A193550 O.g.f.: 1/(1 - x/(1 - x/(1 - 9*x/(1 - 9*x/(1 - 25*x/(1 - 25*x/(1 - 49*x/(1 - 49*x/(1-...))))))))), a continued fraction involving the odd squares.
1, 1, 2, 13, 206, 5794, 252068, 15663997, 1316280854, 143694972886, 19764465724412, 3343418236081618, 682133942067492236, 165161123584687819684, 46817735849074712020808, 15358634840651231221695517, 5772973821383087169122348774
Offset: 0
Keywords
Examples
O.g.f.: A(x) = 1 + x + 2*x^2 + 13*x^3 + 206*x^4 + 5794*x^5 + 252068*x^6 +... Let A(x) = 1 + x*B(x)^2, then B(x) appears to be an integer series: B(x) = 1 + x + 6*x^2 + 97*x^3 + 2782*x^4 + 122670*x^5 + 7687932*x^6 + 649446621*x^7 + 71136143478*x^8 + 9806113041658*x^9 +... the coefficients of B(x) are integer for at least the initial 500 terms.
Links
- Vaclav Kotesovec, Table of n, a(n) for n = 0..240
Crossrefs
Cf. A005439.
Programs
-
Mathematica
nmax = 20; CoefficientList[Series[1/Fold[(1 - #2/#1) &, 1, Reverse[(2*Range[nmax + 1] - 2*Floor[Range[nmax + 1]/2] - 1)^2*x]], {x, 0, nmax}], x] (* Vaclav Kotesovec, Aug 25 2017 *) -
PARI
/* Continued fraction: */ {a(n)=local(A=1+x, CF); CF=1+x; for(k=0, n, CF=1/(1-(2*((n-k)\2)+1)^2*x*CF+x*O(x^n))); A=CF; polcoeff(A, n)}
Formula
G.f.: 1/Q(0), where Q(k) = 1 -(2*k+1)^2*x/(1 -(2*k+1)^2*x/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Sep 17 2013
G.f.: Q(0), where Q(k) = 1 - x*(2*k+1)^2/( x*(2*k+1)^2 - 1/(1 - x*(2*k+1)^2/( x*(2*k+1)^2 - 1/Q(k+1)))); (continued fraction). - Sergei N. Gladkovskii, Oct 09 2013
a(n) ~ 2^(4*n+2) * n^(2*n-1/2) / (Pi^(2*n+1/2) * exp(2*n)). - Vaclav Kotesovec, Aug 25 2017
Comments