A183878 Number of arrangements of n+2 numbers in 0..3 with each number being the sum mod 4 of two others.

4, 35, 446, 2827, 13686, 59859, 250198, 1023347, 4140830, 16663627, 66867438, 267922683, 1072654438, 4292666147, 17175010598, 68709242467, 274856398542, 1099466524251, 4397952122110, 17591988892331, 70368333113174

Offset: 1

Author

R. H. Hardin, Jan 07 2011

Keywords

• nonn

Comments

Column 3 of A183884.

Examples

			Some solutions for n=2:
..1....2....0....0....2....2....3....1....2....2....2....0....3....0....2....2
..3....0....2....2....1....3....1....3....2....1....0....2....2....0....2....2
..3....0....2....2....1....1....1....1....2....3....2....0....1....2....0....0
..2....2....2....0....3....3....2....2....0....1....2....2....3....2....2....0
		

Links

• R. H. Hardin, Table of n, a(n) for n = 1..199

Crossrefs

Cf. A183884.

Formula

Empirical (for n>=3): 4^(n+2) - (2*n+7)*2^(n+2) - 2*n^3 - 9*n^2 - 10*n + 3. - Vaclav Kotesovec, Nov 27 2012

Conjectures from Colin Barker, Apr 05 2018: (Start)

G.f.: x*(4 - 13*x + 258*x^2 - 1087*x^3 + 1318*x^4 + 444*x^5 - 2040*x^6 + 1536*x^7 - 384*x^8) / ((1 - x)^4*(1 - 2*x)^2*(1 - 4*x)).

a(n) = 12*a(n-1) - 58*a(n-2) + 148*a(n-3) - 217*a(n-4) + 184*a(n-5) - 84*a(n-6) + 16*a(n-7) for n>9.

(End)