%I #19 Jan 15 2018 08:52:24

%S 1,1,2,6,24,118,674,4308,30062,225266,1791964,15009118,131566314,

%T 1201452248,11389283418,111761444078,1132680800640,11834071103246,

%U 127261591139010,1406778021294220,15967144849210158,185897394076705298

%N G.f. A(x) satisfies A(x/A(x)) = 1/(1-x).

%H Vaclav Kotesovec, <a href="/A088713/b088713.txt">Table of n, a(n) for n = 0..300</a>

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

%F G.f.: A(x*g(x)) = g(x) = (1-1/A(x))/x where g(x) is the g.f. of A088714.

%F From _Paul D. Hanna_, Dec 06 2009: (Start)

%F G.f. satisfies: A(x) = 1 + A(x)*Series_Reversion(x/A(x)).

%F G.f. satisfies: A( (x/(1+x)) / A(x/(1+x)) ) = 1 + x.

%F (End)

%F Logarithmic derivative: given g.f. A(x), let G(x) = A(x*G(x)) be the g.f. of A088714, then A'(x)/A(x) = (G(x) + x*G'(x)) / (1 - x*G(x)).

%e G.f.: A(x) = 1 + x + 2*x^2 + 6*x^3 + 24*x^4 + 118*x^5 + 674*x^6 +...

%e Illustration of logarithmic derivation.

%e If we form an array of coefficients of x^k in A(x)^n, n>=1, like so:

%e A^1: [1],1, 2, 6, 24, 118, 674, 4308, ...;

%e A^2: [1, 2], 5, 16, 64, 308, 1716, 10724, ...;

%e A^3: [1, 3, 9], 31, 126, 600, 3278, 20070, ...;

%e A^4: [1, 4, 14, 52], 217, 1032, 5560, 33440, ...;

%e A^5: [1, 5, 20, 80, 345], 1651, 8820, 52270, ...;

%e A^6: [1, 6, 27, 116, 519, 2514],13385, 78420, ...;

%e A^7: [1, 7, 35, 161, 749, 3689, 19663], 114269, ...; ...

%e then the sums of the coefficients of x^k, k=0..n-1, in A(x)^n (shown above in brackets) begin:

%e 1 = 1;

%e 1 + 2 = 3;

%e 1 + 3 + 9 = 13;

%e 1 + 4 + 14 + 52 = 71;

%e 1 + 5 + 20 + 80 + 345 = 451;

%e 1 + 6 + 27 + 116 + 519 + 2514 = 3183;

%e 1 + 7 + 35 + 161 + 749 + 3689 + 19663 = 24305; ...

%e and equal the coefficients in log(A(x)):

%e log(A(x)) = x + 3*x^2/2 + 13*x^3/3 + 71*x^4/4 + 451*x^5/5 + 3183*x^6/6 + 24305*x^7/7 + 197551*x^8/8 +...

%e The main diagonal in the above table forms the g.f. G(x) of A088714:

%e [1/1, 2/2, 9/3, 52/4, 345/5, 2514/6, 19663/7, ...]

%e where G(x) = 1 + x + 3*x^2 + 13*x^3 + 69*x^4 + 419*x^5 + 2809*x^6 +...

%e satisfies A'(x)/A(x) = (G(x) + x*G'(x)) / (1 - x*G(x)).

%t terms = 22; A[_] = 1; Do[A[x_] = 1 + x*A[x]*A[1 - 1/A[x]] + O[x]^j // Normal, {j, terms}]; CoefficientList[A[x], x] (* _Jean-François Alcover_, Jan 15 2018 *)

%o (PARI) a(n)=local(A=1+x);for(i=1,n,A=(1+A*serreverse(x/(A+x*O(x^n))))^1);polcoeff(A,n)

%o for(n=0,30,print1(a(n),", ")) \\ _Paul D. Hanna_, Dec 06 2009

%o (PARI) {a(n)=local(A=1+x);if(n==0,1,for(i=1,n,

%o A=1+x*exp(sum(k=1,n-1,sum(j=0,k,polcoeff(A^k+x*O(x^j),j))*x^k/k)+x*O(x^n))));

%o polcoeff(A+x*O(x^n),n)}

%o for(n=0,30,print1(a(n),", ")) \\ _Paul D. Hanna_, Dec 09 2013

%Y Cf. A088714.

%Y Variants: A154677, A168448, A168449, A168478, A168479.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Oct 12 2003