-1}}(s)B_i(s), \end{aligned} \end{align} $$
where
$$ \begin{align*}B_i(s)=(1-q^{-s})^{2^{i-1}-1}\prod_{\genfrac{(}{)}{}{}{Q}{T}=-1}\left(1-|Q|^{-2s}\right)^{2^{i-1}-1}A_i(s).\end{align*} $$
Note that
$B_i(s)$
is non-singular for
$\mathrm {Re}(s)\geq 1.$
Now, we will use the L-series
$L_i(s)$
to prove Theorem 1.3.
Proof of Theorem 1.3
We note from (4.4) that
$$ \begin{align*} \begin{aligned} L_i(s) & =\zeta_q^{2^{i-1}}(s)B_i(s)\\ & = \frac{B_i(s)}{(1-qu)^{2^{i-1}}}, \end{aligned} \end{align*} $$
where
$u=q^{-s}.$
Abusing the use of notations, we will write
$\mathcal {L}_i(u)$
and
$\mathcal {B}_i(u)$
in place of
$L_i(s)$
and
$B_i(s)$
. Now, by Perron’s formula
$$ \begin{align*}\sum_{f\in\mathcal{M}_{n,q}}\delta_2^i(f)=\frac{1}{2\pi i}\oint_{|u|=q^{-1-\epsilon}}\frac{\mathcal{B}_i(u)}{(1-qu)^{2^{i-1}}u^{n+1}}du.\end{align*} $$
We have a pole of order
$2^{i-1}$
at
$u=q^{-1}$
. Hence we enlarge our contour to collect the residue at the pole. Hence
$$ \begin{align} \sum_{f\in\mathcal{M}_{n,q}}\delta_2^i(f)=\frac{1}{2\pi i}\oint_{|u|=q^{-\epsilon}}\frac{\mathcal{B}_i(u)}{(1-qu)^{2^{i-1}}u^{n+1}}du - \text{Res}\left(\frac{\mathcal{B}_i(u)}{(1-qu)^{2^{i-1}}u^{n+1}};u=q^{-1}\right). \end{align} $$
We know that
$\mathcal {B}_i$
converges absolutely when
$|u|<q^{-1}$
. Also
$$ \begin{align*}|qu-1|>q^{1-\epsilon}-1 \gg 1.\end{align*} $$
Hence
$$ \begin{align*}\frac{1}{(1-qu)^{2^{i-1}}}\ll 1.\end{align*} $$
So we get
$$ \begin{align*}\frac{1}{2\pi i}\oint_{|u|=q^{-\epsilon}}\frac{\mathcal{B}_i(u)}{(1-qu)^{2^{i-1}}u^{n+1}}du\ll \oint_{|u|=q^{-\epsilon}}\frac{1}{u^{n+1}}du\ll q^{n\epsilon}.\end{align*} $$
For the next step, we have to compute the residue in (4.5). To do just that, we make a note that for a memomorphic function
$f(z)$
with a pole of order r at
$z=z_0$
, the residue is given by
$$ \begin{align*}\text{Res}(f(z),z_0)=\frac{1}{(r-1)!}\lim_{z\rightarrow z_0}\frac{d^{r-1}}{dz^{r-1}}\left((z-z_0)^rf(z)\right).\end{align*} $$
Let
$$ \begin{align*}g(u)=\frac{\mathcal{B}_i(u)}{(1-qu)^{2^{i-1}}u^{n+1}}.\end{align*} $$
Hence using the above formula for the residue, we get
$$ \begin{align*} \begin{aligned} \text{Res}(g(u),q^{-1})& = \frac{1}{(2^{i-1}-1)!}\lim_{u\rightarrow q^{-1}}\frac{d^{2^{i-1}-1}}{du^{2^{i-1}-1}}\left((u-q^{-1})^{2^{i-1}}g(u)\right) \\ & = \frac{1}{(2^{i-1}-1)!}\lim_{u\rightarrow q^{-1}}\frac{1}{q^{2^{i-1}}}\frac{d^{2^{i-1}-1}}{du^{2^{i-1}-1}}\left(\frac{\mathcal{B}_i(u)}{u^{n+1}}\right)\\ & = \frac{1}{(2^{i-1}-1)!}\lim_{u\rightarrow q^{-1}}\sum_{j=0}^{2^{i-1}-1}\frac{1}{q^{2^{i-1}}}\frac{d^j}{du^j}\left(\frac{1}{u^{n+1}}\right)\frac{d^{2^{i-1}-1-j}}{du^{2^{i-1}-1-j}}\mathcal{B}_i(u). \end{aligned} \end{align*} $$
