A338633 G.f. A(x) satisfies: 1 = A(x) - x/(A(x) - 2^3*x/(A(x) - 3^3*x/(A(x) - 4^3*x/(A(x) - 5^3*x/(A(x) - 6^3*x/(A(x) - ...)))))), a continued fraction relation.
1, 1, 7, 250, 21867, 3725702, 1096355494, 513875333940, 361121449989171, 362961084011245198, 502496711191618404882, 929337000359116522329132, 2238572532534241145084855934, 6875030222633195280825967544508, 26436454884630260855874989243890732
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + x + 7*x^2 + 250*x^3 + 21867*x^4 + 3725702*x^5 + 1096355494*x^6 + 513875333940*x^7 + 361121449989171*x^8 + 362961084011245198*x^9 + ... where 1 = A(x) - x/(A(x) - 2^3*x/(A(x) - 3^3*x/(A(x) - 4^3*x/(A(x) - 5^3*x/(A(x) - 6^3*x/(A(x) - 7^3*x/(A(x) - 8^3*x/(A(x) - 9^3*x/(A(x) - ...))))))))), a continued fraction relation.
Links
- Paul D. Hanna, Table of n, a(n) for n = 0..150
Programs
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PARI
{a(n) = my(A=[1],CF=1); for(i=1,n, A=concat(A,0); for(i=1,#A, CF = Ser(A) - (#A-i+1)^3*x/CF ); A[#A] = -polcoeff(CF,#A-1) );A[n+1] } for(n=0,20,print1(a(n),", "))
Formula
For n > 0, a(n) is odd iff n is a power of 2 (conjecture).
From Vaclav Kotesovec, Nov 12 2020: (Start)
a(n) ~ sqrt(3/(2*Pi)) * (6*Gamma(2/3)/Gamma(1/3)^2)^(3*n + 3/2) * (n!)^3 / sqrt(n).
a(n) ~ 2^(6*n + 4) * 3^(3*n/2 + 5/4) * Pi^(3*n + 5/2) * n^(3*n + 1) / Gamma(1/3)^(9*(n + 1/2)) / exp(3*n). (End)
Comments