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.

A098479 Expansion of 1/sqrt((1-x)^2 - 4*x^3).

Original entry on oeis.org

1, 1, 1, 3, 7, 13, 27, 61, 133, 287, 633, 1407, 3121, 6943, 15517, 34755, 77959, 175213, 394499, 889461, 2007963, 4538485, 10269247, 23258881, 52726599, 119627977, 271624315, 617180533, 1403272799, 3192557561, 7267485523, 16552454205, 37718893317, 85992506271
Offset: 0

Views

Author

Paul Barry, Sep 10 2004

Keywords

Comments

1/sqrt((1-x)^2-4*r*x^3) expands to Sum_{k=0..floor(n/2)} binomial(n-k,k)*binomial(n-2*k,k)*r^k.
Hankel transform is A120580. - Paul Barry, Sep 19 2008
From Joerg Arndt, Jul 01 2011: (Start)
Apparently the number of lattice paths from (0,0) to (n,n) using steps (3,0), (0,3), (1,1).
It appears that 1/sqrt((1-x)^2-4*x^s) is the g.f. for lattice paths from (0,0) to (n,n) using steps (s,0), (0,s), (1,1).
Apparently the number of lattice paths from (0,0) to (n,n) using steps (1,2), (2,1), (1,1). (End)
Diagonal of rational functions 1/(1 - (x*y + x*y^2 + x^2*y)), 1/(1 - (x*y + x^3 + y^3)). - Gheorghe Coserea, Aug 31 2018
Diagonal of the rational function 1 / ((1-x)*(1-y) - x^2*y^3). - Seiichi Manyama, Apr 29 2025

Examples

			From _Joerg Arndt_, Jul 01 2011: (Start)
The triangle of lattice paths from (0,0) to (n,k) using steps (1,2), (2,1), (1,1) begins
  1;
  0, 1;
  0, 1, 1;
  0, 0, 2, 3;
  0, 0, 1, 3, 7;
  0, 0, 0, 3, 7, 13;
  0, 0, 0, 1, 6, 17, 27;
  0, 0, 0, 0, 4, 14, 36, 61;
The triangle of lattice paths from (0,0) to (n,k) using steps (3,0), (0,3), (1,1) begins
  1;
  0, 1;
  0, 0, 1;
  1, 0, 0, 3;
  0, 2, 0, 0,  7;
  0, 0, 3, 0,  0, 13;
  1, 0, 0, 7,  0,  0, 27;
  0, 3, 0, 0, 17,  0,  0, 61;
The diagonals of both appear to be this sequence.  (End)
		

Crossrefs

Programs

  • Mathematica
    a[n_] := Sum[ Binomial[n-k, k]*Binomial[n-2k, k], {k, 0, n/2}]; Table[a[n], {n, 0, 31}] (* Jean-François Alcover, Jan 07 2013, from 1st formula *)
    CoefficientList[Series[1/Sqrt[(1-x)^2-4x^3],{x,0,40}],x] (* Harvey P. Dale, Aug 13 2024 *)
  • PARI
    /* as lattice paths, assuming the first comment is true */
    /* same as in A092566 but use either of the following */
    steps=[[3,0], [0,3], [1,1]];
    steps=[[1,1], [1,2], [2,1]];
    /* Joerg Arndt, Jul 01 2011 */
    
  • Python
    from sympy import binomial
    def a(n): return sum(binomial(n - k, k) * binomial(n - 2*k, k) for k in range(n//2 + 1))
    print([a(n) for n in range(31)]) # Indranil Ghosh, Apr 18 2017

Formula

a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*binomial(n-2*k, k).
D-finite with recurrence: n*a(n) + (-2*n+1)*a(n-1) + (n-1)*a(n-2) + 2*(-2*n+3)*a(n-3) = 0. - R. J. Mathar, Nov 30 2012
G.f.: 1/(1 - x - 2*x^3/(1 - x - x^3/(1 - x - x^3/(1 - x - x^3/(1 - ...))))), a continued fraction. - Ilya Gutkovskiy, Nov 19 2021
a(n) ~ 1 / (sqrt((1-r)*(3-r)) * sqrt(Pi*n) * r^n), where r = 0.432040800333095... is the real root of the equation -1 + 2*r - r^2 + 4*r^3 = 0. - Vaclav Kotesovec, Jun 05 2022