A361712 a(n) = Sum_{k = 0..n-1} binomial(n,k)^2*binomial(n+k,k)*binomial(n+k-1,k).

0, 1, 25, 649, 16921, 448751, 12160177, 336745053, 9513822745, 273585035755, 7988828082775, 236367018090017, 7072779699975601, 213701611408357567, 6511338458568750853, 199850727914988936149, 6173376842290368719385, 191776434791965521115235, 5987554996434696230487955

Offset

0,3

Comments

Conjecture 1: the supercongruence a(p) == a(1) (mod p^5) holds for all primes p >= 7 (checked up to p = 199).

Conjecture 2: for r >= 2, the supercongruence a(p^r) == a(p^(r-1)) (mod p^(3*r+3) holds for all primes p >= 5.

Compare with the Apéry numbers A005259(n) = Sum_{k = 0..n} binomial(n,k)^2 * binomial(n+k,k)^2, which satisfy the weaker supercongruences A005259(p^r) == A005259(p^(r-1)) (mod p^(3*r)) for all primes p >= 5.

Links

Table of n, a(n) for n=0..18.

Peter Bala, Recurrence equation for A361712

Formula

a(n) = (1/12)*(7*A005259(n) + A005259(n-1)) - (1/2)*binomial(2*n,n)^2.

a(n) ~ 2^(1/4)*(1 + sqrt(2))^(4*n+1)/(4*Pi^(3/2)*n^(3/2)).

a(n) = hypergeom([-n, -n, n, n + 1], [1, 1, 1], 1) - binomial(2*n, n)*binomial(2*n - 1, n) = A361878(n) - A361877(n). - Peter Luschny, Mar 27 2023

Example

a(7) - a(1) = (2^2)*(7^5)*5009 == 0 (mod 7^5)
a(11) - a(1) = (2^5)*(11^5)*45864163 == 0 (mod 11^5)
a(7^2) - a(7) = (2*3)*(7^9)*377052719*240136524699189343838527* 17965610580703155723668147409587 == 0 (mod 7^9)

Maple

seq(add(binomial(n, k)^2*binomial(n+k, k)*binomial(n+k-1, k), k = 0..n-1), n = 0..25);
# Alternative:
A361712 := n -> hypergeom([-n, -n, n, n + 1], [1, 1, 1], 1) - binomial(2*n, n)*binomial(2*n-1, n): seq(simplify(A361712(n)), n = 0..18); # Peter Luschny, Mar 27 2023

Crossrefs

Cf. A005259, A212334, A361713, A361714, A361715, A3617157.

Cf. A361877, A361878.

Sequence in context: A260048 A152256 A153111 * A097194 A180811 A318183

Adjacent sequences: A361709 A361710 A361711 * A361713 A361714 A361715

Keyword

nonn,easy

Author

Peter Bala, Mar 21 2023

Status

approved