A201640 a(n) = Sum_{k=1..n} k*binomial(n, k)^3*(n^2 + n - k*n - k + k^2)/((n - k + 1)^2*n).

1, 8, 54, 368, 2550, 17952, 128086, 924128, 6729858, 49395440, 364979560, 2712343680, 20257516240, 151957919232, 1144281700110, 8646263301056, 65531851263978, 498047725561104, 3794627850238756, 28976634967413920, 221728252767064596, 1699859618636556608

Offset: 1

Author

Peter Luschny, Dec 19 2011

Keywords

• nonn

Comments

Let a meander be defined as in the link and m = 3. Then a(n) counts the invertible meanders of length m*(n + 1) built from arcs with central angle 360/m.

Examples

			a(2) = 8 because (the binary representations of) the invertible meanders of length 9 and central angle 120 degree are {100100100, 110110110, 100011000, 111001110, 110001000, 111011100, 111000000, 111111000}.
		

Links

• Peter Luschny, Meanders.

Crossrefs

Row sums of A202409.

Cf. A000079 (m=1), A001791 (m=2), this sequence (m=3).

Programs

Maple

A201640 := proc(n) add(k*binomial(n,k)^3*(n^2+n-k*n-k+k^2)/((n-k+1)^2*n),k=1..n) end; seq(A201640(n,k),n=1..22);
# Alternative:
h := (a, b, n) -> hypergeom([-n, -n, 1 - n], [a, b], -1):
a := n -> h(1, 1, n) + (n - 1)*h(1, 2, n):
seq(simplify(a(n)), n = 1..22); # Peter Luschny, Mar 16 2023
# Recurrence:
a := proc(n) option remember; if n < 3 then return [1, 8][n] fi;
(8*n*(3*n - 1)*(n - 1)*(n - 2)*a(n-2) + n*(21*n^3 - 25*n^2 - 2*n + 8)*a(n-1)) /
((n - 1)*(n + 1)^2*(3*n - 4)) end: seq(a(n), n = 1..22); # Peter Luschny, Mar 16 2023

Mathematica

Table[Sum[k*Binomial[n,k]^3*(n^2+n-k*n-k+k^2)/((n-k+1)^2*n),{k,1,n}],{n,1,20}] (* Vaclav Kotesovec, Oct 24 2012 *)

Formula

Recurrence: (n+1)^2*a(n) = n*(13*n+2)*a(n-1) - 6*(2*n-1)*(3*n-5)*a(n-2) - 2*(17*n^2-62*n+32)*a(n-3) + 16*(n-4)*(n-3)*a(n-4). - Vaclav Kotesovec, Oct 24 2012

a(n) ~ sqrt(3)*8^n/(Pi*n). - Vaclav Kotesovec, Oct 24 2012

From Peter Luschny, Mar 16 2023: (Start)

a(n) = h(1, 1, n) + (n - 1)*h(1, 2, n), where h(a, b, n) = hypergeom([-n, -n, 1 - n], [a, b], -1).

a(n) = (8*n*(3*n - 1)*(n - 1)*(n - 2)*a(n - 2) + n*(21*n^3 - 25*n^2 - 2*n + 8)* a(n-1)) / ((n - 1)*(n + 1)^2*(3*n - 4)) for n >= 3. (End)