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 1-3 of 3 results.

A025236 Duplicate of A014432.

Original entry on oeis.org

1, 3, 3, 12, 30, 111, 363, 1353, 4917, 18777, 71769, 280506, 1103556, 4395009, 17622309
Offset: 1

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A025237 Expansion of (1 -x -sqrt(1-2*x-11*x^2))/(6*x^2).

Original entry on oeis.org

1, 1, 4, 10, 37, 121, 451, 1639, 6259, 23923, 93502, 367852, 1465003, 5874103, 23740276, 96503554, 394542379, 1620716251, 6687296308, 27700303510, 115152607831, 480244735171, 2008802728819, 8425318166635, 35425680021397, 149296062114181, 630526903497706, 2668194946794124, 11311786743536125
Offset: 0

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Comments

a(n) = (1/3)*s(n+2), where s = A014432.
Also, number of walks within N^2 (the first quadrant of Z^2) starting at (0,0), ending on the vertical axis and consisting of n steps taken from {(-1, -1), (-1, 0), (-1, 1), (0, 1), (1, 1)}. - Manuel Kauers, Nov 18 2008
Also, number of walks within N^3 (the first octant of Z^3) starting at (0,0,0) and consisting of n steps taken from {(-1, -1, 1), (-1, 0, 1), (-1, 1, 1), (0, 0, -1), (1, 1, 0)}. - Manuel Kauers, Nov 18 2008
Reversion of x/(1+x+3x^2). Hankel transform is 3^C(n+1,2) [A047656(n+1)]. - Paul Barry, Sep 07 2009

Examples

			G.f.: 1 + x + 4*x^2 + 10*x^3 + 37*x^4 + 121*x^5 + 451*x^6 + 1639*x^7 + ...
		

Crossrefs

Cf. A217275.

Programs

  • Mathematica
    CoefficientList[Series[(1 - x - Sqrt[1 - 2*x - 11*x^2])/(6*x^2), {x, 0, 50}], x] (* G. C. Greubel, Feb 07 2017 *)
  • PARI
    {a(n) = polcoeff((1 - x - sqrt(1 - 2*x - 11*x^2 + x^3*O(x^n))) / (6*x^2), n)}; /* Michael Somos, Sep 23 2003 */

Formula

From Paul Barry, Sep 07 2009: (Start)
G.f.: 1/(1-x-3x^2/(1-x-3x^2/(1-x-3x^2/(1-... (continued fraction);
a(n) = Sum_{k=0..floor(n/2)} binomial(n,2k)*3^k*A000108(k). (End)
D-finite with recurrence: (n+2)*a(n) - (2*n+1)*a(n-1) + 11*(1-n)*a(n-2) = 0. - R. J. Mathar, Nov 15 2011
a(n) ~ (1+2*sqrt(3))^(n+3/2)/(2*sqrt(Pi)*3^(3/4)*n^(3/2)). - Vaclav Kotesovec, Sep 29 2012
G.f. A(x) satisfies: A(x) = 1 + x * (1 + 3*x*A(x)^2) / (1 - x). - Ilya Gutkovskiy, Jun 30 2020

Extensions

Edited by N. J. A. Sloane, Nov 28 2008

A374488 Expansion of 1/(1 - 2*x - 11*x^2)^(3/2).

Original entry on oeis.org

1, 3, 24, 100, 555, 2541, 12628, 59004, 281655, 1315765, 6171132, 28692456, 133315273, 616780815, 2848833120, 13124483344, 60364983987, 277142478921, 1270586298520, 5817063737100, 26600252408961, 121501917998263, 554429553154044, 2527595449990500
Offset: 0

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Author

Seiichi Manyama, Jul 09 2024

Keywords

Crossrefs

Programs

  • Mathematica
    Module[{x}, CoefficientList[Series[1/(1 - (11*x + 2)*x)^(3/2), {x, 0, 25}], x]] (* Paolo Xausa, Aug 25 2025 *)
  • PARI
    a(n) = binomial(n+2, 2)*sum(k=0, n\2, 3^k*binomial(n, 2*k)*binomial(2*k, k)/(k+1));

Formula

a(0) = 1, a(1) = 3; a(n) = ((2*n+1)*a(n-1) + 11*(n+1)*a(n-2))/n.
a(n) = binomial(n+2,2) * A025237(n).
From Seiichi Manyama, Aug 20 2025: (Start)
a(n) = ((n+2)/2) * Sum_{k=0..floor(n/2)} 3^k * binomial(n+1,n-2*k) * binomial(2*k+1,k).
a(n) = Sum_{k=0..n} (1/2)^k * (11/2)^(n-k) * (2*k+1) * binomial(2*k,k) * binomial(k,n-k). (End)
a(n) ~ sqrt(n) * (1 + 2*sqrt(3))^(n + 3/2) / (4 * 3^(3/4) * sqrt(Pi)). - Vaclav Kotesovec, Aug 21 2025
Showing 1-3 of 3 results.