A261682 - OEIS

%I #27 May 03 2021 17:27:37

%S 1,4,6,16,25,61,98,232,381,889,1485,3433,5811,13339,22818,52072,89845,

%T 204001,354521,801421,1401291,3155299,5546381,12444841,21977515,

%U 49155331,87167163,194392627,345994215,769547191,1374282018,3049104232,5461770405,12090343921

%N a(n) = 2^n+(1+(n mod 2)/2)*C(n+1,floor((n+(n mod 2))/2))-1.

%H Alois P. Heinz, <a href="/A261682/b261682.txt">Table of n, a(n) for n = 0..1000</a>

%H Riccardo Biagioli, Frédéric Jouhet, and Philippe Nadeau, <a href="http://arxiv.org/abs/1411.4561">Combinatorics of fully commutative involutions in classical Coxeter groups</a>, arXiv preprint arXiv:1411.4561 [math.CO] (2014). See Prop. 2.1.

%H Riccardo Biagioli, Frédéric Jouhet, and Philippe Nadeau, <a href="http://dx.doi.org/10.1016/j.disc.2015.05.023">Combinatorics of fully commutative involutions in classical Coxeter groups</a>, Discrete Math., 338 (2015), 2242-2259. See Prop. 2.1.

%F 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.

%F 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

%F 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

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

%p seq(a(n), n=0..40);  # _Alois P. Heinz_, Sep 05 2015

%t 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 *)

%o (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

%K nonn

%O 0,2

%A _N. J. A. Sloane_, Sep 04 2015

%E New name from _Wesley Ivan Hurt_, May 02 2021