The last line follows by using the product rule. Noting that the
$j=0$
term corresponds to the leading order term of the residue, we get
$$ \begin{align*} \begin{aligned} \text{Res}(g(u),q^{-1})& = \frac{1}{(2^{i-1}-1)!}\frac{1}{q^{2^{i-1}}}\left(\mathcal{B}_i(q^{-1}) \frac{-(n+2^{i-1}-1)!}{n!}q^{n+2^{i-1}}+O(q^{n+2^{i-1}-1})\right)\\ & = -\mathcal{B}_i(q^{-1})\binom{n+2^{i-1}-1}{n}q^n+O(q^{n-1}), \end{aligned} \end{align*} $$
where
$\binom {s}{t}$
is the binomial coefficent s choose t. Plugging it back into (4.5) and using the relation
$r_2^*(f)=2\delta (f)$
gives us the result.
5 Higher moments of
$r_2^*$
in the large finite field limit
In this section, we use another approach to estimate the moments. We define the set
$\mathcal {F}_n\subset \mathcal {M}_{n,q}$
consisting of all
$f\in \mathcal {M}_{n,q}$
such that
$b_{q}(f)=1.$
By the definition of
$\mathcal {F}_n$
, it is clear that
$$ \begin{align*}\sum_{f\in\mathcal{M}_{n,q}}(r_2^*)^i(f)=\sum_{f\in\mathcal{F}_{n}}(r_2^*)^i(f).\end{align*} $$
for each
$i\in \mathbb {N}$
. Partitioning
$\mathcal {F}_{n}$
as in [Reference Bary-Soroker, Smilansky and Wolf1], we define
$$ \begin{align*}\begin{aligned} &\mathcal{F}_{1, n}=\left\{f \in \mathcal{F}_n: f=P_1 P_2 \ldots P_r, P_i \neq P_j \hspace{2mm}\forall i \neq j, r \geq 0\right\}, \\ &\mathcal{F}_{2, n}=\left\{f \in \mathcal{F}_n: f=T P_1 P_2 \ldots P_r, P_i \neq P_j\hspace{2mm} \forall i \neq j, r \geq 0\right\}, \\ &\mathcal{F}_{3, n}=\left\{f \in \mathcal{F}_n: f=P_1 P_2 \ldots P_r Q_1^2, \operatorname{deg} Q_1=1, P_i \neq P_j \hspace{2mm}\forall i \neq j, r \geq 0\right\}, \\ &\mathcal{F}_{4, n}=\left\{f \in \mathcal{F}_n: f=P_1^2 P_2 \ldots P_r, \operatorname{deg} P_1=1, P_i \neq P_j \hspace{2mm}\forall i \neq j, r \geq 0\right\}, \\ &\mathcal{F}_{5, n}=\left\{f \in \mathcal{F}_n: f=TP_1 P_2 \ldots P_r Q_1^2, \operatorname{deg} Q_1=1, P_i \neq P_j \hspace{2mm}\forall i \neq j, r \geq 0\right\}, \\ &\mathcal{F}_{6, n}=\mathcal{F}_n \backslash \bigcup_{i=1}^5 \mathcal{F}_{i, n}. \end{aligned} \end{align*} $$
The motive behind the partitioning is that the major contribution to the moments of
$r_2^*(f)$
comes from specific types of polynomials described in the partitions. To show this, we will estimate the sum for each of the subsets of
$\mathcal {F}_n$
. Before the estimation, we state the version of the Chebotarev density theorem we will be using.
