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Expansion of 1/sqrt((1-x)^2 - 4*x^3).
12

%I #66 Aug 13 2024 12:15:08

%S 1,1,1,3,7,13,27,61,133,287,633,1407,3121,6943,15517,34755,77959,

%T 175213,394499,889461,2007963,4538485,10269247,23258881,52726599,

%U 119627977,271624315,617180533,1403272799,3192557561,7267485523,16552454205

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

%C 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.

%C Hankel transform is A120580. - _Paul Barry_, Sep 19 2008

%C From _Joerg Arndt_, Jul 01 2011: (Start)

%C Apparently the number of lattice paths from (0,0) to (n,n) using steps (3,0), (0,3), (1,1).

%C 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).

%C Apparently the number of lattice paths from (0,0) to (n,n) using steps (1,2), (2,1), (1,1). (End)

%C 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

%H Michael De Vlieger, <a href="/A098479/b098479.txt">Table of n, a(n) for n = 0..2749</a>

%H Paul Barry, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL12/Barry3/barry93.html">Continued fractions and transformations of integer sequences</a>, JIS 12 (2009) 09.7.6.

%H Hacène Belbachir, Abdelghani Mehdaoui, and László Szalay, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL22/Szalay/szalay42.html">Diagonal Sums in the Pascal Pyramid, II: Applications</a>, J. Int. Seq., Vol. 22 (2019), Article 19.3.5.

%H J. Cigler, <a href="http://arxiv.org/abs/1109.1449">Some nice Hankel determinants</a>, arXiv preprint arXiv:1109.1449 [math.CO], 2011.

%H Steffen Eger, <a href="http://arxiv.org/abs/1511.00622">On the Number of Many-to-Many Alignments of N Sequences</a>, arXiv:1511.00622 [math.CO], 2015.

%H Steffen Eger, <a href="https://arxiv.org/abs/1704.04964">The Combinatorics of Weighted Vector Compositions</a>, arXiv:1704.04964 [math.CO], 2017.

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

%F 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

%F 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

%F 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

%e From _Joerg Arndt_, Jul 01 2011: (Start)

%e The triangle of lattice paths from (0,0) to (n,k) using steps (1,2), (2,1), (1,1) begins

%e 1;

%e 0, 1;

%e 0, 1, 1;

%e 0, 0, 2, 3;

%e 0, 0, 1, 3, 7;

%e 0, 0, 0, 3, 7, 13;

%e 0, 0, 0, 1, 6, 17, 27;

%e 0, 0, 0, 0, 4, 14, 36, 61;

%e The triangle of lattice paths from (0,0) to (n,k) using steps (3,0), (0,3), (1,1) begins

%e 1;

%e 0, 1;

%e 0, 0, 1;

%e 1, 0, 0, 3;

%e 0, 2, 0, 0, 7;

%e 0, 0, 3, 0, 0, 13;

%e 1, 0, 0, 7, 0, 0, 27;

%e 0, 3, 0, 0, 17, 0, 0, 61;

%e The diagonals of both appear to be this sequence. (End)

%t 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 *)

%t CoefficientList[Series[1/Sqrt[(1-x)^2-4x^3],{x,0,40}],x] (* _Harvey P. Dale_, Aug 13 2024 *)

%o (PARI) /* as lattice paths, assuming the first comment is true */

%o /* same as in A092566 but use either of the following */

%o steps=[[3,0], [0,3], [1,1]];

%o steps=[[1,1], [1,2], [2,1]];

%o /* _Joerg Arndt_, Jul 01 2011 */

%o (Python)

%o from sympy import binomial

%o def a(n): return sum(binomial(n - k, k) * binomial(n - 2*k, k) for k in range(n//2 + 1))

%o print([a(n) for n in range(31)]) # _Indranil Ghosh_, Apr 18 2017

%Y Cf. A098480, A098481.

%K easy,nonn

%O 0,4

%A _Paul Barry_, Sep 10 2004