A361763 Expansion of g.f. A(x) satisfying A(x)^3 = A( x^3/(1 - 3*x)^3 ).
1, 3, 9, 28, 93, 333, 1271, 5064, 20673, 85460, 355659, 1486719, 6238608, 26278281, 111114558, 471608944, 2008906581, 8586410085, 36816550550, 158332335279, 682843960665, 2952865525730, 12802463157570, 55646477022330, 242465061290160, 1059022767175173, 4636452916770489
Offset: 1
Keywords
Examples
G.f.: A(x) = x + 3*x^2 + 9*x^3 + 28*x^4 + 93*x^5 + 333*x^6 + 1271*x^7 + 5064*x^8 + 20673*x^9 + 85460*x^10 + 355659*x^11 + 1486719*x^12 + ... where A( x^3/(1 - 3*x)^3 ) = x^3 + 9*x^4 + 54*x^5 + 273*x^6 + 1269*x^7 + 5670*x^8 + 24957*x^9 + 109593*x^10 + 482598*x^11 + 2133082*x^12 + ... which equals A(x)^3. RELATED SERIES. Notice that the following cube root is an integer series ( A(x)/x )^(1/3) = 1 + x + 2*x^2 + 5*x^3 + 15*x^4 + 52*x^5 + 197*x^6 + 779*x^7 + 3135*x^8 + 12709*x^9 + 51757*x^10 + ... + A361762(n)*x^n + ... Also, let B(x) satisfy A(x/B(x)) = x and B(A(x)) = A(x)/x, then B(x) = x/Series_Reversion(A(x)) is the g.f. of A107092, B(x) = 1 + 3*x + x^3 - x^6 + 2*x^9 - 4*x^12 + 9*x^15 - 22*x^18 + 55*x^21 - 142*x^24 + 376*x^27 - 1011*x^30 + ... such that B(x)^3 = B(x^3) + 3*x, as shown by the series B(x)^(1/3) = 1 + x - x^2 + 2*x^3 - 4*x^4 + 9*x^5 - 22*x^6 + 55*x^7 - 142*x^8 + 376*x^9 - 1011*x^10 + ... SPECIFIC VALUES. A(1/5) = A(1/8)^(1/3) = 0.586384210523490911367880492498... A(1/5) = (1/5) * (1 - 3/5)^(-1) * (1 - 3/8)^(-1/3) * (1 - 3/125)^(-1/9) * (1 - 3/1815848)^(-1/27) * ... A(1/6) = A(1/27)^(1/3) = 0.346688997573685318336777346240... A(1/6) = (1/6) * (1 - 3/6)^(-1) * (1 - 3/27)^(-1/3) * (1 - 3/13824)^(-1/9) * (1 - 3/2640087986661)^(-1/27) * ... A(1/9) = A(1/216)^(1/3) = 0.16744549995321182031691216552466... A(1/12) = A(1/729)^(1/3) = 0.11126394649161862248626102306202...
Links
- Paul D. Hanna, Table of n, a(n) for n = 1..500
Crossrefs
Programs
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PARI
{a(n) = my(A=x); for(i=1, #binary(n+1), A = ( subst(A, x, x^3/(1 - 3*x +x*O(x^n))^3 ) )^(1/3) ); polcoeff(A, n)} for(n=1, 30, print1(a(n), ", "))
Formula
G.f. A(x) = Sum_{n>=1} a(n)*x^n satisfies:
(1) A(x)^3 = A( x^3/(1 - 3*x)^3 ).
(2) A(x^3) = A( x/(1 + 3*x) )^3.
(3) A(x) = x * Product_{n>=0} 1/(1 - 3/F(n,x))^(1/3^n), where F(0,x) = 1/x, F(m,x) = (F(m-1,x) - 3)^3 for m > 0.
(4) x/Series_Reversion(A(x)) = B(x) such that B(x)^3 = B(x^3) + 3*x (cf. A107092).
Comments