A288470 a(n) = Sum_{k=0..n} binomial(n,k)*binomial(2*n,2*k).
1, 2, 14, 92, 646, 4652, 34124, 253528, 1901638, 14368844, 109208164, 833981128, 6394017436, 49185717752, 379438594136, 2934361958192, 22741538394694, 176582855512588, 1373431963785332, 10698376362421096, 83447762846703796, 651690159076273192, 5095051571420324264, 39874449115469939152, 312350761370734541596
Offset: 0
Links
- Seiichi Manyama, Table of n, a(n) for n = 0..1109
Programs
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Maple
f:= gfun:-rectoproc({n*(2*n-1)*a(n) = (32*(n-2))*(2*n-5)*a(n-3)+(8*(9*n^2-31*n+28))*a(n-2)+(2*(3*n^2+7*n-9))*a(n-1), a(0)=1,a(1)=2, a(2)=14},a(n),remember): map(f, [$0..30]); -
Mathematica
Table[Sum[Binomial[n, k] Binomial[2 n, 2 k], {k, 0, n}], {n, 0, 24}] (* Michael De Vlieger, Jun 09 2017 *) -
PARI
{a(n) = polcoef((-1+(1+x+1/x)^2)^n, 0)} \\ Seiichi Manyama, Nov 21 2019
Formula
a(n) = hypergeom([-n,-n,1/2-n],[1/2,1],-1).
n*(2*n-1)*a(n) = (32*(n-2))*(2*n-5)*a(n-3)+(8*(9*n^2-31*n+28))*a(n-2)+(2*(3*n^2+7*n-9))*a(n-1).
G.f.: sqrt((1-2*x+sqrt(1-8*x))/(2*(1-7*x-8*x^2))).
a(n) ~ 8^n / sqrt(3*Pi*n). - Vaclav Kotesovec, Nov 27 2017
a(n) = Sum_{k=0..floor(n/2)} binomial(2*n,k) * binomial(3*n-2*k-1,n-2*k). - Seiichi Manyama, Feb 13 2024
From Peter Bala, Aug 30 2025: (Start)
n*(2*n - 1)*(3*n - 4)*a(n) = 2*(21*n^3 - 49*n^2 + 33*n - 6)*a(n-1) + 8*(n - 1)*(3*n - 1)*(2*n - 3)*a(n-2) with a(0) = 1 and a(1) = 2.
a(n) = Sum_{0 <= i, j <= n/2} binomial(2*n, j)*binomial(2*n+i-1, i)*binomial(2*n, n- 2*i-2*j) (verified to satisfy the above second-order recurrence using the MulZeil procedure in Doron Zeilberger's MultiZeilberger Maple package).
Equivalently, a(n) = [x^n] ( (1 + x + x^2 + x^3)/(1 - x^2) )^(2*n). Cf. A240688.
The Gauss congruences hold: a(n*p^k) == a(n*p^(k-1)) (mod p^k) for all primes p and positive integers n and k.
Conjecture: the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) hold for primes p >= 5 and positive integers n and k. (End)
Comments