Lemma 5.1 [Reference Bary-Soroker, Smilansky and Wolf1, Lemma 3.2] Let
$$ \begin{align*}\pi_{q; \alpha}(n):=\#\left\{R \in \mathcal{M}_{n, q}:\left(\frac{R}{T}\right)=\alpha, \alpha=\pm1, R \text { irreducible }\right\}=\frac{q^n}{2 n}+O\left(q^{\lfloor n / 2\rfloor}\right). \end{align*} $$
Then, we have
$$ \begin{align*}\begin{aligned} &\pi_{q; \alpha}(1)=\frac{q-1}{2}, \\ &\pi_{q; \alpha}(2)=\frac{1}{4} q^2-\frac{1+\alpha}{4} q+\frac{\alpha}{4},\textrm{and} \\ &\pi_{q; \alpha}(n)=\frac{q^n}{2 n}+O\left(q^{n-2}\right) \text {, } n \geq 3 \text {. } \end{aligned} \end{align*} $$
Let us introduce some notations before we proceed. Let
$\lambda \vdash n$
denote a partition
$(\lambda _1,\lambda _2,...,\lambda _n)$
of n such that
$\sum _{j=1}^n j\lambda _j=n$
, i.e.,
$\lambda _j$
is the number of times j appears in the partition. In the next few results, certain coefficients show up often. We define
$$ \begin{align*} \gamma_{i,n}=\sum_{\lambda\vdash n}\prod_{j=1}^n\frac{2^{(i-1)\lambda_j}}{j^{\lambda_j}\lambda_j!}. \end{align*} $$
Proposition 5.1 We have
$$ \begin{align*}&\sum_{f\in\mathcal{F}_{1,n}}(r_2^*)^i(f) \\ &\quad =2^i\gamma_{i,n}q^n-2^i\left(2^{(i-1)}\gamma_{i,n-1}+\left(2^{(i-1)}+2^{2(i-1)}\right)\gamma_{i,n-2}\right)q^{n-1}+O(q^{n-2}).\end{align*} $$
Proof Say
$f\in \mathcal {F}_{1,n}.$
We have
$f=P_1P_2....P_k$
for primes
$P_j$
,
$j\leq k$
. The degrees of
$P_i$
form a partition of n say
$(\lambda _1,\lambda _2,\ldots ,\lambda _n)$
, where
$\lambda _j$
is the number of primes the factorization with degree j. Since every prime in the factorization has exponent 1, Theorem 1.1 gives
$$ \begin{align*}r_2^*(f)= 2\cdot2^{\lambda_1+\lambda_2+\ldots+\lambda_n}.\end{align*} $$
Hence, we have
$$ \begin{align*} \sum_{f\in\mathcal{F}_{1,n}}(r_2^*)^i(f)=2^i\sum_{\lambda\vdash n}\prod_{j=1}^n\binom{\pi_{q;1}(j)}{\lambda_j }2^{i\lambda_j}. \end{align*} $$
Note that the term
$\binom {\pi _{q;1}(j)}{\lambda _j}$
denotes the number of ways to choose
$\lambda _j$
primes of degree j. From Lemma 5.1, we have
