A361713 a(n) = Sum_{k = 0..n-1} binomial(n,k)^2 * binomial(n+k-1,k)^2.

0, 1, 17, 406, 10257, 268126, 7213166, 198978074, 5609330705, 161095277710, 4700175389142, 138986764820410, 4157185583199534, 125568602682092818, 3825026187780837266, 117376010145070696906, 3625095243230562818065, 112596592142021739522670, 3514965607470183733302470

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^(4*r+1)) holds for all primes p >= 7.

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 positive integers r and all primes p >= 5.

Links

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

Peter Bala, Recurrence equation for A361713

Formula

a(n) = (1/3)*(A005259(n) + A005259(n-1)) - (1/4)*binomial(2*n,n)^2 = A177316(n) - A060150(n).

a(n) ~ C*(12*sqrt(2) + 17)^n/n^(3/2), where C = 1/(2^(5/4)*Pi^(3/2)).

a(n) = hypergeom([-n, -n, n, n], [1, 1, 1], 1) - binomial(2*n-1, n)^2. This is another way to write the first formula. - Peter Luschny, Mar 27 2023

Maple

seq(add(binomial(n, k)^2*binomial(n+k-1, k)^2, k = 0..n-1), n = 0..25);
# Alternative:
A361713 := n -> hypergeom([-n, -n, n, n], [1, 1, 1], 1) - binomial(2*n - 1, n)^2:
seq(simplify(A361713(n)), n = 0..18); # Peter Luschny, Mar 27 2023

Crossrefs

Cf. A005259, A060150, A177316, A212334, A361712, A361714, A361715, A361717.

Sequence in context: A340353 A007925 A097201 * A053106 A158541 A114357

Adjacent sequences: A361710 A361711 A361712 * A361714 A361715 A361716

Keyword

nonn,easy

Author

Peter Bala, Mar 21 2023

Status

approved