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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

%I

%S 1,1,2,13,206,5794,252068,15663997,1316280854,143694972886,

%T 19764465724412,3343418236081618,682133942067492236,

%U 165161123584687819684,46817735849074712020808,15358634840651231221695517,5772973821383087169122348774

%N 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.

%C Conjecture: given o.g.f. A(x), then sqrt( (A(x) - 1)/x ) is an integer series.

%H Vaclav Kotesovec, <a href="/A193550/b193550.txt">Table of n, a(n) for n = 0..240</a>

%F 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

%F 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

%F a(n) ~ 2^(4*n+2) * n^(2*n-1/2) / (Pi^(2*n+1/2) * exp(2*n)). - _Vaclav Kotesovec_, Aug 25 2017

%e O.g.f.: A(x) = 1 + x + 2*x^2 + 13*x^3 + 206*x^4 + 5794*x^5 + 252068*x^6 +...

%e Let A(x) = 1 + x*B(x)^2, then B(x) appears to be an integer series:

%e 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 +...

%e the coefficients of B(x) are integer for at least the initial 500 terms.

%t 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 *)

%o (PARI) /* Continued fraction: */

%o {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)}

%Y Cf. A005439.

%K nonn

%O 0,3

%A _Paul D. Hanna_, Jul 30 2011

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Last modified September 23 11:18 EDT 2021. Contains 347612 sequences. (Running on oeis4.)