OFFSET
1,1
LINKS
R. H. Hardin, Table of n, a(n) for n = 1..210
FORMULA
Empirical: a(n) = 2*a(n-1) - a(n-3) - 2*a(n-5) + 2*a(n-6) + a(n-8) - 2*a(n-10) + a(n-11).
Empirical for n mod 12 = 0: a(n) = n^4 + n^3 + (25/6)*n^2 - (5/3)*n.
Empirical for n mod 12 = 1: a(n) = n^4 + n^3 + (25/6)*n^2 + (4/3)*n + (5/2).
Empirical for n mod 12 = 2: a(n) = n^4 + n^3 + (25/6)*n^2 - (5/3)*n - (4/3).
Empirical for n mod 12 = 3: a(n) = n^4 + n^3 + (25/6)*n^2 + (4/3)*n - (3/2).
Empirical for n mod 12 = 4: a(n) = n^4 + n^3 + (25/6)*n^2 - (5/3)*n.
Empirical for n mod 12 = 5: a(n) = n^4 + n^3 + (25/6)*n^2 + (4/3)*n + (7/6).
Empirical for n mod 12 = 6: a(n) = n^4 + n^3 + (25/6)*n^2 - (5/3)*n.
Empirical for n mod 12 = 7: a(n) = n^4 + n^3 + (25/6)*n^2 + (4/3)*n - (3/2).
Empirical for n mod 12 = 8: a(n) = n^4 + n^3 + (25/6)*n^2 - (5/3)*n - (4/3).
Empirical for n mod 12 = 9: a(n) = n^4 + n^3 + (25/6)*n^2 + (4/3)*n + (5/2).
Empirical for n mod 12 = 10: a(n) = n^4 + n^3 + (25/6)*n^2 - (5/3)*n.
Empirical for n mod 12 = 11: a(n) = n^4 + n^3 + (25/6)*n^2 + (4/3)*n - (17/6).
Empirical g.f.: 2*x*(5 + 8*x + 38*x^2 + 47*x^3 + 69*x^4 + 48*x^5 + 42*x^6 + 17*x^7 + 14*x^8) / ((1 - x)^5*(1 + x)^2*(1 + x^2)*(1 + x + x^2)). - Colin Barker, Nov 08 2018
EXAMPLE
Some solutions for n=6:
..2....1....0....2....3....2....2....4....4....2....1....6....0....4....1....1
..0....4....0....6....2....3....5....4....5....5....0....3....6....6....3....5
..6....4....3....2....0....5....6....1....5....0....1....6....1....1....4....5
..4....1....4....3....6....2....6....0....0....6....1....2....4....2....6....1
CROSSREFS
KEYWORD
nonn
AUTHOR
R. H. Hardin, Oct 06 2014
STATUS
approved