This question concerns a combinatorial identity obeyed by power series coefficients. Throughout we let $[x^{M}]\{\phi(x)\}$ denote the coefficient of $x^{M}$ in a power series $\phi(x)$.

Let $k$ be a positive integer, and consider the function $F(k,x)$ defined as the following power series in $x$:

\begin{equation} F(k,x)=\sum_{s=1}^{\infty} \frac{(-1)^{s-1}}{s^{2}}\binom{s \ k}{s+1}(s+1)\ x^{s}. \end{equation}

I am interested in the series coefficients of the function $\exp(N F (k,x))$ for positive integer $N.$

Through comparison of various formulas that arose in a research project, I have been lead to the following identity for the case $N=M+1$:

\begin{equation} [x^{M}]\{e^{(M+1)F(k,x)}\}= \frac{k(M+1)}{k+(k-1)M}\binom{(k-1)^{2}M+k(k-1)}{M}~. \end{equation}

Although I am convinced that this identity is true, I have no idea how to demonstrate it, nor do I have any idea why this power series coefficient has such a simple expression. Thus, my main question is how can this identity be motivated and proven ?

More generally, can we determine the coefficient $[x^{M}]\{e^{N F(k,x)}\}?$

I am also interested in a generalization which depends on an additional positive integer $j$. Specifically, set
\begin{equation} F(k,j,x)=\sum_{s=1}^{\infty} \frac{(-1)^{s-1}}{s^{2}}\binom{s \ k}{s\ j+1}(s\ j+1)\ x^{s}~. \end{equation} The previous function is recovered for the special case $j=1.$

Can the coefficients $[x^{M}]\{e^{NF(k,j,x)}\}$ be similarly determined?

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    $\begingroup$ What's the connection with representation theory? $\endgroup$ – Noam D. Elkies Jan 25 '15 at 3:42

Seems that a general formula for the $x^M$ coefficient of $\exp NF(k,x)$ is $$ \frac{N}{M} (k^2-k) \left( {(k^2-k) N - (k-1) M - 1 \atop M-1} \right), $$ which agrees with your formula when $N=M+1$. This should follow from an explicit formula for $dF(k,x)/dx$ as a degree-$k$ algebraic function of $x$ that's closely related with the inverse function of $y(1-y)^{k-1}$, for which there's a closed-form power series expansion of $y^\beta$ for all $\beta$; see for instance these two "one-page papers" on my math webpage.

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  • $\begingroup$ Thanks Noam! I will look into this idea. Does anything similar work in the case j>1? $\endgroup$ – Clay Cordova Jan 25 '15 at 13:35
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    $\begingroup$ You're welcome. For other $j$ there doesn't seem to be such a nice formula, except in the roughly-complementary case of $j=k-1$. Otherwise the power-series coefficients, written as polynomials in $N$, are generally irreducible once you remove the factor of $N$. (For $j=1$ they factor completely, which soon led me to surmise the formula above.) $\endgroup$ – Noam D. Elkies Jan 25 '15 at 16:07

To elaborate on Noam's answer, $$F(k, y(y+1)^{k-1}) = k(k-1)\log(1+y),$$ as can be proved by Lagrange inversion or in other ways, as in Noam's papers. So if $G(k, x) = e^{F(k,x)}$ then $$G(k, y(y+1)^{k-1})=(1+y)^{k(k-1)}.$$

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    $\begingroup$ Thanks for providing the explicit formula Ira! That helps tremendously in making the problem more concrete. $\endgroup$ – Clay Cordova Jan 29 '15 at 22:50

Let $\mathscr{B}_t(z)$ be a generalized binomial series $$\mathscr{B}_t(z)=\sum\limits_{k=0}^{\infty}{tk+1\choose k} \dfrac{1}{tk+1}z^k.$$ The answer follows from these two formulae \begin{gather} \tag{1}\mathscr{B}_t^r(z)=\sum\limits_{k=0}^{\infty}{tk+r\choose k} \dfrac{r}{tk+r}z^k,\\ \tag{2}\log\mathscr{B}_t(z)=\sum\limits_{k=1}^{\infty}{t k\choose k}\frac{z^k}{tk} \end{gather} because $$F(k,x)=k(1-k)\log\mathscr{B}_k(-x)=k(k-1)\log\mathscr{B}_{1-k}(x).$$

You can find (1) (and some other formulae) in (see Ch.5.4) Graham, R. L.; Knuth, D. E. & Patashnik, O. Concrete mathematics. Addison-Wesley Publishing Company, 1994.

Formula (2) follows from calculations performed in Bizley, M. Derivation of a new formula for the number of minimal lattice paths from $(0,0)$ to $(k m, k n)$ having just $t$ contacts with the line $m y = n x$ and having no points above this line; and a proof of Grossman's formula for the number of paths which may touch but do not rise above this line. J. Inst. Actuaries 80, 55-62 (1954).

The proof of (1) (see Concrete Mathematics) is also based on combinatorics of paths. Probably it can give some tips for $j>1$.

See also Donald Knuth's 20th Annual Christmas Tree Lecture: (3/2)-ary Trees for additional connections and for the history of (2).

I applied these formulae in the theory of formal groups. Can you give us some background information about your question?

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    $\begingroup$ (2) can be derived from (1) by subtracting 1, dividing by $r$ and taking the limit as $r\to 0$. $\endgroup$ – Ira Gessel Feb 1 '15 at 14:07
  • $\begingroup$ Yes, and it is much more shoter than Bizley's arguments! $\endgroup$ – Alexey Ustinov Feb 2 '15 at 1:57

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