# p-adic valuation of coefficients of generating function

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?

• By Lagrange inversion, $$g(t) = \sum_{j=0}^\infty \frac{1}{p^j(pj+1)}\binom{pj+1}{j} t^{(p-1)j+1}.$$ – Ira Gessel May 13 at 4:47
• @IraGessel Thanks, if you post that as an answer I will accept it. – Hood May 13 at 22:35

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