$$ \begin{align*}\begin{aligned} &\binom{\pi_{q;1}(1)}{\lambda_1}2^{i\lambda_1} =\frac{2^{(i-1)\lambda_1}}{\lambda_1!}q^{\lambda_1}-\frac{2^{(i-1)\lambda_1}\lambda_1^2}{\lambda_1!}q^{\lambda_1-1}+O\left(q^{\lambda_1-2}\right),\\ &\binom{\pi_{q;1}(2)}{\lambda_2}2^{i\lambda_2} =\frac{2^{(i-1)\lambda_2}}{2^{\lambda_2}\lambda_2!}q^{2\lambda_2}-\frac{2^{(i-1)\lambda_2}2\lambda_2}{2^{\lambda_2}\lambda_2!}q^{2\lambda_2-1}+O\left(q^{2\lambda_2-2}\right), \textrm{and}\\ &\binom{\pi_{q;1}(j)}{\lambda_j}2^{i\lambda_j} =\frac{2^{(i-1)\lambda_j}}{j^{\lambda_j}\lambda_j!}q^{j\lambda_j}+O\left(q^{j\lambda_j-2}\right).\\ \end{aligned} \end{align*} $$
Hence, we get
$$ \begin{align*} &\prod_{j=1}^n\binom{\pi_{q;1}(j)}{\lambda_j}2^{i\lambda_j} \\ &\quad =\prod_{j=1}^n\frac{2^{(i-1)\lambda_j}}{j^{\lambda_j}\lambda_j!}q^n-\left(\lambda_1^2\prod_{j=1}^n\frac{2^{(i-1)\lambda_j}}{j^{\lambda_j}\lambda_j!}+2\lambda_2\prod_{j=1}^n\frac{2^{(i-1)\lambda_j}}{j^{\lambda_j}\lambda_j!}\right)q^{n-1}+O(q^{n-2}).\end{align*} $$
Hence
$$ \begin{align} & \sum_{\lambda\vdash n}\prod_{j=1}^n\binom{\pi_{q;1}(j)}{\lambda_j}2^{i\lambda_j} \notag \\ &\quad =\gamma_{i,n}q^n-\sum_{\lambda\vdash n}\left(\lambda_1^2\prod_{j=1}^n\frac{2^{(i-1)\lambda_j}}{j^{\lambda_j}\lambda_j!}+2\lambda_2\prod_{j=1}^n\frac{2^{(i-1)\lambda_j}}{j^{\lambda_j}\lambda_j!}\right)q^{n-1}+O(q^{n-2}). \end{align} $$
Define
$C_\lambda $
and
$D_{\lambda }$
as
$$ \begin{align*}C_\lambda=\left(\lambda_1^2\prod_{j=1}^n\frac{2^{(i-1)\lambda_j}}{j^{\lambda_j}\lambda_j!}\right), \hspace{1mm}D_\lambda=\left(2\lambda_2\prod_{j=1}^n\frac{2^{(i-1)\lambda_j}}{j^{\lambda_j}\lambda_j!}\right).\end{align*} $$
Also, define the partitions
$\mu $
and
$\nu $
as follows:
$\mu $
is a partition of
$n-1$
given by
$(\lambda _1-1,\lambda _2,..,\lambda _{n-1})$
and
$\nu $
is a partition of
$n-2$
given by
$(\lambda _1-2,\lambda _2,..,\lambda _{n-2})$
. If
$\lambda _1=0$
, then
$C_\lambda =0$
and hence such partitions do not contribute to our sum. If
$\lambda _1=1$
, then
$\lambda _n=0$
and
$$ \begin{align*}C_\lambda=2^{(i-1)}\left(\prod_{j=1}^{n-1}\frac{2^{(i-1)\mu_j}}{j^{\mu_j}\mu_j!}\right),\end{align*} $$
where
$\mu $
is a partition of
$n-1$
. So, when
$\lambda $
runs through all the partitions of n with
$\lambda _1=1$
,
$\mu $
runs through all partitions of
$n-1$
with
$\mu _1=0$
. If
$\lambda _1\geq 2$
,
$\lambda _{n-1}=\lambda _n=0$
and
$$ \begin{align*}C_\lambda=\left(\lambda_1^2\prod_{j=1}^n\frac{2^{(i-1)\lambda_j}}{j^{\lambda_j}\lambda_j!}\right)=\left(\lambda_1\prod_{j=1}^n\frac{2^{(i-1)\lambda_j}}{j^{\lambda_j}\lambda_j!}\right)+\left(\lambda_1(\lambda_1-1)\prod_{j=1}^n\frac{2^{(i-1)\lambda_j}}{j^{\lambda_j}\lambda_j!}\right).\end{align*} $$
