A091713 G.f. satisfies A(x) = x + x*A(A(A(x))).
1, 1, 3, 15, 99, 781, 7001, 69253, 742071, 8506775, 103411463, 1324477033, 17785238513, 249432247233, 3642471258187, 55246757713367, 868523130653947, 14127076257342933, 237386074703124457, 4115341407421082869, 73516094755096807279, 1351801707136238290351
Offset: 1
Keywords
Examples
G.f.: A(x) = x + x^2 + 3*x^3 + 15*x^4 + 99*x^5 + 781*x^6 +...
From _Paul D. Hanna_, Apr 16 2007:
G.f. A(x) is the unique solution to variable A in the infinite system of simultaneous equations:
A = 1 + xC;
B = A*(1 + xD);
C = B*(1 + xE);
D = C*(1 + xF);
E = D*(1 + xG) ; ...
The solution to the variables in the system of equations are
A=A(x), B=A(A(x)), C=A(A(A(x))), D=A(A(A(A(x)))), etc., where:
A(x) = x + x^2 + 3*x^3 + 15*x^4 + 99*x^5 + 781*x^6 + 7001*x^7 +...
A(A(x)) = x + 2*x^2 + 8*x^3 + 46*x^4 + 330*x^5 + 2756*x^6 +...
A(A(A(x))) = x + 3*x^2 + 15*x^3 + 99*x^4 + 781*x^5 + 7001*x^6 +...
A(A(A(A(x)))) = x + 4*x^2 + 24*x^3 + 180*x^4 + 1564*x^5 +...
A(A(A(A(A(x))))) = x + 5*x^2 + 35*x^3 + 295*x^4 + 2815*x^5 +...
ALTERNATE GENERATING METHOD.
The g.f. A(x) equals the sum of products of even iterations of A(x):
A(x) = x + x*A_2(x) + x*A_2(x)*A_4(x) + x*A_2(x)*A_4(x)*A_6(x) + x*A_2(x)*A_4(x)*A_6(x)*A_8(x) +...+ Product_{k=0..n} A_{2*k}(x) +...
where A_n(x) = A_{n-1}(A(x)) is the n-th iteration of A(x) with A_0(x)=x.
Links
- Paul D. Hanna, Table of n, a(n) for n = 1..400
Programs
-
Mathematica
Nest[x + x (# /. x -> # /. x -> #) &, O[x], 30][[3]] (* Vladimir Reshetnikov, Aug 08 2019 *)
-
PARI
{a(n)=local(A);A=x+x^2;for(i=3,n, A=x+x*subst(A,x,subst(A,x,A))+x*O(x^n)); polcoeff(A,n,x)} -
PARI
/* Define the n-th iteration of F: */ {ITERATE(F,n,p)=local(G=x);for(i=1,n,G=subst(F,x,G+x*O(x^p)));G} /* A(x) equals the sum of products of even iterations of A(x): */ {a(n)=local(A=x);for(i=1,n,A=sum(m=0,n-1,prod(k=0,m,ITERATE(A,2*k,n)+x*O(x^n))));polcoeff(A,n)}
Formula
G.f. A(x) = F(x,1) where F(x,n) satisfies: F(x,n) = F(x,n-1)*(1 + x*F(x,n+2)) for n>0 with F(x,0)=1. - Paul D. Hanna, Apr 16 2007
G.f.: A(x) = G(x)/[1 + G(G(x))] where G(x) = A(A(x)) = g.f. of A141117.
G.f.: A(x) = Series_Reversion[ x/(1 + A(A(x))) ].
G.f. satisfies: A(x) = Sum_{n>=0} Product_{k=0..n} A_{2*k}(x), where A_n(x) denotes the n-th iteration of A(x) with A_0(x)=x. - Paul D. Hanna, Jul 21 2011
Comments