A168490 Sequence with Hankel transform equal to 2^floor(n^2/2).
1, 1, 2, 6, 24, 112, 560, 2888, 15136, 80160, 427968, 2300736, 12445440, 67702272, 370205184, 2033976960, 11224014336, 62186741248, 345825348608, 1929744008192, 10802203119616, 60644473282560, 341383505977344, 1926554113200128, 10897582426832896, 61775951537520640, 350901756369403904
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..300
Programs
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Mathematica
CoefficientList[Series[(1-2*x-Sqrt[(1-2*x)(1-10*x+24*x^2)])/(4x*(1-2*x)), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 20 2012 *)
Formula
G.f.: 1/(1-x/(1-x/(1-2*x/(1-2*x/(1-x/(1-x/(1-2*x/(1-2*x/(1-x/(1-x/(1-2*x/(1-... (continued fraction);
G.f.: 1/(1-x-x^2/(1-3*x-4*x^2/(1-3*x-x^2/(1-3*x-4*x^2/(1-3*x-x^2/(1-3*x-4*x^2/(1-... (continued fraction);
G.f.: (1-2*x-sqrt((1-2*x)*(1-10*x+24*x^2)))/(4*x*(1-2*x)).
Recurrence: (n+1)*a(n) = 4*(3*n-1)*a(n-1) - 4*(11*n-17)*a(n-2) + 24*(2*n-5)*a(n-3). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ 2^(n-5/2)*3^(n+1)/(sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 20 2012
a(n) = Sum_{k=0..n} A168511(n,k)*2^(n-k). - Philippe Deléham, Mar 19 2013
Comments