OFFSET
1,1
LINKS
R. H. Hardin, Table of n, a(n) for n = 1..210
FORMULA
Empirical: a(n) = 3*a(n-1) - 3*a(n-2) + 2*a(n-3) - 2*a(n-4) + 2*a(n-6) - 2*a(n-7) + 3*a(n-8) - 3*a(n-9) + a(n-10).
Empirical for n mod 12 = 0: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n + 1
Empirical for n mod 12 = 1: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n - (53/12)
Empirical for n mod 12 = 2: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n + (23/3)
Empirical for n mod 12 = 3: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n - (31/4)
Empirical for n mod 12 = 4: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n + (43/3)
Empirical for n mod 12 = 5: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n - (133/12)
Empirical for n mod 12 = 6: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n + 1
Empirical for n mod 12 = 7: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n + (67/12)
Empirical for n mod 12 = 8: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n + (23/3)
Empirical for n mod 12 = 9: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n - (71/4)
Empirical for n mod 12 = 10: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n + (43/3)
Empirical for n mod 12 = 11: a(n) = (5/4)*n^4 + (5/3)*n^3 + (5/2)*n^2 + n - (13/12)
Empirical g.f.: x*(2 + 47*x + 11*x^2 + 148*x^3 - 19*x^4 + 172*x^5 - 31*x^6 + 32*x^7 - 3*x^8 + x^9) / ((1 - x)^5*(1 + x)*(1 + x^2)*(1 + x + x^2)). - Colin Barker, Nov 09 2018
EXAMPLE
Some solutions for n=6:
..3....0....5....2....2....2....3....4....2....5....4....0....6....6....0....3
..5....0....4....1....3....3....0....6....2....1....3....3....4....2....0....3
..3....6....6....3....2....6....6....6....4....6....1....3....3....4....0....3
..4....6....3....2....1....3....2....2....2....3....2....3....6....6....1....0
..0....3....2....2....2....1....4....2....0....0....5....6....1....2....4....6
CROSSREFS
KEYWORD
nonn
AUTHOR
R. H. Hardin, Nov 03 2014
STATUS
approved