A361713 - OEIS

%I #23 Jul 11 2024 05:11:06

%S 0,1,17,406,10257,268126,7213166,198978074,5609330705,161095277710,

%T 4700175389142,138986764820410,4157185583199534,125568602682092818,

%U 3825026187780837266,117376010145070696906,3625095243230562818065,112596592142021739522670,3514965607470183733302470

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

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

%C 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.

%C 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.

%H Paolo Xausa, <a href="/A361713/b361713.txt">Table of n, a(n) for n = 0..650</a>

%H Peter Bala, <a href="/A361713/a361713.pdf">Recurrence equation for A361713</a>

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

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

%F 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

%p seq(add(binomial(n,k)^2*binomial(n+k-1,k)^2, k = 0..n-1), n = 0..25);

%p # Alternative:

%p A361713 := n -> hypergeom([-n, -n, n, n], [1, 1, 1], 1) - binomial(2*n - 1, n)^2:

%p seq(simplify(A361713(n)), n = 0..18); # _Peter Luschny_, Mar 27 2023

%t A361713[n_] := HypergeometricPFQ[{-n, -n, n, n}, {1, 1, 1}, 1] - Binomial[2*n-1, n]^2; Array[A361713, 20, 0] (* _Paolo Xausa_, Jul 11 2024 *)

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

%K nonn,easy

%O 0,3

%A _Peter Bala_, Mar 21 2023