On the arithmetic of powers of subseries of the exponential series

Question

Let $p$ be a prime number and $q=p-1$. I’m trying to prove that the nonzero coefficients $a_{qk}$ ($k\ge1$) of the power series $$ \sum_{k\ge1} a_{qk} z^{qk} := \left( \sum_{k\ge0} \frac{z^{qk+1}}{(qk+1)!} \right)^q $$ satisfy the congruence $$ a_{qk} \cdot (qk)! \equiv -1 \mod p. $$

I’ve managed to work out that: There is a combinatorial formula for this power series. First of all, the series being raised to the $q$th power can be written $$ f(z) = \frac1q \sum_{1\le i\le q} \zeta^{-i} e^{\zeta^iz} = \sum_{k\ge0} \frac{z^{qk+1}}{(qk+1)!} $$ where $\zeta$ is a primitive $q$th root of unity. And its $q$th power can be expressed in terms of the derivative $$ f'(z) = \frac1q \sum_{1\le i\le q} e^{\zeta^iz} = \sum_{k\ge0} \frac{z^{qk}}{(qk)!} $$ by the formula $$ f(z)^q = \frac1{q^{q}} \sum {q\choose i_1,\dots,i_{q}} {\big|C_q\cdot(i_1,\dots,i_q)\big|} \zeta^{-\big(1\cdot i_1+2\cdot i_2+\cdots+q\cdot i_{q}\big)} f'\Big(\big(i_1\zeta^1+\cdots+i_{q}\zeta^{q}\big)z\Big) $$ where the sum ranges over $C_{q}$-orbits of weak compositions $i_1+\cdots+i_{q}=q$, $i_k\ge0$, and where $\big|C_q\cdot(i_1,\dots,i_q)\big|$ denotes the size of the $C_q$-orbit of the weak composition.

It is straightforward to show that Fermat's little theorem extends to $\mathbf{F}_p[\zeta]/(\zeta^q-1)$ in the sense that the Frobenius is the identity, and it follows that $q$th powers in this ring are fixed by exponentiation by any positive integer power.

This reduces proving the claim to verifying it for the coefficient of $z^q$, i.e. verifying the congruence $$ -1 \equiv \sum {q\choose i_1,\dots,i_{q}} {\big|C_q\cdot(i_1,\dots,i_q)\big|} \cdot \zeta^{-\big(1\cdot i_1+2\cdot i_2+\cdots+q\cdot i_{q}\big)} \Big(i_1\zeta^1+\cdots+i_{q}\zeta^{q}\Big)^q \mod p. $$ I’ve done this by computer for the primes up to $p=17$ but haven’t found a general argument.

Answer 1

Your argument is in fact almost complete. It reduces the problem to checking the normalized coefficient $a_q$ of $z^q$ is congruent to $-1$ mod $p$, but since $$ f = z + \frac{z^{q+1}}{(q+1)!} + \dots ,$$ we have $$f^q = z^q + \frac{ q z^{2q}}{ (q+1)!} + \dots$$ and so $$a_q = q! \equiv -1 \mod p.$$

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Here is an alternate "bijective" presentation of the same argument. We can interpret $a_n$ combinatorially as the number of $q$-colorings of the numbers from $1$ to $n$ such that each color is used a number of times congruent to $1$ mod $q$.

For $n \geq p$, there is an action of $\mathbb Z/p$ on this set of colorings by rotating the colorings of the numbers $n+1-p, n+2-p, \dots, n$. The number of colorings is congruent mod $p$ to the number of fixed points of this action. A coloring is fixed if the last $p$ numbers have the same color.

Since $q=p-1$, removing the last $q$ numbers in such a coloring doesn't affect the congruent mod $q$ of the number of times each coloring is used, and this gives a bijection between fixed points and colorings of the first $n-q$ numbers satisfying the same congruence condition.

So $a_n \equiv a_{n-q} \mod p$ for $n>q$ and $a_q$ is the number of $q$-colorings of $\{1,\dots, q\}$ with all colors occurring exactly once, which is $q! \equiv -1 \mod p$.

Answer 2

Something much more general is true.

Let $p$ be a prime. All congruences in what follows are modulo $p$. A Hurwitz series is a power series of the form $\sum_{n=0}^\infty a_n z^n/n!$ where the $a_n$ are integers.

Suppose that $A(z)=\sum_{n=0}^\infty a_n z^n/n!$ and $B(z)=\sum_{n=0}^\infty b_n z^n/n!$ are Hurwitz series satisfying $a_{n+p-1}\equiv a_n$ and $b_{n+p-1}\equiv b_n$ for all $n\ge 1$. We will show that $c_{n+p-1}\equiv c_n$ for all $n\ge1$, where $\sum_{n=0}^\infty c_n z^n/n! = A(z) B(z)$.

Given this, it follows that if $a_{n+p-1}\equiv a_n$ for all $n\ge1$ then the same congruence holds for coefficients of powers $A(z)^m$. In particular, to obtain the proposer's result we may take $A(z)$ to be $\sum_{k\ge0} z^{qk+1}/(qk+1)!$ and $m=q$, where $q=p-1$.

Proof of the general result. Let $D=d/dz$. Then $D^i A(z) = \sum_{n=0}^\infty a_{n+i}z^n/n!$. Thus what we want to prove is that if $(D^p-D)A(z) \equiv (D^p-D)B(z)\equiv0$ then $(D^p-D)A(z)B(z) \equiv0$, where congruence of Hurwitz series is coefficientwise (as exponential generating functions).

We first show that $D^p$ is a derivation on Hurwitz series modulo $p$; that is, $D^p A(z) B(z)\equiv A(z)\cdot D^p B(z) + D^p A(z) \cdot B(z)$. This follows from Leibniz's formula $$D^p A(z) B(z) =\sum_{i=0}^p \binom{p}{i}D^{i}A(z) \cdot D^{p-i}B(z)$$ since all but the $i=0$ and $i=p$ terms are 0 modulo $p$. Since $D$ is a derivation, it follows that $D^p-D$ is a mod $p$ derivation. So $$(D^p-D)A(z)B(z) \equiv A(z)\cdot (D^p-D)B(z) +(D^p-D)A(z)\cdot B(z)$$ and the result follows.