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

0, 1, 9, 244, 9065, 389376, 18188478, 897376152, 46011772521, 2427553965160, 130930630643384, 7186614533569296, 400132290102421214, 22543708920891189136, 1282873288801683197250, 73628947696550668509744, 4257138240245923453355625, 247733479854085081062353400, 14498252738780732999484606360

Offset: 0

Author

Peter Bala, Aug 03 2024

Keywords

• nonn
• easy

Comments

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.

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).

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.

Further, we conjecture that for r >= 2 and prime p >= 5, a(p^r) == a(p^(r-1)) (mod p^(3*r+3)).

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.

Essentially a duplicate of A112028.

Examples

			Examples of supercongruences:
a(7) - a(1) = 897376152 - 1 = (7^5)*107*499 == 0 (mod 7^5)
a(11) - a(1) = 7186614533569296 - 1 = 5*(11^5)*8924644409 == 0 (mod 11^5).
		

Links

• Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2012), arXiv:1111.3057 [math.NT], (2011).

Crossrefs

Cf. A000984, A010763, A112028, A176335, A375179, A375180.

Programs

Maple

seq(add( binomial(n+k-1, k)^3, k = 0..n-1), n = 0..20);

Formula

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

a(n) ~ 2^(6*n-3)/(7*Pi^(3/2)*n^(3/2)). - Vaclav Kotesovec, Aug 03 2024