A061709 Consider a (hollow) triangle with n cells on each edge, for a total of 3(n-1) cells if n>1, or 1 cell if n=1; a(n) is number of ways of labeling cells with 0's and 1's; triangle may be rotated and turned over.

1, 4, 20, 104, 752, 5600, 44224, 350592, 2800384, 22377984, 178990080, 1431721984, 11453509632, 91626496000, 733009854464, 5864066220032, 46912512917504, 375300002545664, 3002399885885440, 24019198281252864, 192153585175232512, 1537228674957312000

Offset: 1

Author

N. J. A. Sloane, Jun 20 2001

Keywords

• nonn
• easy
• nice

Examples

			a(2) = 4, the labelings being {000}, {001}, {011}, {111}. Some of the 20 solutions for n=3 are as follows:
..0......1.......0......1.......1.......1.......0
.0.0....0.0.....1.0....1.0.....0.0.....0.0.....1.1
0.0.0..0.0.0...0.0.0..0.0.0...1.0.0...0.1.0...0.0.0
The first solution for n = 4 is
...0
..0.0
.0...0
0.0.0.0
		

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (10,-8,-80,128).

Crossrefs

Cf. A061348.

Programs

Mathematica

Join[{1},Table[((2^(3(n-1)))+2^n+3*2^Floor[(3n-1)/2])/6,{n,2,30}]] (* or *) Join[{1},LinearRecurrence[{10,-8,-80,128},{4,20,104,752},30]] (* Harvey P. Dale, Apr 22 2013 *)

PARI

Vec(-x*(64*x^4+16*x^3-12*x^2-6*x+1)/((2*x-1)*(8*x-1)*(8*x^2-1)) + O(x^100)) \\ Colin Barker, Mar 17 2015

Formula

a(n) = (1/6)*(2^(3*(n-1))+2^n+3*2^(floor((3*n-1)/2))) for n>1.

a(2)=4, a(3)=20, a(4)=104, a(5)=752, a(n)=10*a(n-1)-8*a(n-2)- 80*a(n-3)+ 128*a(n-4). - Harvey P. Dale, Apr 22 2013

G.f.: -x*(64*x^4+16*x^3-12*x^2-6*x+1) / ((2*x-1)*(8*x-1)*(8*x^2-1)). - Colin Barker, Mar 17 2015