# §24.15(i) Genocchi Numbers

 24.15.1 $\displaystyle\frac{2t}{e^{t}+1}$ $\displaystyle=\sum_{n=1}^{\infty}G_{n}\frac{t^{n}}{n!},$ 24.15.2 $\displaystyle G_{n}$ $\displaystyle=2(1-2^{n})\mathop{B_{n}\/}\nolimits.$

See Table 24.15.1.

# §24.15(ii) Tangent Numbers

 24.15.3 $\mathop{\tan\/}\nolimits t=\sum_{n=0}^{\infty}T_{n}\frac{t^{n}}{n!},$
 24.15.4 $T_{2n-1}=(-1)^{n-1}\frac{2^{2n}(2^{2n}-1)}{2n}\mathop{B_{2n}\/}\nolimits,$ $n=1,2,\dots$,
 24.15.5 $T_{2n}=0,$ $n=0,1,\dots$. Symbols: $n$: integer and $T_{n}$: tangent numbers Permalink: http://dlmf.nist.gov/24.15.E5 Encodings: TeX, pMML, png

# §24.15(iii) Stirling Numbers

The Stirling numbers of the first kind $\mathop{s\/}\nolimits\!\left(n,m\right)$, and the second kind $\mathop{S\/}\nolimits\!\left(n,m\right)$, are as defined in §26.8(i).

 24.15.6 $\displaystyle\mathop{B_{n}\/}\nolimits$ $\displaystyle=\sum_{k=0}^{n}(-1)^{k}\frac{k!\mathop{S\/}\nolimits\!\left(n,k% \right)}{k+1},$ 24.15.7 $\displaystyle\mathop{B_{n}\/}\nolimits$ $\displaystyle=\sum_{k=0}^{n}(-1)^{k}\binomial{n+1}{k+1}\mathop{S\/}\nolimits\!% \left(n+k,k\right)\bigg/\binomial{n+k}{k},$ 24.15.8 $\displaystyle\sum_{k=0}^{n}(-1)^{n+k}\mathop{s\/}\nolimits\!\left(n+1,k+1% \right)\mathop{B_{k}\/}\nolimits$ $\displaystyle=\frac{n!}{n+1}.$

In (24.15.9) and (24.15.10) $p$ denotes a prime. See Horata (1991).

 24.15.9 $p\frac{\mathop{B_{n}\/}\nolimits}{n}\equiv\mathop{S\/}\nolimits\!\left(p-1+n,p% -1\right)\;\;(\mathop{{\rm mod}}p^{2}),$ $1\leq n\leq p-2$,
 24.15.10 $\frac{2n-1}{4n}p^{2}\mathop{B_{2n}\/}\nolimits\equiv{\mathop{S\/}\nolimits\!% \left(p+2n,p-1\right)\;\;(\mathop{{\rm mod}}p^{3})},$ $2\leq 2n\leq p-3$.

# §24.15(iv) Fibonacci and Lucas Numbers

The Fibonacci numbers are defined by $u_{0}=0$, $u_{1}=1$, and $u_{n+1}=u_{n}+u_{n-1}$, $n\geq 1$. The Lucas numbers are defined by $v_{0}=2$, $v_{1}=1$, and $v_{n+1}=v_{n}+v_{n-1}$, $n\geq 1$.

 24.15.11 $\displaystyle\sum_{k=0}^{\left\lfloor\ifrac{n}{2}\right\rfloor}{n\choose 2k}% \left(\frac{5}{9}\right)^{k}\mathop{B_{2k}\/}\nolimits u_{n-2k}$ $\displaystyle=\frac{n}{6}v_{n-1}+\frac{n}{3^{n}}v_{2n-2},$ 24.15.12 $\displaystyle\sum_{k=0}^{\left\lfloor\ifrac{n}{2}\right\rfloor}{n\choose 2k}% \left(\frac{5}{4}\right)^{k}\mathop{E_{2k}\/}\nolimits v_{n-2k}$ $\displaystyle=\frac{1}{2^{n-1}}.$

For further information on the Fibonacci numbers see §26.11.