A361688 a(n) = A361540(2*n,n) / binomial(2*n,n) for n >= 0.

1, 2, 71, 10915, 4063645, 2842101221, 3255178907803, 5605980824208871, 13710496284516264953, 45746570903514799640905, 202291094041887013214628871, 1160411497892246920315488823067, 8496377826955803443098054623140629, 78398366060939693412478828210386035725

Offset: 0

Author

Paul D. Hanna, Mar 20 2023

Keywords

• nonn

Comments

E.g.f. F(x,y) of triangle A361540 satisfies the following.

(1) F(x,y) = Sum_{n>=0} (F(x,y)^n + y)^n * x^n/n!.

(2) F(x,y) = Sum_{n>=0} F(x,y)^(n^2) * exp(y*x*F(x,y)^n) * x^n/n!.

This sequence equals the central terms of triangle A361540 divided by the central binomial coefficients A000984.

Examples

			E.g.f. A(x) = 1 + 2*x + 71*x^2/2! + 10915*x^3/3! + 4063645*x^4/4! + 2842101221*x^5/5! + 3255178907803*x^6/6! + 5605980824208871*x^7/7! + 13710496284516264953*x^8/8! + ... + a(n)*x^n/n! + ...
		

Links

• Paul D. Hanna, Table of n, a(n) for n = 0..25

Crossrefs

Cf. A361540, A000984.

Programs

PARI

/* E.g.f. of triangle A361540 is F(x,y) = Sum_{n>=0} (F(x,y)^n + y)^n * x^n/n! */
{A361540(n,k) = my(F = 1); for(i=1,n, F = sum(m=0, n, (F^m + y +x*O(x^n))^m * x^m/m! )); n!*polcoeff(polcoeff(F, n,x),k,y)}
{a(n) = A361540(2*n,n)/binomial(2*n,n)}
for(n=0, 15, print1(a(n), ", "))