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Let $p$ be a prime and suppose that $g(t)$ is the generating function $g(t) = 1/p\,g(t)^p + t$ with low order terms $g(t) = t + O(t^p)$. An easy induction shows that the coefficient of $t^n$ in $g(t)$ is zero unless $n\equiv 1\pmod{p-1}$. I need to show that the $p$-adic valuation of $[t^n] g(t)$ is equal to $-v(n!)$ when $n\equiv 1\pmod{p-1}$.

Expanded out, the recurrence says that

$\displaystyle f(n) = 1/p \sum_{a_1+\cdots + a_p = n} \prod_{i} f(i)$

When $p=3$, this sequence starts

$\displaystyle t+\frac{t^3}{3}+\frac{t^5}{3}+\frac{4 \, t^7}{9}+\frac{55 \,t^9}{81}+\frac{91 \,t^{11}}{81}+\frac{476 \,t^{13}}{243}+\frac{2584 \,t^{15}}{729}+\frac{4807 \,t^{17}}{729}+O(t^{19})$.

Does anyone have any suggestions for how to check this?

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    $\begingroup$ By Lagrange inversion, $$g(t) = \sum_{j=0}^\infty \frac{1}{p^j(pj+1)}\binom{pj+1}{j} t^{(p-1)j+1}.$$ $\endgroup$ – Ira Gessel May 13 at 4:47
  • $\begingroup$ @IraGessel Thanks, if you post that as an answer I will accept it. $\endgroup$ – Hood May 13 at 22:35
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Let $g(t) = th(t^{p-1}/p)$. Then the functional equation for $g(t)$ gives $h(z) =1+zh(z)^p$. It is well known that the coefficients of $h(z)$ are given by $$h(z) = \sum_{j=0}^\infty \frac{1}{pj+1}\binom{pj+1}{j}z^j.$$ This can be proved by Lagrange inversion or in other ways. These number are sometimes called Fuss-Catalan numbers and they can be found in the OEIS as A062993 or A070914.

It follows that $$g(t) = \sum_{j=0}^\infty \frac{1}{p^j(pj+1)}\binom{pj+1}{j}t^{(p-1)j+1}.$$

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