A091190 G.f. A(x) satisfies x*A(x)^3 = B(x*A(x^3)) where B(x) = x/(1 - 3*x).
1, 1, 2, 5, 13, 35, 97, 273, 778, 2240, 6499, 18976, 55703, 164243, 486130, 1443620, 4299365, 12836825, 38413933, 115184282, 346005073, 1041072108, 3137060983, 9465689545, 28596915843, 86492865522, 261876842801, 793661873276
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + x + 2*x^2 + 5*x^3 + 13*x^4 + 35*x^5 + 97*x^6 + 273*x^7 + 778*x^8 + 2240*x^9 + 6499*x^10 + 18976*x^11 + 55703*x^12 + ... where A(x)^3 = A(x^3) / (1 - 3*x*A(x^3)). RELATED SERIES. A(x)^3 = 1 + 3*x + 9*x^2 + 28*x^3 + 87*x^4 + 270*x^5 + 839*x^6 + 2607*x^7 + 8100*x^8 + 25169*x^9 + 78207*x^10 + 243009*x^11 + 755095*x^12 + ... Also, D(x) = x*A(D(x)) is the g.f. of A370441, which begins D(x) = x + x^2 + 3*x^3 + 12*x^4 + 54*x^5 + 261*x^6 + 1324*x^7 + 6952*x^8 + 37461*x^9 + ... + A370441(n)*x^n + ... such that D(x)^3 = D( x^3 + 3*D(x)^4 ).
Links
- Paul D. Hanna, Table of n, a(n) for n = 0..900
Programs
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Mathematica
m = 28; B[x_] = x/(1 - 3 x); A[_] = 1; Do[A[x_] = (B[x A[x^3]]/x)^(1/3) + O[x]^m // Normal, {m}]; CoefficientList[A[x], x] (* Jean-François Alcover, Oct 29 2019 *)
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PARI
{a(n) = my(A,p=3,m=1); if(n<0,0, m=1; A=1+O(x); while(m<=n, m*=p; A = x*subst(A,x,x^p); A = (A/(1-p*A)/x)^(1/p)); polcoeff(A,n))} for(n=0,30, print1(a(n),", "))
Formula
From Paul D. Hanna, Mar 09 2024: (Start)
G.f. A(x) = Sum_{n>=0} a(n)*x^n satisfies the following formulas.
(1) A(x)^3 = A(x^3) / (1 - 3*x*A(x^3)).
(2) A(x) = x/Series_Reversion(D(x)) where D(x) = x*A(D(x)) is the g.f. of A370441.
(End)
Extensions
Corrected by T. D. Noe, Oct 25 2006
Comments