Hence
$$ \begin{align*}C_\lambda=2^{(i-1)}\left(\prod_{j=1}^{n-1}\frac{2^{(i-1)\mu_j}}{j^{\mu_j}\mu_j!}\right)+2^{2(i-1)}\left(\prod_{j=1}^{n-2}\frac{2^{(i-1)\nu_j}}{j^{\nu_j}\nu_j!}\right),\end{align*} $$
where
$\mu $
is a partition of
$n-1$
and
$\nu $
is a partition of
$n-2$
. In this case, when
$\lambda $
runs through all partitions of n with
$\lambda _1\geq 2$
,
$\mu $
runs through all partitions of
$n-1$
with
$\mu _1\geq 1$
and
$\nu $
run through all partitions of
$n-2$
. So overall, as
$\lambda $
runs through all partitions of n,
$\mu $
and
$\nu $
runs through all the partitions of
$n-1$
and
$n-2$
. Hence
$$ \begin{align} \begin{aligned} \sum_{\lambda\vdash n}C_\lambda&=2^{(i-1)}\sum_{\mu\vdash (n-1)} \left(\prod_{j=1}^{m-1}\frac{2^{(i-1)\mu_j}}{j^{\mu_j}\mu_j!}\right)+2^{2(i-1)}\sum_{\nu\vdash (n-2)}\left(\prod_{j=1}^{n-2}\frac{2^{(i-1)\nu_j}}{j^{\nu_j}\nu_j!}\right)\\ &=2^{(i-1)}\gamma_{i,n-1}+2^{2(i-1)}\gamma_{i,n-2}. \end{aligned} \end{align} $$
If
$\lambda _2=0$
, then
$D_\lambda =0$
and hence such partitions do not contribute to the sums. If
$\lambda _2\geq 1, \lambda _{n-1}=\lambda _n=0.$
Hence
$$ \begin{align*}D_{\lambda}=\left(2\lambda_2\prod_{j=1}^n\frac{2^{(i-1)\lambda_j}}{j^{\lambda_j}\lambda_j!}\right)=2^{(i-1)}\prod_{j=1}^{n-2}\frac{2^{(i-1)\nu_j}}{j^{\nu_j}\nu_j!},\end{align*} $$
where
$\nu $
is a partition of
$n-2$
given by
$(\lambda _1,\lambda _2-1,..,\lambda _{n-2})$
. We see that as
$\lambda $
runs through all partitions of n,
$\nu $
runs through all partitions of
$n-2$
. Hence
$$ \begin{align} \sum_{\lambda\vdash n}D_\lambda=2^{(i-1)}\sum_{\nu\vdash (n-2)}\left(\prod_{j=1}^{n-2}\frac{2^{(i-1)\nu_j}}{j^{\nu_j}\nu_j!}\right)=2^{(i-1)}\gamma_{i,n-2}. \end{align} $$
Proposition 5.2 We have
$$ \begin{align*}\sum_{f\in\mathcal{F}_{2,n}}(r_2^*)^i(f)=2^i\gamma_{i,n-1}q^{n-1}+O(q^{n-2}),\end{align*} $$
Proof We exploit the one to one correspondence between
$\mathcal {F}_{2,n}$
and
$\mathcal {F}_{1,n-1}$
. For
$f\in \mathcal {F}_{2,n}$
, the degrees of
$P_i$
form a partition of
$n-1$
and so we have
$$ \begin{align*}\sum_{f\in\mathcal{F}_{2,n}}(r_2^*)^i(f)=2^i\sum_{\lambda\vdash n-1}\prod_{j=1}^{n-1}\binom{\pi_{q;1}(j)}{\lambda_j}2^{i\lambda_j}.\end{align*} $$
Using (5.1), we have the result.
