A338635 G.f. A(x) satisfies: 1 = A(x) - x/(A(x) - 3*x/(A(x) - 6*x/(A(x) - 10*x/(A(x) - 15*x/(A(x) - 21*x/(A(x) - 28*x/(A(x) - ... - (n*(n+1)/2)*x/(A(x) - ...))))))), a continued fraction relation.
1, 1, 2, 17, 274, 6749, 231276, 10465440, 604220826, 43388420549, 3797054582794, 398157728106929, 49311011342018168, 7124133759620985652, 1187818792835133749984, 226420783437860189825400, 48936975180367428260159850, 11904986360488865549641429797, 3238569202146221391019821488694
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + x + 2*x^2 + 17*x^3 + 274*x^4 + 6749*x^5 + 231276*x^6 + 10465440*x^7 + 604220826*x^8 + 43388420549*x^9 + 3797054582794*x^10 + ... where 1 = A(x) - x/(A(x) - 3*x/(A(x) - 6*x/(A(x) - 10*x/(A(x) - 15*x/(A(x) - 21*x/(A(x) - 28*x/(A(x) - 36*x/(A(x) - 45*x/(A(x) - 55*x/(A(x) - ...)))))))))), a continued fraction relation in which the triangular numbers appear as coefficients.
Links
- Paul D. Hanna, Table of n, a(n) for n = 0..200
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)*(#A-i+2)/2*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 = A118113(k) for some k >= 1, where A118113(k) = 2*Fibbinary(k) + 1 (conjecture).
a(n) ~ 2^(3*n + 5) * n^(2*n + 3/2) / (Pi^(2*n + 3/2) * exp(2*n)). - Vaclav Kotesovec, Nov 12 2020