A183068 Central terms of triangle A183065.
1, 26, 3246, 606500, 137915470, 35218238076, 9702014515116, 2818627826459016, 851612982884556750, 265166341958122567820, 84556145346599067308596, 27489903606068331188121816, 9081510922185418532993154796
Offset: 0
Links
- Peter Bala, A recurrence for A183068
Programs
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Maple
seq(simplify(binomial(2*n, n)*hypergeom([-n, -n, n+1, n+1/2], [1, 1, 1], 4)), n = 0..20);
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PARI
{a(n)=polcoeff(polcoeff(sum(m=0,2*n,(4*m)!/m!^4*x^(2*m)*y^m/(1-x-x*y+x*O(x^(2*n)))^(4*m+1)),2*n,x),n,y)}
Formula
a(n) = A183065(2*n,n).
a(n) = [x^(2*n)*y^n] Sum_{m >= 0} (4*m)!/m!^4 * x^(2*m)*y^m/(1-x-x*y)^(4*m+1).
From Peter Bala, Jul 15 2024: (Start)
a(n) = binomial(2*n, n)*Sum_{k = 0..n} binomial(n, k)^2*binomial(2*n+2*k, 2*k)* binomial(2*k, k) = Sum_{k = 0..n} (2*n+2*k)!/(k!^4*(n-k)!^2). Cf. A002897(n) = Sum_{k = 0..n} (2*n+k)!/(k!^3*(n-k)!^2) and A005258(n) = n!*Sum_{k = 0..n} (n+k)!/(k!^3*(n-k)!^2).
a(n) = binomial(2*n, n)*hypergeom([-n, -n, n+1, n+1/2], [1, 1, 1], 4).
Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(2*r)) holds for all primes p and positive integers n and r (End)
a(n) ~ sqrt(3) * (2 + sqrt(6))^(4*n + 3/2) / (16*Pi^2*n^2). - Vaclav Kotesovec, Jul 16 2024