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A261682 If n mod 2 = 0 then a(n) = 2^n+binomial(n+1,n/2)-1 otherwise a(n) = 2^n+(3/2)*binomial(n+1,(n+1)/2)-1. 1
1, 4, 6, 16, 25, 61, 98, 232, 381, 889, 1485, 3433, 5811, 13339, 22818, 52072, 89845, 204001, 354521, 801421, 1401291, 3155299, 5546381, 12444841, 21977515, 49155331, 87167163, 194392627, 345994215, 769547191, 1374282018, 3049104232, 5461770405, 12090343921 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

LINKS

Alois P. Heinz, Table of n, a(n) for n = 0..1000

Riccardo Biagioli, Frédéric Jouhet, and Philippe Nadeau, Combinatorics of fully commutative involutions in classical Coxeter groups, arXiv preprint arXiv:1411.4561 [math.CO] (2014). See Prop. 2.1.

Riccardo Biagioli, Frédéric Jouhet, and Philippe Nadeau, Combinatorics of fully commutative involutions in classical Coxeter groups, Discrete Math., 338 (2015), 2242-2259. See Prop. 2.1.

FORMULA

If n mod 2 = 0, then a(n+2) = a(n+1) + a(n) + A000108(n+1) - 2^n - 1; otherwise, a(n+2) = a(n+1) + a(n) + A046224(n+2) - 2^n - 1. - Eric Werley, Sep 16, 2015

Conjecture: -(n+2)*(15*n^2-29*n-56)*a(n) +9*(5*n^3-3*n^2-24*n-20)*a(n-1) +2*(15*n^3-14*n^2-219*n+158)*a(n-2) +36*(-5*n^3+8*n^2+31*n-38)*a(n-3) +8*(n-2)*(15*n^2+n-70)*a(n-4)=0. - R. J. Mathar, Jan 04 2017

MAPLE

a:= n-> 2^n +((2+irem(n, 2))/2)*binomial(n+1, ceil(n/2))-1:

seq(a(n), n=0..40);  # Alois P. Heinz, Sep 05 2015

MATHEMATICA

Table[If[EvenQ@ n, 2^n + Binomial[n + 1, n/2] - 1, 2^n + (3/2) Binomial[n + 1, (n + 1)/2] - 1], {n, 0, 33}] (* Michael De Vlieger, Sep 24 2015 *)

PROG

(PARI) a(n) = if (n%2, 2^n+(3/2)*binomial(n+1, (n+1)/2)-1, 2^n+binomial(n+1, n/2)-1); \\ Michel Marcus, Sep 05 2015

CROSSREFS

Sequence in context: A242251 A122537 A059736 * A102731 A007179 A112576

Adjacent sequences:  A261679 A261680 A261681 * A261683 A261684 A261685

KEYWORD

nonn

AUTHOR

N. J. A. Sloane, Sep 04 2015

STATUS

approved

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Last modified July 5 23:36 EDT 2020. Contains 335475 sequences. (Running on oeis4.)