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A364510
a(n) = binomial(4*n, n)^2.
4
1, 16, 784, 48400, 3312400, 240374016, 18116083216, 1401950721600, 110634634890000, 8862957169158400, 718528370729238784, 58818762721626513424, 4853704694918904043024, 403242220875862752160000, 33694913171561404510440000, 2829611125043050701300998400
OFFSET
0,2
LINKS
FORMULA
a(n) = Sum_{i = -n..n} (-1)^i * binomial(2*n, n+i)^2 * binomial(4*n, 2*n+i).
Compare with Dixon's identity: Sum_{i = -n..n} (-1)^i * binomial(2*n, n+i)^3 = (3*n)!/n!^3.
a(n) = A005810(n)^2.
P-recursive: a(n) = 16 * ( (4*n - 1)*(4*n - 2)*(4*n - 3)/(3*n*(3*n - 1)*(3*n - 2)) )^2 * a(n-1) with a(0) = 1.
a(n) ~ c^n * 2/(3*Pi*n), where c = (2^16)/(3^6).
a(n) = [x^n] G(x)^(16*n), where the power series G(x) = 1 + x + 9*x^2 + 225*x^3 + 7525*x^4 + 295228*x^5 + 12787152*x^6 + 592477457*x^7 + 28827755219*x^8 + ... appears to have integer coefficients.
exp( Sum_{n > = 1} a(n)*x^n/n ) = F(x)^16, where the power series F(x) = 1 + x + 25*x^2 + 1033*x^3 + 53077*x^4 + 3081944*x^5 + 193543624*x^6 + 12835533333*x^7 + 886092805699*x^8 + ... appears to have integer coefficients.
The supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3^r)) hold for all primes p >= 5 and all positive integers n and r.
a(n) = [x^(3*n)] ( (1 - x)^(6*n)*Legendre_P(2*n, (1 + x)/(1 - x)) ). - Peter Bala, Aug 14 2023
MAPLE
seq( binomial(4*n, n)^2, n = 0..15);
MATHEMATICA
A364510[n_]:=Binomial[4n, n]^2; Array[A364510, 15, 0] (* Paolo Xausa, Oct 05 2023 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, Jul 28 2023
STATUS
approved