A088358 a(n) equals sum of first n terms of A(x)^n for n>=1, with a(0)=1.

1, 1, 3, 16, 127, 1321, 16680, 244518, 4049199, 74404069, 1498276873, 32764372213, 772675039936, 19541627299052, 527590805816280, 15146369004674536, 460804123171138079, 14811876349937896743, 501663013214822053815, 17858867621856721343253, 666744417234185576463077

Offset: 0

Author

Michael Somos and Paul D. Hanna, Sep 27 2003

Keywords

• nonn

Examples

			G.f.: A(x) = 1 + x + 3*x^2 + 16*x^3 + 127*x^4 + 1321*x^5 + 16680*x^6 +...
The coefficients in A(x)^n begin:
n=1: [1, 1,  3,  16,  127,  1321,  16680,  244518,  4049199, ...];
n=2: [1, 2,  7,  38,  295,  2992,  37020,  534386,  8745915, ...];
n=3: [1, 3, 12,  67,  513,  5088,  61716,  877053, 14181891, ...];
n=4: [1, 4, 18, 104,  791,  7696,  91582, 1281160, 20462071, ...];
n=5: [1, 5, 25, 150, 1140, 10916, 127565, 1756710, 27706465, ...];
n=6: [1, 6, 33, 206, 1572, 14862, 170761, 2315256, 36052245, ...];
n=7: [1, 7, 42, 273, 2100, 19663, 222432, 2970108, 45656093, ...];
n=8: [1, 8, 52, 352, 2738, 25464, 284024, 3736560, 56696823, ...];
n=9: [1, 9, 63, 444, 3501, 32427, 357186, 4632138, 69378300, ...]; ...
where the initial terms are derived from the above coefficients like so:
a(1) = 1 = 1;
a(2) = 1 + 2 = 3;
a(3) = 1 + 3 + 12 = 16;
a(4) = 1 + 4 + 18 + 104 = 127;
a(5) = 1 + 5 + 25 + 150 + 1140 = 1321;
a(6) = 1 + 6 + 33 + 206 + 1572 + 14862 = 16680; ...
RELATED EXPANSIONS.
The series B(x) = Series_Reversion(x/A(x)) begins:
B(x) = x + x^2 + 4*x^3 + 26*x^4 + 228*x^5 + 2477*x^6 + 31776*x^7 +...
such that A(x) = 1 + x*B'(x)/(1 - B(x)); also,
B(x) = Sum_{n>=1} b(n)*x^n where b(n) = [x^(n-1)] A(x)^n/n for n>=1:
[1/1, 2/2, 12/3, 104/4, 1140/5, 14862/6, 222432/7, 3736560/8, ...].
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..400

Crossrefs

Cf. A233436.

Programs

PARI

{a(n)=local(A); if(n<2,n>=0,A=1+x; for(i=2,n,A+=x^i*subst(Pol((A+O(x^i))^i),x,1)); polcoeff(A,n))}
for(n=0,25,print1(a(n),", "))

PARI

/* A(x) = 1 + x*B'(x)/(1 - B(x)) where B(x/A(x)) = x: */
{a(n)=local(A=1+x);for(i=1,n,B=serreverse(x/A+x*O(x^n));A=1+x*deriv(B)/(1-B));polcoeff(A,n)}
for(n=0,25,print1(a(n),", "))

Formula

G.f. satisfies: A(x) = 1 + x*B'(x)/(1 - B(x)) where B(x/A(x)) = x. - Paul D. Hanna, Nov 01 2013

a(n) ~ c * n! * n^alpha / LambertW(1)^n, where alpha = (1 + 3*LambertW(1))/(1 + 1/LambertW(1)) and c = 0.192874788982750074134074506494559... - Vaclav Kotesovec, Sep 13 2024