# §24.14 Sums

###### Contents

 24.14.1 $\displaystyle\sum_{k=0}^{n}{n\choose k}\mathop{B_{k}\/}\nolimits\!\left(x% \right)\mathop{B_{n-k}\/}\nolimits\!\left(y\right)$ $\displaystyle=n(x+y-1)\mathop{B_{n-1}\/}\nolimits\!\left(x+y\right)-(n-1)% \mathop{B_{n}\/}\nolimits\!\left(x+y\right),$ 24.14.2 $\displaystyle\sum_{k=0}^{n}{n\choose k}\mathop{B_{k}\/}\nolimits\mathop{B_{n-k% }\/}\nolimits$ $\displaystyle=(1-n)\mathop{B_{n}\/}\nolimits-n\mathop{B_{n-1}\/}\nolimits.$ 24.14.3 $\displaystyle\sum_{k=0}^{n}{n\choose k}\mathop{E_{k}\/}\nolimits\!\left(h% \right)\mathop{E_{n-k}\/}\nolimits\!\left(x\right)$ $\displaystyle=2(\mathop{E_{n+1}\/}\nolimits\!\left(x+h\right)-(x+h-1)\mathop{E% _{n}\/}\nolimits\!\left(x+h\right)),$ 24.14.4 $\displaystyle\sum_{k=0}^{n}{n\choose k}\mathop{E_{k}\/}\nolimits\mathop{E_{n-k% }\/}\nolimits$ $\displaystyle=-2^{n+1}\mathop{E_{n+1}\/}\nolimits\!\left(0\right)$ $\displaystyle=-2^{n+2}(1-2^{n+2})\frac{\mathop{B_{n+2}\/}\nolimits}{n+2}.$ 24.14.5 $\displaystyle\sum_{k=0}^{n}{n\choose k}\mathop{E_{k}\/}\nolimits\!\left(h% \right)\mathop{B_{n-k}\/}\nolimits\!\left(x\right)$ $\displaystyle=2^{n}\mathop{B_{n}\/}\nolimits\!\left(\tfrac{1}{2}(x+h)\right),$ 24.14.6 $\displaystyle\sum_{k=0}^{n}{n\choose k}2^{k}\mathop{B_{k}\/}\nolimits\mathop{E% _{n-k}\/}\nolimits$ $\displaystyle=2(1-2^{n-1})\mathop{B_{n}\/}\nolimits-n\mathop{E_{n-1}\/}\nolimits.$

Let $m+n$ be even with $m$ and $n$ nonzero. Then

 24.14.7 $\sum_{j=0}^{m}\sum_{k=0}^{n}\binom{m}{j}\binom{n}{k}\frac{\mathop{B_{j}\/}% \nolimits\mathop{B_{k}\/}\nolimits}{m+n-j-k+1}=(-1)^{m-1}\frac{m!n!}{(m+n)!}% \mathop{B_{m+n}\/}\nolimits.$

# §24.14(ii) Higher-Order Recurrence Relations

In the following two identities, valid for $n\geq 2$, the sums are taken over all nonnegative integers $j,k,\ell$ with $j+k+\ell=n$.

 24.14.8 $\displaystyle\sum\frac{(2n)!}{(2j)!(2k)!(2\ell)!}\mathop{B_{2j}\/}\nolimits% \mathop{B_{2k}\/}\nolimits\mathop{B_{2\ell}\/}\nolimits$ $\displaystyle=(n-1)(2n-1)\mathop{B_{2n}\/}\nolimits+n(n-\tfrac{1}{2})\mathop{B% _{2n-2}\/}\nolimits,$ 24.14.9 $\displaystyle\sum\frac{(2n)!}{(2j)!(2k)!(2\ell)!}\mathop{E_{2j}\/}\nolimits% \mathop{E_{2k}\/}\nolimits\mathop{E_{2\ell}\/}\nolimits$ $\displaystyle=\tfrac{1}{2}\left(\mathop{E_{2n}\/}\nolimits-\mathop{E_{2n+2}\/}% \nolimits\right).$

In the next identity, valid for $n\geq 4$, the sum is taken over all positive integers $j,k,\ell,m$ with $j+k+\ell+m=n$.

 24.14.10 $\sum\frac{(2n)!}{(2j)!(2k)!(2\ell)!(2m)!}\mathop{B_{2j}\/}\nolimits\mathop{B_{% 2k}\/}\nolimits\mathop{B_{2\ell}\/}\nolimits\mathop{B_{2m}\/}\nolimits=-{2n+3% \choose 3}\mathop{B_{2n}\/}\nolimits-\frac{4}{3}n^{2}(2n-1)\mathop{B_{2n-2}\/}\nolimits.$

For (24.14.11) and (24.14.12), see Al-Salam and Carlitz (1959). These identities can be regarded as higher-order recurrences. Let $\det[a_{r+s}]$ denote a Hankel (or persymmetric) determinant, that is, an $(n+1)\times(n+1)$ determinant with element $a_{r+s}$ in row $r$ and column $s$ for $r,s=0,1,\dots,n$. Then

 24.14.11 $\displaystyle\det[\mathop{B_{r+s}\/}\nolimits]$ $\displaystyle=(-1)^{n(n+1)/2}\left(\prod_{k=1}^{n}k!\right)^{6}\Bigg/\left(% \prod_{k=1}^{2n+1}k!\right),$ 24.14.12 $\displaystyle\det[\mathop{E_{r+s}\/}\nolimits]$ $\displaystyle=(-1)^{n(n+1)/2}\left(\prod_{k=1}^{n}k!\right)^{2}.$