OFFSET
0,2
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..939
FORMULA
a(n) = Sum_{k=0..n} 2^(n-k) * binomial(2*k,k) * binomial(2*n,2*k).
a(0) = 1, a(1) = 4 and n * (2*n-1) * (4*n-5) * a(n) = (4*n-3) * (24*n^2-36*n+8) * a(n-1) - 4 * (n-1) * (2*n-3) * (4*n-1) * a(n-2) for n > 1. - Seiichi Manyama, Aug 28 2020
a(n) ~ 2^(n - 5/4) * (1 + sqrt(2))^(2*n + 1/2) / sqrt(Pi*n). - Vaclav Kotesovec, Aug 31 2020
From Peter Bala, May 02 2022: (Start)
Conjecture: a(n) = [x^n] ( (1 + x^2)*(1 + x)^2/(1 - x)^2 )^n. Equivalently, a(n) = Sum_{k = 0..n} Sum_{j = 0..n-2*k} binomial(n,k)*binomial(2*n,j)*binomial(3*n-2*k-j-1,n-2*k-j).
If the conjecture is true then the Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p and positive integers n and k. Calculation suggests that the supercongruences a(n*p^k) == a(n*p^(k-1)) (mod p^(2*k)) hold for primes p >= 3 and positive integers n and k.
It appears that a(n)^2 = Sum_{k = 0..2*n} (-1)^k*2^(2*n-k)*binomial(2*k,k)^2* binomial(2*n+k,2*k). Compare with the pair of identities: binomial(2*n,n) = Sum_{k = 0..n} 2^(n-2*k)*binomial(2*k,k)*binomial(n,2*k) and binomial(2*n,n)^2 = Sum_{k = 0..2*n} (-1)^k*2^(4*n-2*k)*binomial(2*k,k)^2*binomial(2*n+k,2*k). (End)
MATHEMATICA
a[n_] := Sum[2^(n - k) * Binomial[2*k, k] * Binomial[2*n, 2*k], {k, 0, n}]; Array[a, 21, 0] (* Amiram Eldar, Aug 25 2020 *)
PROG
(PARI) N=40; x='x+O('x^N); Vec(sqrt((1-2*x+sqrt(1-12*x+4*x^2))/(2*(1-12*x+4*x^2))))
(PARI) {a(n) = sum(k=0, n, 2^(n-k)*binomial(2*k, k)*binomial(2*n, 2*k))}
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Seiichi Manyama, Aug 25 2020
STATUS
approved