A183879 Number of arrangements of n+2 numbers in 0..4 with each number being the sum mod 5 of two others.

1, 25, 821, 8361, 57625, 336617, 1817149, 9433849, 48036737, 242284857, 1216409221, 6093687497, 30495285865, 152537717449, 762827288141, 3814448005209, 19072935346513, 95366219539097, 476834503269013

Offset: 1

Author

R. H. Hardin, Jan 07 2011

Keywords

• nonn

Comments

Column 4 of A183884.

Examples

			Some solutions for n=2:
..4....3....4....1....2....3....0....1....1....3....3....2....1....4....1....3
..1....1....1....4....1....1....0....3....2....4....2....4....2....3....4....4
..2....2....3....2....3....4....0....2....3....1....1....3....4....1....3....2
..3....4....2....3....4....2....0....4....4....2....4....1....3....2....2....1
		

Links

• R. H. Hardin, Table of n, a(n) for n = 1..185

Crossrefs

Cf. A183884.

Formula

Empirical (for n>=2): 5^(n+2) - (10*n^2+70*n+124)*2^n + 2*(3*n+8)*(n^2+5*n+8). - Vaclav Kotesovec, Nov 27 2012

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

G.f.: x*(1 + 10*x + 538*x^2 - 1960*x^3 + 701*x^4 + 4706*x^5 - 7204*x^6 + 4312*x^7 - 960*x^8) / ((1 - x)^4*(1 - 2*x)^3*(1 - 5*x)).

a(n) = 15*a(n-1) - 92*a(n-2) + 306*a(n-3) - 609*a(n-4) + 747*a(n-5) - 554*a(n-6) + 228*a(n-7) - 40*a(n-8) for n>9.

(End)