# binomial coefficients

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##### 2: 26.21 Tables
###### §26.21 Tables
Abramowitz and Stegun (1964, Chapter 24) tabulates binomial coefficients $\genfrac{(}{)}{0.0pt}{}{m}{n}$ for $m$ up to 50 and $n$ up to 25; extends Table 26.4.1 to $n=10$; tabulates Stirling numbers of the first and second kinds, $s\left(n,k\right)$ and $S\left(n,k\right)$, for $n$ up to 25 and $k$ up to $n$; tabulates partitions $p\left(n\right)$ and partitions into distinct parts $p\left(\mathcal{D},n\right)$ for $n$ up to 500. … Goldberg et al. (1976) contains tables of binomial coefficients to $n=100$ and Stirling numbers to $n=40$.
##### 3: 24.6 Explicit Formulas
24.6.2 $B_{n}=\frac{1}{n+1}\sum_{k=1}^{n}\sum_{j=1}^{k}(-1)^{j}j^{n}{\genfrac{(}{)}{0.% 0pt}{}{n+1}{k-j}}\Bigg{/}{\genfrac{(}{)}{0.0pt}{}{n}{k}},$
##### 4: 24.5 Recurrence Relations
24.5.1 $\sum_{k=0}^{n-1}{n\choose k}B_{k}\left(x\right)=nx^{n-1},$ $n=2,3,\dots$,
24.5.2 $\sum_{k=0}^{n}{n\choose k}E_{k}\left(x\right)+E_{n}\left(x\right)=2x^{n},$ $n=1,2,\dots$.
24.5.3 $\sum_{k=0}^{n-1}{n\choose k}B_{k}=0,$ $n=2,3,\dots$,
24.5.4 $\sum_{k=0}^{n}{2n\choose 2k}E_{2k}=0,$ $n=1,2,\dots$,
##### 5: 1.2 Elementary Algebra
###### §1.2(i) BinomialCoefficients
See also §26.3(i). … For complex $z$ the binomial coefficient $\genfrac{(}{)}{0.0pt}{}{z}{k}$ is defined via (1.2.6). …
1.2.3 $\genfrac{(}{)}{0.0pt}{}{n}{0}+\genfrac{(}{)}{0.0pt}{}{n}{1}+\dots+\genfrac{(}{% )}{0.0pt}{}{n}{n}=2^{n}.$
1.2.4 $\genfrac{(}{)}{0.0pt}{}{n}{0}-\genfrac{(}{)}{0.0pt}{}{n}{1}+\dots+(-1)^{n}% \genfrac{(}{)}{0.0pt}{}{n}{n}=0.$
##### 6: 24.14 Sums
24.14.3 $\sum_{k=0}^{n}{n\choose k}E_{k}\left(h\right)E_{n-k}\left(x\right)=2(E_{n+1}% \left(x+h\right)-(x+h-1)E_{n}\left(x+h\right)),$
24.14.5 $\sum_{k=0}^{n}{n\choose k}E_{k}\left(h\right)B_{n-k}\left(x\right)=2^{n}B_{n}% \left(\tfrac{1}{2}(x+h)\right),$
##### 7: 17.2 Calculus
###### §17.2(ii) BinomialCoefficients
17.2.27 $\genfrac{[}{]}{0.0pt}{}{n}{m}_{q}=\frac{\left(q;q\right)_{n}}{\left(q;q\right)% _{m}\left(q;q\right)_{n-m}}\\ =\frac{\left(q^{-n};q\right)_{m}(-1)^{m}q^{nm-\genfrac{(}{)}{0.0pt}{}{m}{2}}}{% \left(q;q\right)_{m}},$
