A082590 Expansion of 1/((1 - 2*x)*sqrt(1 - 4*x)).
1, 4, 14, 48, 166, 584, 2092, 7616, 28102, 104824, 394404, 1494240, 5692636, 21785872, 83688344, 322494208, 1246068806, 4825743832, 18726622964, 72798509728, 283443548276, 1105144970992, 4314388905704, 16862208539008, 65972020761116, 258354647959984, 1012627828868072
Offset: 0
Keywords
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..300
- Shalosh B. Ekhad and Doron Zeilberger, Automatic Solution of Richard Stanley's Amer. Math. Monthly Problem #11610 and ANY Problem of That Type, arXiv:1112.6207 [math.CO], 2011. See subpages for rigorous derivations of the g.f., the recurrence, asymptotics for this sequence.
- Alejandro Erickson and Frank Ruskey, Enumerating maximal tatami mat coverings of square grids with v vertical dominoes, arXiv:1304.0070 [math.CO], 2013.
- Y. Kamiyama, On the middle dimensional homology classes of equilateral polygon spaces, arXiv:1507.03161 [math.AT], 2015.
Programs
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Maple
A082590 := proc(n) coeftayl( 1/(1-2*x)/sqrt(1-4*x),x=0,n) ; end proc: # R. J. Mathar, Nov 06 2013 A082590 := n -> 2^n*JacobiP(n, 1/2, -1 - n, 3): seq(simplify(A082590(n)), n = 0..26); # Peter Luschny, Jan 22 2025
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Mathematica
CoefficientList[ Series[ 1/((1 - 2*x)*Sqrt[1 - 4*x]), {x, 0, 25}], x] (* Jean-François Alcover, Mar 26 2013 *) Table[2^(n) JacobiP[n, 1/2, -1-n, 3], {n, 0, 30}] (* Vincenzo Librandi, May 26 2013 *)
Formula
a(n) = 2^n*JacobiP(n, 1/2, -1-n, 3).
a(n) = Sum_{k=0..n+1} binomial(2*n+2, k) * sin((n - k + 1)*Pi/2). - Paul Barry, Nov 02 2004
From Paul Barry, Sep 04 2005: (Start)
a(n) = Sum_{k=0..n} 2^(n-k)*binomial(2*k, k).
a(n) = Sum_{k=0..n} (2*k)! * (2*(n-k))!/(n!*k!*(n-k)!). (End)
a(n) = Sum_{k=0..n} C(2*n, n)*C(n, k)/C(2*n, 2*k). - Paul Barry, Mar 18 2007
G.f.: 1/(1 - 4*x + 2*x^2/(1 + x^2/(1 - 4*x + x^2/(1 + x^2/(1 - 4*x + x^2/(1 + ... (continued fraction). - Paul Barry, Apr 26 2009
D-finite with recurrence: n*a(n) + 2*(-3*n+1)*a(n-1) + 4*(2*n-1)*a(n-2) = 0. - R. J. Mathar, Dec 03 2012
a(n) ~ 2^(2*n + 1)/sqrt(Pi*n). - Vaclav Kotesovec, Aug 15 2013
a(n) = 2^(n + 1)*Pochhammer(1/2, n+1)*hyper2F1([1/2,-n], [3/2], -1)/n!. - Peter Luschny, Aug 02 2014
a(n) - 2*a(n-1) = A000984(n). - R. J. Mathar, Apr 24 2024
a(n) = 2^n*JacobiP(n, 1/2, -1 - n, 3). - Peter Luschny, Jan 22 2025
Comments