cp's OEIS Frontend

A375178 a(n) = Sum_{k = 0..n-1} binomial(n+k-1, k)^3 (same as A112028 with an extra 0 at the start).

%I A375178 #22 Aug 17 2024 22:48:18
%S A375178 0,1,9,244,9065,389376,18188478,897376152,46011772521,2427553965160,
%T A375178 130930630643384,7186614533569296,400132290102421214,
%U A375178 22543708920891189136,1282873288801683197250,73628947696550668509744,4257138240245923453355625,247733479854085081062353400,14498252738780732999484606360
%N A375178 a(n) = Sum_{k = 0..n-1} binomial(n+k-1, k)^3 (same as A112028 with an extra 0 at the start).
%C A375178 Compare with the identity Sum_{k = 0..n-1} binomial(n+k-1, k) = (1/2) * binomial(2*n, n) = (1/2) * A000984(n) for n >= 1.
%C A375178 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 A375178 For prime p, binomial(p+k-1, k) == 0 (mod p) for 1 <= k <= p-1. It follows that a(p) == 1 (mod p^3) for all primes p. We conjecture that, in fact, the stronger congruence a(p) == 1 (mod p^5) holds for all primes p >= 7.
%C A375178 Further, we conjecture that for r >= 2 and prime p >= 5, a(p^r) == a(p^(r-1)) (mod p^(3*r+3)).
%C A375178 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} binomial(n+k-1, k)^(2*m+1). Then the congruence b_m(p) == 1 (mod p^(2*m+1)) clearly holds for all primes p. We conjecture that the stronger 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.
%C A375178 Essentially a duplicate of A112028.
%H A375178 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 A375178 a(n) = Sum_{k = 0..n-1} (-1)^k * binomial(-n, k)^3.
%F A375178 a(n) ~ 2^(6*n-3)/(7*Pi^(3/2)*n^(3/2)). - _Vaclav Kotesovec_, Aug 03 2024
%e A375178 Examples of supercongruences:
%e A375178 a(7) - a(1) = 897376152 - 1 = (7^5)*107*499 == 0 (mod 7^5)
%e A375178 a(11) - a(1) = 7186614533569296 - 1 = 5*(11^5)*8924644409 == 0 (mod 11^5).
%p A375178 seq(add( binomial(n+k-1, k)^3, k = 0..n-1), n = 0..20);
%Y A375178 Cf. A000984, A010763, A112028, A176335, A375179, A375180.
%K A375178 nonn,easy
%O A375178 0,3
%A A375178 _Peter Bala_, Aug 03 2024