# §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 $\displaystyle\sum_{k=0}^{n}{n\choose k}B_{k}B_{n-k}$ $\displaystyle=(1-n)B_{n}-nB_{n-1}.$ Symbols: $B_{\NVar{n}}$: Bernoulli numbers, $\binom{\NVar{m}}{\NVar{n}}$: binomial coefficient, $k$: integer and $n$: integer Permalink: http://dlmf.nist.gov/24.14.E2 Encodings: TeX, pMML, png See also: Annotations for 24.14(i) 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 $\displaystyle\sum_{k=0}^{n}{n\choose k}E_{k}E_{n-k}$ $\displaystyle=-2^{n+1}\mathop{E_{n+1}\/}\nolimits\!\left(0\right)=-2^{n+2}(1-2% ^{n+2})\frac{B_{n+2}}{n+2}.$ 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 $\displaystyle\sum_{k=0}^{n}{n\choose k}2^{k}B_{k}E_{n-k}$ $\displaystyle=2(1-2^{n-1})B_{n}-nE_{n-1}.$ Symbols: $B_{\NVar{n}}$: Bernoulli numbers, $E_{\NVar{n}}$: Euler numbers, $\binom{\NVar{m}}{\NVar{n}}$: binomial coefficient, $k$: integer and $n$: integer Referenced by: §24.14(i) Permalink: http://dlmf.nist.gov/24.14.E6 Encodings: TeX, pMML, png See also: Annotations for 24.14(i)

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

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

## §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 $\displaystyle\sum\frac{(2n)!}{(2j)!(2k)!(2\ell)!}B_{2j}B_{2k}B_{2\ell}$ $\displaystyle=(n-1)(2n-1)B_{2n}+n(n-\tfrac{1}{2})B_{2n-2},$ 24.14.9 $\displaystyle\sum\frac{(2n)!}{(2j)!(2k)!(2\ell)!}E_{2j}E_{2k}E_{2\ell}$ $\displaystyle=\tfrac{1}{2}\left(E_{2n}-E_{2n+2}\right).$

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

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