A177379 E.g.f. satisfies: A(x) = 1/(1-x - x*log(A(x))).

1, 1, 4, 27, 260, 3270, 50904, 946134, 20462896, 505137312, 14020517520, 432340670520, 14667108820704, 542979374426736, 21784934875431168, 941691211940974320, 43634507604383543040, 2157698329617806488320

Offset

0,3

Links

Table of n, a(n) for n=0..17.

Formula

E.g.f.: A(x) = 1/(1 - Series_Reversion(x/(1 - log(1-x)))).

...

Let G(x) = e.g.f. of A138013, then:

. A(x) = exp(G(x) - 1),

. A(x) = 1/(1 - x*G(x))

where G(x) = 1 - log(1 - x*G(x)).

...

Let F(x) = e.g.f. of A177380, then:

. [x^n] A(x)^(-n+1)/(-n+1) = A177380(n)/n! for n>1,

. [x^n] F(x)^(n+1)/(n+1) = a(n)/n! for n>=0,

. A(x) = F(x*A(x)) and A(x/F(x)) = F(x),

. A(x) = (1/x)*Series_Reversion(x/F(x))

where F(x) = 1+x + x*log(F(x)).

Contribution from Paul D. Hanna, Jul 16 2010: (Start)

E.g.f. satisfies: A(x) = (1 + x*A'(x)/A(x))*(1 - x*A(x))/(1-x).

...

Let A_n(x) denote the n-th iteration of x*A(x) with G = x/(1-x), then:

. A(x) = 1 + G + G*Dx(G)/2! + G*Dx(G*Dx(G))/3! + G*Dx(G*Dx(G*Dx(G)))/4! +...

. A_n(x)/x = 1 + n*G + n^2*G*Dx(G)/2! + n^3*G*Dx(G*Dx(G))/3! + n^4*G*Dx(G*Dx(G*Dx(G)))/4! +...

where Dx(F) = d/dx(x*F).

...

Given e.g.f. A(x), the matrix log of the Riordan array (A(x),x*A(x)) equals the matrix L defined by L(n,k)=k+1 and L(n,n)=0, for n>=0, n>k.

(End)

a(n) ~ sqrt(s-1) * n^(n-1) * s^(n+1) / exp(n), where s = -LambertW(-1,-exp(-2)) = 3.14619322062... (see A226572). - Vaclav Kotesovec, Jan 11 2014

Example

E.g.f.: A(x) = 1 + x + 4*x^2/2! + 27*x^3/3! + 260*x^4/4! +...
Log(A(x)) = G(x) - 1 where G(x) = e.g.f. of A138013 begins:
G(x) = 1 + x + 3*x^2/2! + 17*x^3/3! + 146*x^4/4! + 1694*x^5/5! + ...
and satisfies: exp(1 - G(x)) = 1 - x*G(x).
Contribution from Paul D. Hanna, Jul 16 2010: (Start)
Given e.g.f. A(x), and A179424 = Riordan array (A(x),x*A(x)) where the g.f. of column k in A179424 equals A(x)^(k+1):
1;
1, 1;
4/2!, 2, 1;
27/3!, 10/2!, 3, 1;
260/4!, 78/3!, 18/2!, 4, 1;
3270/5!, 832/4!, 159/3!, 28/2!, 5, 1;
...
then the matrix log of A179424 equals the triangular matrix:
0;
1, 0;
1, 2, 0;
1, 2, 3, 0;
1, 2, 3, 4, 0;
1, 2, 3, 4, 5, 0; ...
in which column k equals k+1 in row n for n>k>=0.
(End)

Mathematica

CoefficientList[1/(1-InverseSeries[Series[x/(1-Log[1-x]), {x, 0, 20}], x]), x]*Range[0, 20]! (* Vaclav Kotesovec, Jan 11 2014 *)

Prog

(PARI) {a(n)=n!*polcoeff(1/(1-serreverse(x/(1-log(1-x+x*O(x^n))))), n)}
Contribution from Paul D. Hanna, Jul 16 2010: (Start)
(PARI) /* Using matrix log of Riordan array (A(x), x*A(x)): */
{a(n)=local(L=matrix(n+1, n+1, r, c, if(r>c, c)), M=sum(m=0, #L, L^m/m!)); n!*M[n+1, 1]}
(PARI) /* From A(x) = (1 + x*A'(x)/A(x))*(1 - x*A(x))/(1-x): */
{a(n)=local(A=1+x); for(k=2, n, A=A-polcoeff((1+x*deriv(A)/A)*(1-x*A)/(1-x+x*O(x^n)), k)*x^k/(k-1)); n!*polcoeff(A, n)} (End)

Crossrefs

Cf. A177380, A138013, A226572.

Cf. A179424. [From Paul D. Hanna, Jul 16 2010]

Sequence in context: A302836 A301335 A362701 * A052813 A218653 A359461

Adjacent sequences: A177376 A177377 A177378 * A177380 A177381 A177382

Keyword

nonn

Author

Paul D. Hanna, May 15 2010

Status

approved