A295765 G.f. satisfies: A(x) = Sum_{n>=0} binomial((n+1)^2,n)/(n+1)^2 * x^n/A(x)^n.

1, 1, 3, 25, 369, 7881, 220845, 7677363, 319307665, 15487290535, 859400072837, 53749578759526, 3743585586509849, 287496351622105328, 24143937833744911767, 2201703647718624364913, 216700738558116024114289, 22900073562659910815354339, 2586409916780162599516986945, 310947096149155992699450689912, 39650252031533561961437812566315

Offset

0,3

Links

Paul D. Hanna, Table of n, a(n) for n = 0..300

Formula

G.f. A(x) satisfies: [x^n] A(x)^(n+1) = binomial((n+1)^2,n)/(n+1) for n>=0.

a(n) ~ c * exp(n) * n^(n - 5/2), where c = 1.56162380971247949723297... - Vaclav Kotesovec, Oct 17 2020

Example

G.f.: A(x) = 1 + x + 3*x^2 + 25*x^3 + 369*x^4 + 7881*x^5 + 220845*x^6 + 7677363*x^7 + 319307665*x^8 + 15487290535*x^9 + 859400072837*x^10 + ...
such that
A(x) = 1 + x/A(x) + 4*(x/A(x))^2 + 35*(x/A(x))^3 + 506*(x/A(x))^4 + 10472*(x/A(x))^5 + 285384*(x/A(x))^6 +...+ binomial((n+1)^2,n)/(n+1)^2*(x/A(x))^n + ...
RELATED SERIES.
Define B(x) = A(x*B(x)) and A(x) = B(x/A(x)) then B(x) begins
B(x) = 1 + x + 4*x^2 + 35*x^3 + 506*x^4 + 7881*x^5 + 220845*x^6 + 7677363*x^7 + 319307665*x^8 + 15487290535*x^9 + ... + binomial((n+1)^2,n)/(n+1)^2*x^n + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k in A(x)^(n+1) begins:
  [1, 1, 3, 25, 369, 7881, 220845, 7677363, 319307665, ...];
  [1, 2, 7, 56, 797, 16650, 460291, 15862152, 655825337, ...];
  [1, 3, 12, 94, 1293, 26409, 719922, 24587202, 1010428347, ...];
  [1, 4, 18, 140, 1867, 37272, 1001476, 33887832, 1384043365, ...];
  [1, 5, 25, 195, 2530, 49366, 1306860, 43802060, 1777652015, ...];
  [1, 6, 33, 260, 3294, 62832, 1638166, 54370836, 2192294775, ...];
  [1, 7, 42, 336, 4172, 77826, 1997688, 65638294, 2629075183, ...];
  [1, 8, 52, 424, 5178, 94520, 2387940, 77652024, 3089164371, ...];
  [1, 9, 63, 525, 6327, 113103, 2811675, 90463365, 3573805950, ...]; ...
in which the main diagonal begins:
  [1, 2, 12, 140, 2530, 62832, 1997688, ..., binomial((n+1)^2,n)/(n+1), ...].

Mathematica

terms = 21; A[_] = 1; Do[A[x_] = Sum[Binomial[(n+1)^2, n]/(n+1)^2*x^n/ A[x]^n, {n, 0, terms}] + O[x]^terms // Normal, terms];
CoefficientList[A[x], x] (* Jean-François Alcover, Jan 14 2018 *)

Prog

(PARI) {a(n) = my(A=[1]); for(m=1, n, A = concat(A, 0); V = Vec( Ser(A)^(m+1) ); A[m+1] = (binomial((m+1)^2, m)/(m+1) - V[m+1])/(m+1); ); A[n+1]}
for(n=0, 20, print1(a(n), ", "))

Crossrefs

cf. A295764, A295763, A143669.

Sequence in context: A129506 A143139 A231637 * A012481 A132617 A241703

Adjacent sequences: A295762 A295763 A295764 * A295766 A295767 A295768

Keyword

nonn

Author

Paul D. Hanna, Jan 06 2018

Status

approved