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

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

Offset

0,4

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

Links

Michael De Vlieger, Table of n, a(n) for n = 0..2749

Paul Barry, Continued fractions and transformations of integer sequences, JIS 12 (2009) 09.7.6.

Hacène Belbachir, Abdelghani Mehdaoui, and László Szalay, Diagonal Sums in the Pascal Pyramid, II: Applications, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.

J. Cigler, Some nice Hankel determinants, arXiv preprint arXiv:1109.1449 [math.CO], 2011.

Steffen Eger, On the Number of Many-to-Many Alignments of N Sequences, arXiv:1511.00622 [math.CO], 2015.

Steffen Eger, The Combinatorics of Weighted Vector Compositions, arXiv:1704.04964 [math.CO], 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

Example

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)

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 *)

Prog

(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

Crossrefs

Cf. A098480, A098481.

Sequence in context: A333653 A301594 A080241 * A119445 A146904 A146432

Adjacent sequences: A098476 A098477 A098478 * A098480 A098481 A098482

Keyword

easy,nonn

Author

Paul Barry, Sep 10 2004

Status

approved