A192478 G.f. satisfies: A(x) = x*Product_{n>=1} 1/(1 - x*A(x)^n).

1, 0, 1, 1, 3, 5, 14, 28, 75, 170, 443, 1076, 2795, 7046, 18398, 47458, 124793, 326905, 865853, 2293103, 6114688, 16327765, 43800590, 117720693, 317452606, 857742087, 2323588888, 6306229600, 17151172495, 46725729232, 127522337636, 348562660876

Offset: 1

Author

Paul D. Hanna, Jul 01 2011

Keywords

• nonn

Comments

Related q-series identity (Euler):

Product_{n>=1} 1/(1-x*q^n) = Sum_{n>=0} x^n*q^n / Product_{k=1..n} (1-q^k); here q=A(x).

Examples

			G.f.: A(x) = x + x^3 + x^4 + 3*x^5 + 5*x^6 + 14*x^7 + 28*x^8 + 75*x^9 +...
The g.f. A = A(x) satisfies the relations:
A = x/((1 - x*A)*(1 - x*A^2)*(1 - x*A^3)*(1 - x*A^4)*...),
A = x*(1 + x*A/(1-A) + x^2*A^2/((1-A)*(1-A^2)) + x^3*A^3/((1-A)*(1-A^2)*(1-A^3)) +...).
A = x*(1 + x*A/((1-A)*(1-x*A)) + x^2*A^4/((1-A)*(1-x*A)*(1-A^2)*(1-x*A^2)) + x^3*A^9/((1-A)*(1-x*A)*(1-A^2)*(1-x*A^2)*(1-A^3)*(1-x*A^3)) +...).
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 1..200

Crossrefs

Cf. A145268, A192477.

Programs

Mathematica

nmax = 40; A[] = 0; Do[A[x] = x/Product[1 - x*A[x]^k, {k, 1, nmax}] + O[x]^(nmax + 1) // Normal, nmax + 1]; Rest[CoefficientList[A[x], x]] (* Vaclav Kotesovec, Sep 29 2023 *)
(* Calculation of constants {d,c}: *) {1/r, s*Sqrt[r*((1 - r)*Log[1 - s] + (1 - 2*r)*Log[s] + (1 - r)* QPolyGamma[0, Log[r]/Log[s], s]) / (2*Pi*Log[s]*(2*(1 - r)*r - s^3*Derivative[0, 2][QPochhammer][r, s]))]} /. FindRoot[{(1 - r)*r == s*QPochhammer[r, s], s^2*Derivative[0, 1][QPochhammer][r, s] == (r - 1)*r}, {r, 1/3}, {s, 1/2}, WorkingPrecision -> 120] (* Vaclav Kotesovec, Sep 29 2023 *)

PARI

{a(n) = my(A=x+x^2); for(i=1, n, A = x/prod(m=1, n, (1-x*A^m +x*O(x^n) ))); polcoeff(A, n)}

PARI

{a(n) = my(A=x+x^2); for(i=1, n, A = x*sum(m=0, n, x^m*A^m/prod(k=1, m, (1-A^k +x*O(x^n) )))); polcoeff(A, n)}

PARI

{a(n) = my(A=x+x^2); for(i=1, n, A = x*sum(m=0, sqrtint(n+1), A^(m^2)*x^m/prod(k=1, m, (1-A^k)*(1-x*A^k +x*O(x^n) )))); polcoeff(A, n)}

Formula

G.f. satisfies: A(x) = x*Sum_{n>=0} x^n*A(x)^n / Product_{k=1..n} (1 - A(x)^k).

G.f. satisfies: A(x) = x*Sum_{n>=0} x^n*A(x)^(n^2) / (Product_{k=1..n} (1 - A(x)^k)*(1 - x*A(x)^k)) due to Cauchy's identity.

a(n) ~ c * d^n / n^(3/2), where d = 2.8676696539306775593459072106295374201870092501605360805431348994... and c = 0.1427560675782905659177127310754143738629932908335931082457954696... - Vaclav Kotesovec, Sep 29 2023