A122868 Expansion of 1/sqrt(1-6x-3x^2).
1, 3, 15, 81, 459, 2673, 15849, 95175, 576963, 3523257, 21640365, 133549155, 827418645, 5143397535, 32063180535, 200367960201, 1254816463923, 7873205412825, 49482344889261, 311457546052659, 1963051327342449, 12387750763156227, 78258731003169435
Offset: 0
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1236 (first 201 terms from Vincenzo Librandi)
- 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.
- W. Y. C. Chen, N. Y. Li, L. W. Shapiro and S. H. F. Yan, Matrix identities on weighted partial Motzkin paths, European J. Combinatorics, 28 (2007), 1196-2007.
- M. Dziemianczuk, Counting Lattice Paths With Four Types of Steps, Graphs and Combinatorics, September 2013, Volume 30, Issue 6, pp 1427-1452.
- M. Dziemianczuk, On Directed Lattice Paths With Additional Vertical Steps, arXiv preprint arXiv:1410.5747 [math.CO], 2014.
- J. L. Ramírez and V. F. Sirvent, A Generalization of the k-Bonacci Sequence from Riordan Arrays, The Electronic Journal of Combinatorics, 22(1) (2015), #P1.38.
Crossrefs
Top row of array in A232973.
Programs
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Maple
b:= proc(x, y) option remember; `if`([x, y]=[0$2], 1, `if`(x>0, add(b(x-1, y+j), j=-1..1), 0)+ `if`(y>0, b(x, y-1), 0)+`if`(y<0, b(x, y+1), 0)) end: a:= n-> b(n, 0): seq(a(n), n=0..23); # Alois P. Heinz, Sep 21 2021 r := (-3)^(1/2): seq(simplify(r^n*LegendreP(n, -r)), n=0..10); # Mark van Hoeij, Nov 13 2022
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Mathematica
CoefficientList[Series[1/Sqrt[1-6*x-3*x^2], {x, 0, 20}], x] (* Vaclav Kotesovec, Oct 19 2012 *)
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Maxima
a(n):=coeff(expand((1+3*x+3*x^2)^n),x,n); makelist(a(n),n,0,12);
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PARI
my(x = 'x + O('x^30)); Vec(1/sqrt(1-6*x-3*x^2)) \\ Michel Marcus, Jan 29 2016
Formula
a(n) = Sum_{k=0..floor(n/2)} C(n,2k)*C(2k,k)*3^(n-k).
E.g.f. : exp(3x)*Bessel_I(0,2*sqrt(3)x).
D-finite with recurrence: n*a(n) + 3*(1-2*n)*a(n-1) + 3*(1-n)*a(n-2) = 0. - R. J. Mathar, Nov 14 2011 [proved in Belbachir et al. (see Table 1)]
a(n) ~ (1+sqrt(3))*(3+2*sqrt(3))^n/(2*sqrt(2*Pi*n)). - Vaclav Kotesovec, Oct 19 2012
a(n) = (-3)^(n/2)*LegendreP(n, -(-3)^(1/2)). - Mark van Hoeij, Nov 13 2022
Comments