OFFSET
0,2
FORMULA
a(n) = [x^n] ( (1+x)^4/(1-x)^4 )^n.
The g.f. exp( Sum_{k>=1} a(k) * x^k/k ) has integer coefficients and equals (1/x) * Series_Reversion( x*(1-x)^4/(1+x)^4 ). See A365847.
From Peter Bala, Jul 20 2024: (Start)
a(n) = binomial(5*n-1, n)*hypergeom([-n, -4*n], [1 - 5*n], -1).
For n >=1, a(n) = (4/3) * [x^n] S(x)^(3*n) = (4/5) * [x^n] (1/S(-x))^(5*n), where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. of the sequence of large Schröder numbers A006318.
n*(4*n - 3)*(2*n - 1)*(4*n - 1)*(85*n^4 - 510*n^3 + 1138*n^2 - 1119*n + 409)*a(n) = 2*(29665*n^8 - 237320*n^7 + 794282*n^6 - 1443212*n^5 + 1544750*n^4 - 987560*n^3 + 363568*n^2 - 69168*n + 5040)*a(n-1) + (n - 2)*(4*n - 7)*(2*n - 3)*(4*n - 5)*(85*n^4 - 170*n^3 + 118*n^2 - 33*n + 3)*a(n-2) with a(0) = 1 and a(1) = 8.
The Gauss congruences hold: a(n*p^r) == a(n*p^(r-1)) (mod p^r) for all primes p and positive integers n and r.
Conjecture: the supercongruences a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) hold for all primes p >= 5 and positive integers n and r. (End)
a(n) ~ (349 + 85*sqrt(17))^n / (17^(1/4) * sqrt(Pi*n) * 2^(5*n - 1/2)). - Vaclav Kotesovec, Aug 08 2024
MAPLE
seq(simplify(binomial(5*n-1, n)*hypergeom([-n, -4*n], [1 - 5*n], -1)), n = 0..20); # Peter Bala, Jul 29 2024
PROG
(PARI) a(n) = sum(k=0, n, binomial(4*n, k)*binomial(5*n-k-1, n-k));
KEYWORD
nonn,easy
AUTHOR
Seiichi Manyama, Feb 10 2024
STATUS
approved