A307604 G.f. A(x) satisfies: A(x) = (1/(1 - x)) * A(x^2)^2*A(x^3)^3*A(x^4)^4* ... *A(x^k)^k* ...
1, 1, 3, 6, 17, 28, 72, 122, 282, 493, 1027, 1790, 3673, 6300, 12205, 21117, 39782, 67989, 124937, 212189, 381705, 644625, 1136315, 1905352, 3312916, 5513005, 9443362, 15624026, 26445046, 43451200, 72751824, 118792691, 196966722, 319714816, 525316191, 847734183, 1381904765
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + x + 3*x^2 + 6*x^3 + 17*x^4 + 28*x^5 + 72*x^6 + 122*x^7 + 282*x^8 + 493*x^9 + ...
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 0..5000
Programs
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Mathematica
terms = 36; A[] = 1; Do[A[x] = 1/(1 - x) Product[A[x^k]^k, {k, 2, terms}] + O[x]^(terms + 1) // Normal, terms + 1]; CoefficientList[A[x], x]
Formula
G.f.: Product_{k>=1} 1/(1 - x^k)^(k*A074206(k)).
a(n) ~ (-Gamma(2+r) * zeta(2+r) / zeta'(r))^(1/(50*(2+r))) * exp(12/625 + ((2+r)/(1+r)) * (-Gamma(2+r) * zeta(2+r) / zeta'(r))^(1/(2+r)) * n^((1+r)/(2+r))) / (A^(144/625) * sqrt(2*Pi*(2+r)) * n^(1/2 + 1/(50*(2+r)))), where r = A107311 is the root of the equation zeta(r)=2 and A is the Glaisher-Kinkelin constant A074962. - Vaclav Kotesovec, Mar 18 2021
Comments