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

%I

%S 1,1,4,27,260,3270,50904,946134,20462896,505137312,14020517520,

%T 432340670520,14667108820704,542979374426736,21784934875431168,

%U 941691211940974320,43634507604383543040,2157698329617806488320

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

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

%F ...

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

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

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

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

%F ...

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

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

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

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

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

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

%F Contribution from _Paul D. Hanna_, Jul 16 2010: (Start)

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

%F ...

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

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

%F . 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! +...

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

%F ...

%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.

%F (End)

%F 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

%e E.g.f.: A(x) = 1 + x + 4*x^2/2! + 27*x^3/3! + 260*x^4/4! +...

%e Log(A(x)) = G(x) - 1 where G(x) = e.g.f. of A138013 begins:

%e G(x) = 1 + x + 3*x^2/2! + 17*x^3/3! + 146*x^4/4! + 1694*x^5/5! + ...

%e and satisfies: exp(1 - G(x)) = 1 - x*G(x).

%e Contribution from _Paul D. Hanna_, Jul 16 2010: (Start)

%e 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):

%e 1;

%e 1, 1;

%e 4/2!, 2, 1;

%e 27/3!, 10/2!, 3, 1;

%e 260/4!, 78/3!, 18/2!, 4, 1;

%e 3270/5!, 832/4!, 159/3!, 28/2!, 5, 1;

%e ...

%e then the matrix log of A179424 equals the triangular matrix:

%e 0;

%e 1, 0;

%e 1, 2, 0;

%e 1, 2, 3, 0;

%e 1, 2, 3, 4, 0;

%e 1, 2, 3, 4, 5, 0; ...

%e in which column k equals k+1 in row n for n>k>=0.

%e (End)

%t CoefficientList[1/(1-InverseSeries[Series[x/(1-Log[1-x]),{x,0,20}],x]),x]*Range[0,20]! (* _Vaclav Kotesovec_, Jan 11 2014 *)

%o (PARI) {a(n)=n!*polcoeff(1/(1-serreverse(x/(1-log(1-x+x*O(x^n))))),n)}

%o Contribution from _Paul D. Hanna_, Jul 16 2010: (Start)

%o (PARI) /* Using matrix log of Riordan array (A(x),x*A(x)): */

%o {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]}

%o (PARI) /* From A(x) = (1 + x*A'(x)/A(x))*(1 - x*A(x))/(1-x): */

%o {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)

%Y Cf. A177380, A138013, A226572.

%Y Cf. A179424. [From _Paul D. Hanna_, Jul 16 2010]

%K nonn

%O 0,3

%A _Paul D. Hanna_, May 15 2010

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Last modified April 21 08:13 EDT 2021. Contains 343146 sequences. (Running on oeis4.)