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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

From Seiichi Manyama, Aug 09 2025: (Start)

a(n) = [x^n] (1-x)^(n-1)/(1-2*x)^(4*n).

a(n) = Sum_{k=0..n} 2^k * binomial(4*n,k) * binomial(n-1,n-k).

a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(4*n+k-1,k) * binomial(n-1,n-k). (End)

a(n) = [x^n] ( (1+x)^3/(1-x)^2 )^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)^2/(1+x)^3 ). See A365842.

a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(3*n,k). - Seiichi Manyama, Jul 31 2025

a(n) ~ 3^(3*n + 1/2) / (5 * sqrt(Pi*n) * 2^(n-1)). - Vaclav Kotesovec, Jul 31 2025

a(n) = Sum_{k=0..n} 2^k * binomial(2*n+k-1,k). - Seiichi Manyama, Aug 01 2025

a(n) = [x^n] 1/((1-x) * (1-2*x)^(2*n)). - Seiichi Manyama, Aug 09 2025

a(n) = [x^n] 1/( (1+x)^2 * (1-x)^3 )^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)^2*(1-x)^3 ). See A365854.

a(n) = Sum_{k=0..floor(n/2)} binomial(3*n+k-1,k) * binomial(n,n-2*k).

a(n) = Sum_{k=0..floor(n/2)} binomial(2*n+k-1,k) * binomial(2*n-2*k-1,n-2*k).

a(n) = [x^n] ( (1+x)^2/(1-x)^2 )^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)^2/(1+x)^2 ).

a(n) = 2 * A103885(n) for n >= 1. - Peter Bala, Sep 16 2024

From Seiichi Manyama, Aug 09 2025: (Start)

a(n) = [x^n] (1-x)^(n-1)/(1-2*x)^(2*n).

a(n) = Sum_{k=0..n} 2^k * binomial(2*n,k) * binomial(n-1,n-k).

a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(2*n+k-1,k) * binomial(n-1,n-k). (End)