A098482 Expansion of 1/sqrt((1-x)^2 - 4*x^4).

1, 1, 1, 1, 3, 7, 13, 21, 37, 73, 147, 283, 531, 1007, 1953, 3817, 7423, 14371, 27877, 54333, 106189, 207585, 405743, 793719, 1554889, 3049525, 5984803, 11751067, 23086695, 45388291, 89289765, 175746797, 346077153, 681795925, 1343790319, 2649687079, 5226711507

Offset: 0

Author

Paul Barry, Sep 10 2004

Keywords

• easy
• nonn

Comments

From Joerg Arndt, Jul 01 2011: (Start)

Empirical: Number of lattice paths from (0,0) to (n,n) using steps (4,0), (0,4), (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).

Empirical: Number of lattice paths from (0,0) to (n,n) using steps (3,1), (1,3), (1,1). (End)

1/sqrt((1-x)^2-4*r*x^4) expands to sum(k=0..floor(n/2), binomial(n-2*k,k)*binomial(n-3*k,k)*r^k ).

Diagonal of the rational function 1 / ((1-x)*(1-y) - x^3*y^4). - Seiichi Manyama, Apr 29 2025

a(n) is the number of ways to tile a strip of length n with 1 X 1 squares and 1 X 2 red dominos and 1 X 2 blue dominos, with an equal number of red and blue dominos. - Greg Dresden and Leo Zhang, Jul 08 2025

Examples

			From _Joerg Arndt_, Jul 01 2011: (Start)
The triangle of lattice paths from (0,0) to (n,k) using steps (3,1), (1,3), (1,1) begins
  1;
  0, 1;
  0, 0, 1;
  0, 1, 0, 1;
  0, 0, 2, 0, 3;
  0, 0, 0, 3, 0, 7;
  0, 0, 1, 0, 4, 0, 13;
  0, 0, 0, 3, 0, 8, 0, 21;
  0, 0, 0, 0, 6, 0, 18, 0, 37;
  0, 0, 0, 1, 0, 10, 0, 37, 0, 73;
The triangle of lattice paths from (0,0) to (n,k) using steps (4,0), (0,4), (1,1) begins
  1;
  0, 1;
  0, 0, 1;
  0, 0, 0, 1;
  1, 0, 0, 0, 3;
  0, 2, 0, 0, 0, 7;
  0, 0, 3, 0, 0, 0, 13;
  0, 0, 0, 4, 0, 0, 0, 21;
  1, 0, 0, 0, 8, 0, 0, 0, 37;
  0, 3, 0, 0, 0, 18, 0, 0, 0, 73;
The diagonals of both appear to be this sequence.  (End)
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• Steffen Eger, On the Number of Many-to-Many Alignments of N Sequences, arXiv:1511.00622 [math.CO], 2015.

Crossrefs

Cf. A098479, A098483, A098484.

Programs

Maple

seq(add(binomial(n-3*k,k)*binomial(n-2*k,k),k=0..floor(n/3)),n=0..34); # Zerinvary Lajos, Apr 03 2007

Mathematica

CoefficientList[Series[1/Sqrt[(1-x)^2-4*x^4], {x, 0, 20}], x] (* Vaclav Kotesovec, Jun 23 2014 *)

PARI

/* as lattice paths, assuming the first comment is true */
/* same as in A092566 but use either of */
steps=[[4,0], [0,4], [1,1]];
steps=[[3,1], [1,3], [1,1]];
/* Joerg Arndt, Jul 01 2011 */

Formula

a(n) = Sum_{k=0..floor(n/4)} binomial(n-2*k, k) * binomial(n-3*k, k).

D-finite with recurrence: n*a(n) = (2*n-1)*a(n-1) - (n-1)*a(n-2) + 4*(n-2)*a(n-4). - Vaclav Kotesovec, Jun 23 2014

a(n) ~ 2^(n+1/2) / sqrt(3*Pi*n). - Vaclav Kotesovec, Jun 23 2014

G.f.: 1/(1 - x - 2*x^4/(1 - x - x^4/(1 - x - x^4/(1 - x - x^4/(1 - ...))))), a continued fraction. - Ilya Gutkovskiy, Nov 19 2021

a(n) = Sum_{k=0..floor(n/4)} binomial(n-2*k, 2*k) * binomial(2*k, k). - Greg Dresden and Leo Zhang, Jul 08 2025