A352235 G.f. A(x) satisfies: A(x) = 1 + x*A(x) / (A(x) - 3*x*A'(x)).

1, 1, 3, 24, 309, 5262, 108894, 2618718, 71246145, 2154788970, 71563126710, 2586270267600, 100995812044266, 4237522832234832, 190126298040192912, 9085093650185205498, 460711407231295513689, 24715373661154672634058, 1398648334415007990887454

Offset: 0

Author

Paul D. Hanna, Mar 08 2022

Keywords

• nonn

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 24*x^3 + 309*x^4 + 5262*x^5 + 108894*x^6 + 2618718*x^7 + 71246145*x^8 + ...
such that A(x) = 1 + x*A(x)/(A(x) - 3*x*A'(x)).
Related table.
The table of coefficients of x^k in A(x)^(3*n+2) begins:
n=0: [1,  2,   7,   54,   675,  11286,   230742, ...];
n=1: [1,  5,  25,  190,  2210,  34981,   688635, ...];
n=2: [1,  8,  52,  416,  4642,  69872,  1322848, ...];
n=3: [1, 11,  88,  759,  8349, 120549,  2195886, ...];
n=4: [1, 14, 133, 1246, 13790, 193060,  3391017, ...];
n=5: [1, 17, 187, 1904, 21505, 295154,  5017618, ...];
n=6: [1, 20, 250, 2760, 32115, 436524,  7217250, ...]; ...
in which the following pattern holds:
[x^n] A(x)^(3*n+2) = [x^(n-1)] (3*n+2) * A(x)^(3*n+2), n >= 1,
as illustrated by
[x^1] A(x)^2 = 2 = [x^0] 2*A(x)^2 = 2*1;
[x^2] A(x)^5 = 25 = [x^1] 5*A(x)^5 = 5*5;
[x^3] A(x)^8 = 416 = [x^2] 8*A(x)^8 = 8*52;
[x^4] A(x)^11 = 8349 = [x^3] 11*A(x)^11 = 11*759;
[x^5] A(x)^14 = 193060 = [x^4] 14*A(x)^14 = 14*13790;
[x^6] A(x)^17 = 5017618 = [x^5] 17*A(x)^17 = 17*295154; ...
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..378

Crossrefs

Cf. A088715, A286797, A317352, A352236, A352237, A352238.

Programs

PARI

/* Using A(x) = 1 + x*A(x)/(A(x) - 3*x*A'(x)) */
{a(n) = my(A=1); for(i=1,n, A = 1 + x*A/(A - 3*x*A' + x*O(x^n)) );
polcoeff(A,n)}
for(n=0,20, print1(a(n),", "))

PARI

/* Using [x^n] A(x)^(3*n+2) = [x^(n-1)] (3*n+2)*A(x)^(3*n+2) */
{a(n) = my(A=[1]); for(i=1,n, A=concat(A,0);
A[#A] = polcoeff((x*Ser(A)^(3*(#A-2)+2) - Ser(A)^(3*(#A-2)+2)/(3*(#A-2)+2)),#A-1));A[n+1]}
for(n=0,20, print1(a(n),", "))

Formula

G.f. A(x) satisfies:

(1) [x^n] A(x)^(3*n+2) = [x^(n-1)] (3*n+2) * A(x)^(3*n+2) for n >= 1.

(2) A(x) = 1 + x*A(x)/(A(x) - 3*x*A'(x)).

(3) A'(x) = A(x) * (1 + x/(1 - A(x))) / (3*x).

(4) A(x) = exp( Integral (1 + x/(1 - A(x))) / (3*x) dx ).

a(n) ~ c * 3^n * n! * n^(2/3), where c = 0.09232038797888963484135336... - Vaclav Kotesovec, Nov 16 2023