# A recursion similar to the one for Bernoulli numbers

For the Bernoulli numbers $B_m$, there is a recursion: $B_0=1$ and $\sum_{j=0}^{m-1}\binom{m+1}{j}B_j=-(m+1)B_m$ for $m\ge 1$.

It is known that $B_{m}=0$ when $m\gt 1$ is odd.

Now, define $A_m$ with a similar recursion, also starting with $A_0=1$ $$\sum_{j=0}^{m-1}\binom{m}{j}A_j=-2A_m$$

How can it be shown that $A_m=0$, when $m$ is even?

I know an explicit formula: $A_m=\sum_{j=0}^{m} (-1)^j \frac{j!}{2^j}{m\brace j}$ involving Stirling numbers of second kind, but it does not seem to be readily helpful.

A proof would easily follow from the knowledge of a generating function for $A_m$ (which I don't have). I would also welcome a proof without generating function.

EDIT: from Kelenner answer, a connection can be made with tangent numbers $T_{2k+1}$ - these are integers from http://oeis.org/A000182 $(1,2,16,272,7936,..)$ such that $$\tan(x)=\sum_{k\ge 0} T_{2k+1} \frac{x^{2k+1}}{(2k+1)!}$$ $$A_{2k+1}=\frac{(-1)^{k+1}}{2^{2k+1}}T_{2k+1}$$ and an explicit formula for tangent numbers involving Stirling numbers: $$T_m=(-1)^{\frac {m-1}{2}}\sum_{j=0}^{m} (-2)^{m-j}j!{m\brace j}$$

• In light of Kelenner's answer, this sequence is clearly connected to Genocchi numbers. – Lucian Nov 21 '15 at 22:40

## 1 Answer

You have for $m\geq 1$ $$\sum_{j=0}^m {m \choose j} A_j=-A_m$$ This gives that with $\displaystyle f(x)=\sum_{k=0}^{+\infty}\frac{A_j}{j!} x^j$, if we multiply by $x^m$ and sum for $m\geq 1$, we have $$f(x)\exp(x)-1=-(f(x)-1)$$ hence $\displaystyle f(x)=\frac{2}{\exp(x)+1}$. Now we find easily that $f(x)+f(-x)=2$, and we are done.

• Very nice proof. This generating function shows the connection of the $A_n$ with the tangent numbers. – René Gy Nov 21 '15 at 21:10