フィボナッチ数から円周率を作る自作の式たち

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{F_{n+3}^2-F_n{F}_{n+6}}{{\log }_{\phi }({\phi }^{2n-1}-F_{2n}-F_{2n-2})}=\pi }$

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\left\{\frac{10i}{3}(\log 2-\log (\sqrt{\frac{2F_{n+1}^2-3F_n^2}{F_{n+1}{F}_n}}+\frac{i{F}_{n+1}}{F_n}))\right\}=\pi }$

${\displaystyle 10\sum _{n=0}^{\mathrm{\infty }}\frac{(2n-1)!!({\phi }^{2n+1}-F_{2n+2}-F_{2n})}{(2n)!!2^{2n+1}(2n+1)}=\pi }$

${\displaystyle \sum _{n=0}^{\mathrm{\infty }}(\genfrac{}{}{0ex}{}{2n}{n})\frac{10}{{16}^n(2n+1)(F_{2n}+F_{2n+2}+F_{2n+1}\sqrt{5})}=\pi }$

${\displaystyle 4\sqrt{5}\sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n{F}_n(\frac{1}{2}{\left(\frac{1}{4\phi }\right)}^n+\frac{1}{3}{\left(\frac{1}{9\phi }\right)}^n)}{2n+1}+2\phi \log \frac{11\phi +7}{\phi +7}=\pi }$

${\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{4(-1{)}^n{F}_{n+1}}{(2n+1)(2n+3)(F_{n-1}+F_{n+1}+\sqrt{5}{F}_n)}(\frac{(10n+21)+(2n+1)\sqrt{5}}{2^{2n+3}}+\frac{(30n+51)+(2n+1)\sqrt{5}}{3^{2n+3}})=\pi }$

${\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{3(2n-1)!!(\phi F_{n+1}+F_n)}{(8n)!!!!!!!!{\phi }^{n+1}(2n+1)}=\pi }$

${\displaystyle \sum _{k=1}^{\mathrm{\infty }}\sum _{j=1}^{\mathrm{\infty }}\frac{4(-1{)}^{j-1}}{2j-1}{\left(\frac{1}{F_{2k+1}}\right)}^{2j-1}=\pi }$

フィボナッチ数列の奇数項から円周率
${\displaystyle \frac{5\sqrt{5}}{2}\sum _{n=0}^{\mathrm{\infty }}\frac{(2n-1)!!(2n-1)!!F_{2n+1}}{(2n+1)!2^{2n+1}}=\pi }$

フィボナッチ数列の偶数項から円周率
${\displaystyle 5\sqrt{\frac{5-\sqrt{5}}{2}}\sum _{n=0}^{\mathrm{\infty }}\frac{(2n)!!(2n+2)!!F_{2n+2}}{(2n+2)!2^{2n+2}}=\pi }$

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}4{\tan }^{-1}\frac{F_{4n-1}+F_{4n+1}}{F_{4n}^2}=\pi }$

${\displaystyle \sum _{n=0}^{\mathrm{\infty }}(-1{)}^{n}4{\tan }^{-1}\frac{F_{4n+1}+F_{4n+5}}{F_{4n+3}^2}=\pi }$

${\displaystyle \sum _{n=0}^{\mathrm{\infty }}4{\tan }^{-1}\frac{3(F_{8n+5}+F_{8n+7}+F_{8n+9}+F_{8n+11})}{F_{8n+8}^2}=\pi }$

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}4{\tan }^{-1}\frac{F_{2m}(F_{2m(2k+1)}-F_{2m(2n-1)})}{F_{4mn}^2}=\pi }$

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}(\frac{4{({\tan }^{-1}\frac{F_{4n+3}+F_{4n+5}}{F_{4n+4}^2})}^2}{{\tan }^{-1}\frac{F_{4n+3}+F_{4n+5}}{F_{4n+4}^2}-{\tan }^{-1}\frac{F_{4n+7}+F_{4n+9}}{F_{4n+8}^2}}+\sum _{k=1}^{n}4{\tan }^{-1}\frac{F_{4k-1}+F_{4k+1}}{F_{4k}^2})=\pi }$

${\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{2\sqrt{2}(2n-1)!!(\sqrt{\phi }{F}_{n+1}+\sqrt{{\phi }^{-1}}{F}_n){\left(\sqrt{{\phi }^{-1}}\right)}^{2n+1}}{(4n)!!!!(2n+1)}=\pi }$

${\displaystyle \frac{1}{\phi }(5+10\sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^n}{1-(10n{)}^2})=\pi }$

$\{a_n\}$ を次の漸化式で定義される数列とします。

　　$\{\begin{array}{l}{a}_0=2\\ a_1=10\\ a_n=10a_{n-1}-5a_{n-2}(n\ge 2)\end{array}$

最初の数項はこのようになります。
　　$\{a_n\}=\{2,10,90,850,8050,76250,\dots \}$

また、記号$\{\cdot \}$で小数部分を表すものとします。
　　　　$\text{e.g.}\{1.618\}=0.618$

このとき、
${\displaystyle \sum _{n=0}^{\mathrm{\infty }}\frac{(-1{)}^n\cdot 20\sqrt{5-2\sqrt{5}}}{(2\sqrt{5}{a}_n-5\cdot \left\{\frac{(5+2\sqrt{5})a_n}{10}\right\})(2n+1)}=\pi }$