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

1,2

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).

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

Example

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

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 *)

Prog

(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

Crossrefs

Cf. A061348.

Sequence in context: A026305 A195256 A131786 * A254537 A135159 A190724

Adjacent sequences: A061706 A061707 A061708 * A061710 A061711 A061712

Keyword

nonn,easy,nice

Author

N. J. A. Sloane, Jun 20 2001

Status

approved