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 *)
CoefficientList[Series[1/Sqrt[(1-x)^2-4x^3], {x, 0, 40}], x] (* Harvey P. Dale, Aug 13 2024 *)
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
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Sep 10 2004
STATUS
approved