This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A361714 #22 Jul 11 2024 05:11:21
%S A361714 0,1,7,82,1063,14376,199204,2806770,40053031,577468684,8397778882,
%T A361714 123029274666,1814016998116,26898142793068,400836647993292,
%U A361714 5999796281063082,90162110212198695,1359731143731297396,20571691450059355174,312134224830052880826,4748435338386591995938
%N A361714 a(n) = Sum_{k = 0..n-1} (-1)^(n+k+1)*binomial(n,k)*binomial(n+k-1,k)^2.
%C A361714 Conjecture 1: the supercongruence a(p) == a(1) (mod p^5) holds for all primes p >= 7 (checked up to p = 199).
%C A361714 Conjecture 2: for r >= 2, the supercongruence a(p^r) == a(p^(r-1)) (mod p^(3*r+3)) holds for all primes p >= 7.
%C A361714 Compare with the Apéry numbers A005258(n) = Sum_{k = 0..n} (-1)^(n+k) * binomial(n,k) * binomial(n+k,k)^2, which satisfy the weaker supercongruences A005258(p^r) == A005258(p^(r-1)) (mod p^(3*r)) for positive integer r and all primes p >= 5.
%H A361714 Paolo Xausa, <a href="/A361714/b361714.txt">Table of n, a(n) for n = 0..800</a>
%F A361714 a(n) = binomial(2*n-1,n)^2 - (1/5)*(A005258(n) - 3*A005258(n-1)) for n >= 1.
%F A361714 P-recursive:
%F A361714 (395*n^10 - 6083*n^9 + 39816*n^8 - 144606*n^7 + 318639*n^6 - 436307*n^5 + 362870*n^4 - 167820*n^3 + 33096*n^2)*a(n) = (10665*n^10 - 174906*n^9 + 1243697*n^8 - 5033114*n^7 + 12789951*n^6 - 21235254*n^5 + 23221451*n^4 - 16437246*n^3 + 7182940*n^2 - 1753656*n + 185472)*a(n-1) - (69125*n^10 - 1202775*n^9 + 9159576*n^8 - 40005738*n^7 + 110271201*n^6 - 198723383*n^5 + 234346978*n^4 - 175661976*n^3 + 78402944*n^2 - 18529392*n + 1901088)*a(n-2) - 4*(n - 3)^2*(1580*n^8 - 19592*n^7 + 101515*n^6 - 284307*n^5 + 464411*n^4 - 444309*n^3 + 236490*n^2 - 62500*n + 7000)*a(n-3) with a(0) = 0, a(1) = 1 and a(2) = 7.
%F A361714 a(n) = binomial(2*n-1, n)^2 - (-1)^n*hypergeom([-n, n, n], [1, 1], 1). This is another way to write the first formula. - _Peter Luschny_, Mar 27 2023
%e A361714 Examples of supercongruence:
%e A361714 a(11) - a(1) = 23029274666 - 1 = 5*(11^5)*152783 == 0 (mod 11^5).
%e A361714 a(13) - a(1) = 26898142793068 - 1 = (3^2)*7*(13^5)*1149913 == 0 (mod 13^5).
%e A361714 a(5^2) - a(5) = 3994642669575050040375014376 - 14376 = (2^6)*(3^6)*(5^9)*103* 425601520324429 == 0 (mod 5^9).
%p A361714 seq(add((-1)^(n+k+1)*binomial(n,k)*binomial(n+k-1,k)^2, k = 0..n-1), n = 0..20);
%p A361714 # Alternative:
%p A361714 A361714 := n -> binomial(2*n-1, n)^2 - (-1)^n*hypergeom([-n, n, n], [1, 1], 1):
%p A361714 seq(simplify(A361714(n)), n = 0..20); # _Peter Luschny_, Mar 27 2023
%t A361714 A361714[n_] := Binomial[2*n-1, n]^2 - (-1)^n*HypergeometricPFQ[{-n, n, n}, {1, 1}, 1]; Array[A361714, 20, 0] (* _Paolo Xausa_, Jul 11 2024 *)
%Y A361714 Cf. A005258, A352654, A361712, A361713, A361715, A361717.
%K A361714 nonn,easy
%O A361714 0,3
%A A361714 _Peter Bala_, Mar 21 2023