カタラン数と中心二項係数の等式一覧(結果のみ)

\begin{aligned} \frac{1}{\sqrt{1-4x}}&=\sum_{n=0}^{\infty}\binom{2n}{n}x\\ \frac{1-\sqrt{1-4x}}{2x}&=\sum_{n=0}^{\infty}\frac{1}{n+1}\binom{2n}{n}x^n \end{aligned}

\begin{aligned} \frac{1}{\sqrt{1-4x}}\bigg(\frac{1-\sqrt{1-4x}}{2x}\bigg)^k&=\sum_{n=0}^{\infty}\binom{2n+k}{n}x^n\\ \bigg(\frac{1-\sqrt{1-4x}}{2x}\bigg)^k&=\sum_{n=0}^{\infty}\frac{k}{n+k}\binom{2n+k-1}{n}x^n \end{aligned}

\begin{aligned} \sum_{k=0}^{n}\binom{2k}{k}\binom{2(n-k)}{n-k}&=4^n\\ \sum_{k=0}^{n}\binom{2k}{k}\binom{2(n-k)+1}{n-k}&=\frac{1}{2}\bigg(4^{n+1}-\binom{2(n+1)}{n+1}\bigg)\\ \sum_{k=0}^{n}\binom{2k}{k}\binom{2(n-k)+2}{n-k}&=4^{n+1}-\frac{1}{2}\binom{2(n+2)}{n+2}\\ \sum_{k=0}^{n}\binom{2k+1}{k}\binom{2(n-k)+1}{n-k}&=4^{n+1}-\frac{1}{2}\binom{2(n+2)}{n+2}\\ \end{aligned}

\begin{aligned} \sum_{k=0}^{n}\bigg(\binom{2k+1}{k}+\binom{2k+2}{k}\bigg)&=\frac{1}{2}\binom{2(n+2)}{n+2}-1\\ \sum_{k=0}^{n}2^{n-k}\binom{2k+2}{k}&=\frac{1}{2}\binom{2(n+2)}{n+2}-2^{n+1}\\ \sum_{k=0}^{n}3^{n-k}\binom{2k+3}{k}&=\frac{1}{2}\binom{2(n+3)}{n+3}-3^{n+2}\\ \sum_{k=0}^{n}a_{n-k}\binom{2k+4}{k}&=\frac{1}{2}\bigg(\binom{2(n+4)}{n+4}-a_{n+4}+2a_{n+3}\bigg)\\ \sum_{k=0}^{n}b_{n-k}\binom{2k+5}{k}&=\frac{1}{2}\bigg(\binom{2(n+5)}{n+5}-b_{n+5}+3b_{n+4}-b_{n+3}\bigg)\\ \end{aligned}
\begin{aligned} a_n&:=\frac{(2+\sqrt{2})^{n+1}-(2-\sqrt{2})^{n+1}}{2\sqrt{2}}\\ b_n&:=\frac{\displaystyle\bigg(\frac{5+\sqrt{5}}{2}\bigg)^{n+1}-\bigg(\frac{5-\sqrt{5}}{2}\bigg)^{n+1}}{\sqrt{5}} \end{aligned}

\begin{aligned} \sum_{k=0}^{n}\frac{3}{k+3}\binom{2k+2}{k}&=\frac{1}{n+3}\binom{2(n+2)}{n+2}-1\\ \sum_{k=0}^{n}2^{n-k}\frac{4}{k+4}\binom{2k+3}{k}&=\frac{1}{n+4}\binom{2(n+3)}{n+3}-2^{n+2}\\ \sum_{k=0}^{n}a_{n-k}\frac{5}{k+5}\binom{2k+4}{k}&=\frac{1}{n+5}\binom{2(n+4)}{n+4}+\frac{a_{n+5}-5a_{n+4}+5a_{n+3}}{2}\\ \sum_{k=0}^{n}\frac{3^{n-k+1}-1}{2}\frac{6}{k+6}\binom{2k+5}{k}&=\frac{1}{n+6}\binom{2(n+5)}{n+5}-\frac{3^{n+4}+1}{2}\\ \end{aligned}
\begin{aligned} a_{n}:=\frac{\displaystyle\bigg(\frac{3+\sqrt{5}}{2}\bigg)^{n+1}-\displaystyle\bigg(\frac{3-\sqrt{5}}{2}\bigg)^{n+1}}{\sqrt{5}} \end{aligned}

\begin{aligned} \sum_{k=0}^{n}k2^k\frac{1}{n}\binom{2n-1-k}{n-1}&=\binom{2n}{n}\\ \sum_{k=0}^{n}k(k+1)2^k\frac{1}{n}\binom{2n-1-k}{n-1}&=4^n\\ \sum_{k=0}^{n}2^k\binom{2n-k}{n}&=4^n \end{aligned}

\begin{aligned} \sum_{k=0}^{n}\frac{2k+1}{n+k+1}\binom{2n}{n-k}&=\binom{2n}{n}\\ \sum_{k=0}^{n}\frac{2(k+1)^2}{n+k+2}\binom{2n+1}{n-k}&=4^n \end{aligned}

\begin{aligned} &\lbrack x^ny^m \rbrack \frac{1}{\sqrt{1-4x}\sqrt{1-4y}(\sqrt{1-4x}+\sqrt{1-4y})}=\frac{1}{4}\binom{2(n+m+1)}{n+m+1}\\ &\sum_{k=0}^{\lfloor \frac{n}{2}\rfloor}\frac{1}{(k!)^2}\frac{2^{n-2k}}{(n-2k)!}=\frac{\displaystyle\binom{2n}{n}}{n!}\\ &\sum_{k=0}^n\frac{\displaystyle\binom{2k}{k}}{k!}\cdot \frac{\displaystyle(-1)^{n-k}\binom{2(n-k)}{n-k}}{(n-k)!}= \begin{cases} 0&(n\equiv 1 \pmod{2})\\\\ \displaystyle\frac{\displaystyle\binom{n}{n/2}^2}{n!}&(n\equiv 0 \pmod{2})\\\\ \end{cases}\\ &\bigg(\frac{3}{4}\bigg)^n\sum_{k=0}^{n}\bigg(-\frac{1}{3}\bigg)^k\binom{2k}{k}\binom{2(n-k)}{n-k}=\lbrack x^0 \rbrack (x+1+x^{-1})^n\\ \end{aligned}