A338636 G.f. A(x) satisfies: 1 = A(x) - x/(A(x) - 3^2*x/(A(x) - 5^2*x/(A(x) - 7^2*x/(A(x) - 9^2*x/(A(x) - ...))))), a continued fraction relation.
1, 1, 8, 272, 19480, 2353568, 429016872, 110046546096, 37825128764472, 16793443888112960, 9358539226503013960, 6397425528561882140240, 5264539843826571207135320, 5134140710880677886077086432, 5855644914993764696284947092840
Offset: 0
Keywords
Examples
G.f. A(x) = 1 + x + 8*x^2 + 272*x^3 + 19480*x^4 + 2353568*x^5 + 429016872*x^6 + 110046546096*x^7 + 37825128764472*x^8 + 16793443888112960*x^9 + ... where 1 = A(x) - x/(A(x) - 3^2*x/(A(x) - 5^2*x/(A(x) - 7^2*x/(A(x) - 9^2*x/(A(x) - 11^2*x/(A(x) - 13^2*x/(A(x) - 15^2*x/(A(x) - 17^2*x/(A(x) - 19^2*x/(A(x) - ...)))))))))), a continued fraction relation.
Links
- Paul D. Hanna, Table of n, a(n) for n = 0..150
Programs
-
PARI
{a(n) = my(A=[1],CF=1); for(i=1,n, A=concat(A,0); for(i=1,#A, CF = Ser(A) - (2*(#A-i)+1)^2*x/CF ); A[#A] = -polcoeff(CF,#A-1) );A[n+1] } for(n=0,20,print1(a(n),", "))
Formula
a(n) = 0 (mod 8) for n > 1 (conjecture).
For n > 0, a(n) = 1 (mod 3) iff n = A191107(k) for some k >= 1 (conjecture).
For n > 0, a(n) = 2 (mod 3) iff n = A186776(k) for some k >= 2 where A186776 is the Stanley sequence S(0,2) (conjecture).
a(n) ~ 2^(6*n + 1) * n^(2*n - 1/2) / (Pi^(2*n + 1/2) * exp(2*n)). - Vaclav Kotesovec, Nov 12 2020