cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-2 of 2 results.

A166229 Expansion of (1-2x-sqrt(1-8x+8x^2))/(2x).

Original entry on oeis.org

1, 2, 8, 36, 176, 912, 4928, 27472, 156864, 912832, 5394176, 32282240, 195264000, 1191825920, 7331457024, 45406194944, 282896763904, 1771868302336, 11150040870912, 70461597988864, 446971590516736, 2845144452292608
Offset: 0

Views

Author

Paul Barry, Oct 09 2009

Keywords

Comments

Binomial transform of A166228. Hankel transform is A166231.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[(1-2*x-Sqrt[1-8*x+8*x^2])/(2*x), {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 20 2012 *)

Formula

a(n) = 0^n + Sum_{k = 0..n} C(n-1,k-1)*A006318(k). - Paul Barry, Nov 04 2009
G.f.: 1/(1-2x/(1-x-x/(1-2x/(1-x-x/(1-2x/(1-x-x/(1-... (continued fraction). - Paul Barry, Dec 10 2009
Recurrence: (n+1)*a(n) = 4*(2*n-1)*a(n-1) - 8*(n-2)*a(n-2). - Vaclav Kotesovec, Oct 20 2012
a(n) ~ sqrt(1+sqrt(2))*(4+2*sqrt(2))^n/(sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Oct 20 2012
From Peter Bala, May 01 2024: (Start)
O.g.f.: A(x) = x*S(x/(1 - x)), where S(x) = (1 - x - sqrt(1 - 6*x + x^2))/(2*x) is the o.g.f. for the large Schröder numbers A006318.
a(n) = A174347(n+1) - A174347(n).
The g.f. satisfies x^2*A(x)^2 - x*(1 - 2*x)*A(x) + x*(1 - x) = 0.
A(x) = (1 - x)/(1 - 2*x - x*(1 - x)/(1 - 2*x - x*(1 - x)/(1 - 2*x - ...))). (End)

A166231 a(n) = 2^C(n+1,2)*A006012(n).

Original entry on oeis.org

1, 4, 48, 1280, 69632, 7602176, 1660944384, 725849473024, 634418209226752, 1109011408239984640, 3877275019992826380288, 27111105978154633171828736, 379138479844261543254652092416, 10604214183559196000870296488771584
Offset: 0

Views

Author

Paul Barry, Oct 09 2009

Keywords

Comments

Hankel transform of A166228, A166229.

Crossrefs

Programs

  • Mathematica
    Table[2^(Binomial[n + 1, 2])*Sum[Binomial[n, 2 k]*2^(n - k), {k, 0, Floor[n/2]}], {n, 0, 25}] (* G. C. Greubel, May 07 2016 *)

Formula

a(n) = 2^C(n+1,2)*Sum_{k=0..floor(n/2)} C(n,2k)*2^(n-k).
a(n) = 4^n*A166232(n).
Showing 1-2 of 2 results.