cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A375179 a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(2*n, k)^3.

This page as a plain text file.
%I A375179 #12 Aug 14 2024 08:36:54
%S A375179 0,1,63,3160,154175,7623126,385867944,20012582304,1061140695615,
%T A375179 57352579248250,3150527165580938,175455318597459264,
%U A375179 9885323482971453608,562487110567071566800,32279596239888192636000,1866149091367606758188160,108583784076103277424118335,6354063452012073380091209250
%N A375179 a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(2*n, k)^3.
%C A375179 Compare with the identity Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(2*n, k) = (1/2) * binomial(2*n, n) = (1/2) * A000984(n) for n >= 1.
%C A375179 The central binomial coefficients satisfy the supercongruence (1/2) * binomial(2*p, p) == 1 (mod p^3) for all primes p >= 5 (Wolstenholme's theorem).
%C A375179 We conjecture that for the present sequence the stronger supercongruence a(p) == 1 (mod p^5) holds for all primes p >= 7.
%C A375179 Further, we conjecture that for r >= 2 and prime p >= 5, a(p^r) == a(p^(r-1)) (mod p^(3*r+3)).
%C A375179 More generally, for a positive integer m, define a sequence {b_m(n) : n >= 0} by setting b_m(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(2*n, k)^(2*m+1). We conjecture that the supercongruence b_m(p) == 1 (mod p^(2*m+3)) holds for all primes p >= 2*m + 5, and for r >= 2, the supercongruence b_m(p^r) == b_m(p^(r-1)) (mod p^(3*r+2*m+1)) also holds for all primes p >= 2*m + 5.
%H A375179 Romeo Meštrović, <a href="https://arxiv.org/abs/1111.3057">Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2012)</a>, arXiv:1111.3057 [math.NT], (2011).
%F A375179 a(n) ~ 2^(6*n-1) / (Pi^(3/2) * n^(3/2)). - _Vaclav Kotesovec_, Aug 08 2024
%e A375179 Examples of supercongruences:
%e A375179 a(7) - a(1) = 20012582304 - 1 = (7^5)*389*3061 == 0 (mod 7^5);
%e A375179 a(17) - a(1) = 6354063452012073380091209250 - 1 = (17^5)*6007*15725921* 47373254231 == 0 (mod 17^5).
%p A375179 seq(add( (-1)^(n+k+1)*binomial(2*n, k)^3, k = 0..n-1), n = 0..20);
%Y A375179 Cf. A000984, A375178, A375180.
%K A375179 nonn,easy
%O A375179 0,3
%A A375179 _Peter Bala_, Aug 04 2024