17.2.28 $\lim_{q\to 1}\genfrac{[}{]}{0.0pt}{}{n}{m}_{q}=\genfrac{(}{)}{0.0pt}{}{n}{m}=% \frac{n!}{m!(n-m)!},$
17.2.30 $\genfrac{[}{]}{0.0pt}{}{-n}{m}_{q}=\genfrac{[}{]}{0.0pt}{}{m+n-1}{m}_{q}(-1)^{% m}q^{-mn-\genfrac{(}{)}{0.0pt}{}{m}{2}},$
##### 8: 17.3 $q$-Elementary and $q$-Special Functions
17.3.2 $E_{q}\left(x\right)=\sum_{n=0}^{\infty}\frac{(1-q)^{n}q^{\genfrac{(}{)}{0.0pt}% {}{n}{2}}x^{n}}{\left(q;q\right)_{n}}=\left(-(1-q)x;q\right)_{\infty}.$
17.3.7 $\beta_{n}\left(x,q\right)=(1-q)^{1-n}\sum_{r=0}^{n}(-1)^{r}\genfrac{(}{)}{0.0% pt}{}{n}{r}\frac{r+1}{(1-q^{r+1})}q^{rx}.$
17.3.8 $A_{m,s}\left(q\right)=q^{\genfrac{(}{)}{0.0pt}{}{s-m}{2}+\genfrac{(}{)}{0.0pt}% {}{s}{2}}\sum_{j=0}^{s}(-1)^{j}q^{\genfrac{(}{)}{0.0pt}{}{j}{2}}\genfrac{[}{]}% {0.0pt}{}{m+1}{j}_{q}\frac{(1-q^{s-j})^{m}}{(1-q)^{m}}.$
##### 9: 12.13 Sums
12.13.2 $U\left(a,x+y\right)=e^{-\frac{1}{2}xy-\frac{1}{4}y^{2}}\sum_{m=0}^{\infty}% \genfrac{(}{)}{0.0pt}{}{-a-\tfrac{1}{2}}{m}y^{m}U\left(a+m,x\right),$
12.13.3 $V\left(a,x+y\right)=e^{\frac{1}{2}xy+\frac{1}{4}y^{2}}\sum_{m=0}^{\infty}% \genfrac{(}{)}{0.0pt}{}{a-\tfrac{1}{2}}{m}y^{m}V\left(a-m,x\right),$
12.13.5 $U\left(a,x\cos t+y\sin t\right)\\ =e^{\frac{1}{4}(x\sin t-y\cos t)^{2}}\*\sum_{m=0}^{\infty}\genfrac{(}{)}{0.0pt% }{}{-a-\tfrac{1}{2}}{m}(\tan t)^{m}U\left(m+a,x\right)U\left(-m-\tfrac{1}{2},y% \right),$ $\Re a\leq-\tfrac{1}{2},0\leq t\leq\tfrac{1}{4}\pi$.
##### 10: 26.9 Integer Partitions: Restricted Number and Part Size
26.9.4 $\genfrac{[}{]}{0.0pt}{}{m}{n}_{q}=\prod_{j=1}^{n}\frac{1-q^{m-n+j}}{1-q^{j}},$ $n\geq 0$,
is the Gaussian polynomial (or $q$-binomial coefficient); see also §§17.2(i)17.2(ii). …
26.9.5 $\sum_{n=0}^{\infty}p_{k}\left(n\right)q^{n}=\prod_{j=1}^{k}\frac{1}{1-q^{j}}=1% +\sum_{m=1}^{\infty}\genfrac{[}{]}{0.0pt}{}{k+m-1}{m}_{q}q^{m},$
26.9.6 $\sum_{n=0}^{\infty}p_{k}\left(\leq m,n\right)q^{n}=\genfrac{[}{]}{0.0pt}{}{m+k% }{k}_{q}.$
26.9.7 $\sum_{m,n=0}^{\infty}p_{k}\left(\leq m,n\right)x^{k}q^{n}=1+\sum_{k=1}^{\infty% }\genfrac{[}{]}{0.0pt}{}{m+k}{k}_{q}x^{k}=\prod_{j=0}^{m}\frac{1}{1-x\,q^{j}}.$