Proposition 5.3 We have
$$ \begin{align*}\sum_{f\in\mathcal{F}_{3,n}}(r_2^*)^i(f)=2^{(i-1)}\gamma_{i,n-2}q^{n-1}+O(q^{n-2}).\end{align*} $$
Proof Say
$f=P_1P_2...P_rQ_1^2\in \mathcal {F}_{3,n}$
with degree of
$Q_1$
being 1. Let Q be fixed. Then the degrees of
$P_i$
form a partition of
$n-2$
and for a fixed Q, we have a correspondence between
$\mathcal {F}_{3,n}$
and
$\mathcal {F}_{1,n-2}$
. So we have
$$ \begin{align*}\sum_{f\in\mathcal{F}_{3,n},Q\hspace{1mm}\text{fixed}}(r_2^*)^i(f)=2^i\gamma_{i,n-2}q^{n-2}+O(q^{n-3}).\end{align*} $$
Note that the choice of Q does not influence the function
$r_2^*(f)$
. We have
$\frac {q-1}{2}$
choices for Q. Hence
$$ \begin{align*}\sum_{f\in\mathcal{F}_{3,n}}(r_2^*)^i(f)=\frac{q-1}{2}\left(2^i\gamma_{i,n-2}q^{n-2}+O(q^{n-3})\right)=2^{(i-1)}\gamma_{i,n-2}q^{n-1}+O(q^{n-2}).\end{align*} $$
Proposition 5.4 We have
$$ \begin{align*}\sum_{f\in\mathcal{F}_{4,n}}(r_2^*)^i(f)=\frac{6^i}{2}\gamma_{i,n-2}q^{n-1}+O(q^{n-2}).\end{align*} $$
Proof For
$f\in \mathcal {F}_{4,n}$
, say
$f=P_1^2P_2..P_r$
with degree of
$P_1$
being 1, the degrees of
$P_2,P_3,..P_r$
form a partition
$\lambda =(\lambda _1,\lambda _2,...,\lambda _{n-2})$
of
$n-2$
where
$\lambda _1$
is the number of linear terms in the factorization excluding
$P_1$
. Also,
$$ \begin{align*}r_2^*(f)=6\cdot2^{\lambda_1+\lambda_2+\ldots\lambda_{n-2}}.\end{align*} $$
Hence
$$ \begin{align*}\sum_{f\in\mathcal{F}_{4,n}}(r_2^*)^i(f)=\left(\frac{q-1}{2}\right)6^i\sum_{\lambda\vdash n-2}\binom{\pi_{q;1}(1)-1}{\lambda_1}2^{i\lambda_1}\prod_{j=2}^{n-2}\binom{\pi_{q;1}(j)}{\lambda_j}2^{i\lambda_j}.\end{align*} $$
Similar analysis gives the result.
Proposition 5.5 We have
$$ \begin{align*}\sum_{f\in\mathcal{F}_{5,n}\cup\mathcal{F}_{6,n}}(r_2^*)^i(f)=O(q^{n-2}).\end{align*} $$
Proof Exploiting the one to one correspondence between
$\mathcal {F}_{5,n}$
and
$\mathcal {F}_{3,n-1}$
, we have from Proposition 5.3 that
$$ \begin{align*}\sum_{f\in\mathcal{F}_{5,n}}(r_2^*)^i(f)=O(q^{n-2}).\end{align*} $$
Also, from [Reference Bary-Soroker, Smilansky and Wolf1, Lemma 3.1], the size of the set
$\mathcal {F}_{6,n}$
is bounded above by
$\frac {7}{3}q^{n-2}.$
Hence, the result is evident.
The coefficients
$\gamma _{i,n}$
also arise naturally in the study of Ewens sampling formula which describes the probabilities associated with the number of different alleles observed in a sample. Using the sampling formula, we get that
$$ \begin{align*}\gamma_{i,n}=\sum_{\lambda\vdash n}\prod_{j=1}^{n}\frac{2^{(i-1)\lambda_j}}{j^{\lambda_j}\lambda_j!}=\frac{(n+2^{(i-1)}-1)!}{n!(2^{(i-1)}-1)!}=\binom{n+2^{(i-1)}-1}{n}.\end{align*} $$
Refer to [Reference Karlin and McGregor5] for more details. Finally, we prove Theorem 1.4.
Proof of Theorem 1.4
From the definition of
$\mathcal {F}_n$
,
$$ \begin{align*}\sum_{f\in\mathcal{M}_{n,q}}(r_2^*)^i(f)=\sum_{f\in\mathcal{F}_{n}}(r_2^*)^i(f).\end{align*} $$
Splitting the sum over partitions of
$\mathcal {F}_n$
and using the estimates obtained in Proposition 5.1–5.5, we get the result.
