Binomial sum formula for $(n+1)^{n-1}$

Question

Has anybody seen a proof for $$ (n+1)^{n-1}=\frac{1}{2^n}\sum_{k=0}^n C_n^k(2k+1)^{k-1}(2(n-k)+1)^{n-k-1} ? $$ There are lots of reasons to think that this is true. In particular the formula holds for $n=0,1,2,3,4,5$.

Answer 1

A (not straight) proof.

Using Bürmann-Lagrange formula we expand the regular at $z=0$ solution $w(z)$ of the equation $w=ze^{aw^2}$ as the following series $$ w(z)=\sum_{n=0}^{\infty}\frac{a^n(2n+1)^{n-1}} {n!}z^{2n+1} $$ (see problem 26.07 in СБОРНИК ЗАДАЧ ПО ТЕОРИИ АНАЛИТИЧЕСКИХ ФУНКЦИЙ, Под редакцией М. А ЕВГРАФОВА (Russian), -- any other reference - for this particular expansion, not for Bürmann-Lagrange formula - will be greatly appreciated). Using Cauchy formulas for the coefficients of the series products, we get from this expansion $$ w^2=\sum_{n=0}^{\infty}\frac{a^{n}} {n!}z^{2n+2} \sum_{k=0}^{n}C_n^k(2k+1)^{k-1}(2(n-k)+1)^{n-k-1}. $$ On the other hand, since $w^2=z^2e^{2aw^2}$, applying the expansion in the problem 26.06 (from the same problem-book), we get that $$ w^2=\sum_{n=0}^{\infty}\frac{a^{n}2^n(n+1)^{n-1}} {n!}z^{2n+2}. $$ Equating to each other the coefficients for $z^{2n+2}$ in both series we obtain the desired equality.

The first mentioned problem 26.07 is for any $m\geq1$ and states that the regular at $z=0$ solution $w(z)$ of the equation $w=ze^{aw^m}$ has the expansion $$ w(z)=\sum_{n=0}^{\infty}\frac{a^n(mn+1)^{n-1}} {n!}z^{mn+1} $$ (above we have used it only for the value $m=2$).

Based on the same two mentioned problems, the formula can be generalized for any $m\in\Bbb{N}$ as follows.

Let $\alpha_k^{n}\in\Bbb{N}\cup\{0\}; \;\; 0\leq\alpha_k^{n}\leq n; \;\; 0\leq k\leq m.$ Denote by $A_{m}^{n}: =\{\alpha_1^{n},\alpha_2^{n},\ldots,\alpha_m^{n}\}$ an $n$-tuple of such numbers such that $\sum_{k=0}^m\alpha_k^{n}=n$. Further denote by $ (A_{m}^{n})!:=\alpha_1^{n}!\alpha_2^{n}!\cdots \alpha_m^{n}!$, and by $ C_n^{A_{m}^{n}}:=\binom{{n}}{A_m^n} := \frac{n!}{(A_{m}^{n})!}.$ Then, \begin{gather*} \boxed{\quad(n+1)^{n-1}=\frac{1}{m^n} \sum_{A_{m}^{n}} C_n^{A_{m}^{n}} (m\alpha_1^{n}+1)^{\alpha_1^{n}-1} \cdots (m\alpha_m^{n}+1)^{\alpha_m^{n}-1}\quad}. \end{gather*}

Answer 2

This is a particular case (for $x = 1$, $y = 1$ and $z = -2$) of the following result:

Let $n$ be a nonnegative integer. In the polynomial ring $\mathbb{Z}\left[ x,y,z\right] $, we have $$ \sum\limits_{k=0}^{n}\dbinom{n}{k}xy\left( x-kz\right) ^{k-1}\left( y-\left( n-k\right) z\right) ^{n-k-1}=\left( x+y\right) \left( x+y-nz\right) ^{n-1}. $$

This result is Theorem 2 in MathOverflow post #273459 (where I derive it from the famous Abel identity).