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A078532
Coefficients of power series that satisfies A(x)^3 - 9*x*A(x)^4 = 1, A(0)=1.
10
1, 3, 27, 315, 4158, 59049, 880308, 13586859, 215233605, 3479417370, 57168561996, 951892141473, 16026585711660, 272383068872700, 4666865660812044, 80521573261807755, 1397858693681272230, 24398716826612190447, 427921056863230599900, 7537621933880388620010
OFFSET
0,2
COMMENTS
If A(x) = Sum_{k>=1} a(k)x^k satisfies A(x)^n - (n^2)*x*A(x)^(n+1) = 1, then a(n-1) = n^(2n-3) and a(2n-1) = n^(4n-2) (conjecture).
If A(x) = Sum_{k>=1} a(k)x^k satisfies A(x)^n - (n^2)*x*A(x)^(n+1) = 1, then a(k)=n^(2k)*binomial(k/n+1/n+k-1,k)/(k+1) and, consequently, a(n-1) = n^(2n-3) and a(2n-1) = n^(4n-2). - Emeric Deutsch, Dec 10 2002
A generalization of the Catalan sequence (A000108) since for n = 1 the equation A(x)^n -(n^2)*x*A(x)^(n+1) = 1 reduces to A(x)=1+xA(x)^2. - Emeric Deutsch, Dec 10 2002
Radius of convergence of g.f. A(x) is r = 1/(3*4^(4/3)) where A(r) = 4^(1/3). - Paul D. Hanna, Jul 24 2012
Self-convolution cube yields A214668.
D-finite with recurrence n*(n-1)*(n+1)*a(n) -216*(4*n-5)*(2*n-1)*(4*n-11)*a(n-3)=0. - R. J. Mathar, Mar 24 2023
LINKS
FORMULA
a(n) = 3^(2n)*binomial(4n/3-2/3, n)/(n+1). - Emeric Deutsch, Dec 10 2002
Sequence with offset 1 is expansion of reversion of g.f. x*(1-9*x)^(1/3), which equals x times the g.f. of A004990.
a(n) ~ 2^(8*n/3-5/6) * 3^n / (sqrt(Pi) * n^(3/2)). - Vaclav Kotesovec, Dec 03 2014
EXAMPLE
A(x)^3 - 9x*A(x)^4 = 1 since A(x)^3 = 1 +9x +108x^2 +1458x^3 +21060x^4 +... and A(x)^4 = 1 +12x +162x^2 +2340x^3 +... also a(2)=3^3, a(5)=3^10.
MATHEMATICA
Table[3^(2n) Binomial[(4n-2)/3, n]/(n+1), {n, 0, 20}] (* Harvey P. Dale, Nov 03 2011 *)
PROG
(PARI) for(n=0, 25, print1(9^n * binomial((4*n-2)/3, n)/(n+1), ", ")) \\ G. C. Greubel, Jan 26 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Paul D. Hanna, Nov 28 2002
EXTENSIONS
More terms from Harvey P. Dale, Nov 03 2011
STATUS
approved