$\pi$的幂的新级数及相关同余式
数论
2020-07-17 v9 组合数学
摘要
通过符号计算，我们推导出97个与Ramanujan型级数相关的π \pi π ∑ k = 0 ∞ P ( k ) ( 2 k k ) ( 3 k k ) ( 6 k 3 k ) ( k + 1 ) ( 2 k − 1 ) ( 6 k − 1 ) ( − 640320 ) 3 k = 18 × 557403 3 10005 5 π \sum_{k=0}^\infty \frac{P(k) \binom{2k}k\binom{3k}k \binom{6k}{3k}}{(k+1)(2k-1)(6k-1)(-640320)^{3k}} =\frac{18\times557403^3\sqrt{10005}}{5\pi} k = 0 ∑ ∞ ( k + 1 ) ( 2 k − 1 ) ( 6 k − 1 ) ( − 640320 ) 3 k P ( k ) ( k 2 k ) ( k 3 k ) ( 3 k 6 k ) = 5 π 18 × 55740 3 3 10005 ∑ k = 1 ∞ ( 3 k + 1 ) 16 k ( 2 k + 1 ) 2 k 3 ( 2 k k ) 3 = π 2 − 8 2 , \sum_{k=1}^\infty \frac{(3k+1)16^k}{(2k+1)^2k^3\binom{2k}k^3} = \frac{\pi^2-8}2, k = 1 ∑ ∞ ( 2 k + 1 ) 2 k 3 ( k 2 k ) 3 ( 3 k + 1 ) 1 6 k = 2 π 2 − 8 , ∑ n = 0 ∞ 3 n + 1 ( − 100 ) n ∑ k = 0 n ( n k ) 2 T k ( 1 , 25 ) T n − k ( 1 , 25 ) = 25 8 π , \sum_{n=0}^\infty\frac{3n+1}{(-100)^n} \sum_{k=0}^n{n\choose k}^2T_k(1,25)T_{n-k}(1,25) = \frac{25}{8\pi}, n = 0 ∑ ∞ ( − 100 ) n 3 n + 1 k = 0 ∑ n ( k n ) 2 T k ( 1 , 25 ) T n − k ( 1 , 25 ) = 8 π 25 , T k ( b , c ) T_k(b,c) T k ( b , c ) ( x 2 + b x + c ) k (x^2+bx+c)^k ( x 2 + b x + c ) k x k x^k x k 1 / π 1/\pi 1/ π π \pi π ∑ k = 0 ∞ 39480 k + 7321 ( − 29700 ) k T k ( 14 , 1 ) T k ( 11 , − 11 ) 2 = 6795 5 π . \sum_{k=0}^\infty\frac{39480k+7321}{(-29700)^k}T_k(14,1)T_k(11,-11)^2=\frac{6795\sqrt5}{\pi}. k = 0 ∑ ∞ ( − 29700 ) k 39480 k + 7321 T k ( 14 , 1 ) T k ( 11 , − 11 ) 2 = π 6795 5 .
引用
@article{arxiv.1911.05456,
title = {New series for powers of $\pi$ and related congruences},
author = {Zhi-Wei Sun},
journal= {arXiv preprint arXiv:1911.05456},
year = {2020}
}
备注
70 pages, final published version