A375180 a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(3*n, k)^3.

0, 1, 215, 45928, 10362231, 2450260001, 600869373182, 151570671244560, 39096342054496887, 10267275084850974619, 2736324289110748127715, 738255282011665067114400, 201254884472471159485086750, 55352399437924814524429123488, 15341068552569688728602977821596

Offset: 0

Author

Peter Bala, Aug 05 2024

Keywords

• nonn
• easy

Comments

Compare with the identity Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(3*n, k) = (1/3) * binomial(3*n, n) = (1/3) * A005809(n) for n >= 1.

The binomial coefficients satisfy the supercongruence (1/3) * binomial(3*p, p) == 1 (mod p^3) for all primes p >= 5 (Meštrović, Equation 35).

We conjecture that for the present sequence the stronger supercongruence 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} (-1)^(n+k+1)*binomial(3*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.

Examples

			Examples of supercongruences:
a(7) - a(1) = 151570671244560 - 1 = (7^5)*379*2269*10487 == 0 (mod 7^5);
a(13) - a(1) = 55352399437924814524429123488 - 1 = (13^5)*149080105032749915900459 == 0 (mod 13^5).
		

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. A005809, A375178, A375179.

Programs

Maple

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

Formula

a(n) ~ 3^(9*n - 1/2) / (Pi^(3/2) * n^(3/2) * 2^(6*n+3)). - Vaclav Kotesovec, Aug 08 2024