A105864 Expansion of (1/(1-x^2))*c(x/(1-x^2)), where c(x) is the g.f. of A000108.
1, 1, 3, 7, 21, 65, 215, 735, 2585, 9281, 33883, 125383, 469229, 1772801, 6752623, 25902975, 99978865, 388001025, 1513077235, 5926139207, 23301146501, 91942524481, 363957103303, 1444966207967, 5752187960841, 22955311342145
Offset: 0
Links
- Paul Barry, Continued fractions and transformations of integer sequences, JIS 12 (2009) 09.7.6.
Crossrefs
Partial sums of A128750.
Programs
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Mathematica
a[0] = a[1] = 1; a[2] = 3; a[3] = 7; a[n_] := a[n] = (-((n-3)*a[n-4]) - 2*(2*n-3)*a[n-3] + 2*(n-1)*a[n-2] + 2*(2*n-1)*a[n-1])/(n+1); Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Sep 09 2017, using "FindSequenceFunction" *) -
PARI
{a(n)=polcoeff((1-sqrt(1-4*x/(1-x^2 +O(x^(n+2)))))/(2*x), n)} /* Paul D. Hanna */
Formula
G.f.: (1 - sqrt((1 - 4*x - x^2)/(1 - x^2)))/(2*x).
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k) * A000108(n-2*k).
G.f. satisfies: A(x) = 1/(1-x^2) + x*A(x)^2. - Paul D. Hanna, Sep 06 2011
Conjecture: (n+1)*a(n) + 2*(1-2*n)*a(n-1) + 2*(1-n)*a(n-2) + 2*(2*n-3)*a(n-3) + (n-3)*a(n-4) = 0. - R. J. Mathar, Nov 15 2011
G.f.: (1-1/G(0))/(2*x), where G(k) = 1 + 4*x*(4*k+1)/( (1-x^2)*(4*k+2) - x*(1-x^2)*(4*k+2)*(4*k+3)/(x*(4*k+3) + (1-x^2)*(k+1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 26 2013
a(n) ~ 5^(1/4)*(2+sqrt(5))^(n+1)/(4*sqrt(2*Pi)*n^(3/2)). - Vaclav Kotesovec, Sep 16 2013
G.f.: 1/G(x), where G(x) = 1 - x^2 - (x - x^3)/ G(x) (continued fraction). - Nikolaos Pantelidis, Jan 08 2023
Comments