OFFSET
0,3
LINKS
Andrew Howroyd, Table of n, a(n) for n = 0..200
Alin Bostan, Jordan Tirrell, Bruce W. Westbury and Yi Zhang, On sequences associated to the invariant theory of rank two simple Lie algebras, arXiv:1911.10288 [math.CO], 2019.
Alin Bostan, Jordan Tirrell, Bruce W. Westbury and Yi Zhang, On some combinatorial sequences associated to invariant theory, arXiv:2110.13753 [math.CO], 2021.
Wei Chen, Enumeration of Set Partitions Refined by Crossing and Nesting Numbers, MS Thesis, Department of Mathematics. Simon Fraser University, Fall 2014. See Th. 3.4.1.
Juan B. Gil, David Kenepp and Michael Weiner, Pattern-avoiding permutations by inactive sites, Pennsylvania State University, Altoona (2020).
Eric Marberg, Crossings and nestings in colored set partitions, arXiv preprint arXiv:1203.5738 [math.CO], 2012-2013.
FORMULA
Marberg gives a recurrence and g.f.
a(n) = (2*(5*n^2 + 6*n - 2)*a(n-1) - 9*(n - 2)*(n + 1)*a(n-2))/((n + 2)*(n + 3)) for n >= 2. - Andrew Howroyd, Dec 26 2019, after Marberg theorem 1.7.
a(n) ~ 3^(2*n + 11/2) / (16*Pi*n^4). - Vaclav Kotesovec, Jan 03 2020
a(n) = Sum_{j=0..n} Sum_{k=0..j} C(n-1,j)*C(j+2,k) * C(j+2,k+1) * C(j+2,k+2) / (C(j+2,1) * C(j+2,2)). - Michael D. Weiner, Jan 21 2020
G.f.: hypergeom([1/4, 5/4],[2],-64*x/((9*x-1)^3*(x-1)))/(3*(1-9*x)^(3/4)*x^2*(1-x)^(1/4))-(2*x-1)*(x-1)/(3*x^2). - Mark van Hoeij, Jul 27 2021
a(n) = ((n+3)*hypergeom([1/2, -1-n, -2-n],[2, 2],4)-3*(n+2)*hypergeom([1/2, -1-n, -n],[2, 2],4))/(2*n*(n+2)) for n > 0. - Mark van Hoeij, Nov 06 2023
MAPLE
A216947 := proc(n)
option remember ;
local npr ;
if n <=1 then
1 ;
else
npr := n-2 ;
9*npr*(npr+3)*procname(n-2)-2*(5*npr^2+26*npr+30)*procname(n-1) ;
-%/(npr+4)/(npr+5) ;
end if;
end proc:
seq(A216947(n), n=0..20) ; # R. J. Mathar, Nov 16 2012
MATHEMATICA
a[n_] := a[n] = Module[{npr, s}, If[n <= 1, 1, npr = n-2; s = 9*npr*(npr + 3)*a[n-2] - 2*(5*npr^2 + 26*npr + 30)*a[n-1]; -s/(npr + 4)/(npr + 5)]];
Table[a[n], {n, 0, 22}] (* Jean-François Alcover, Nov 29 2017, after R. J. Mathar *)
PROG
(PARI) seq(n)={my(a=vector(n+1)); a[1]=1; a[2]=1; for(n=2, n, a[n+1] = (2*(5*n^2 + 6*n - 2)*a[n] - 9*(n - 2)*(n + 1)*a[n-1])/((n + 2)*(n + 3))); a} \\ Andrew Howroyd, Dec 26 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
N. J. A. Sloane, Sep 22 2012
STATUS
approved