OFFSET
1,1
COMMENTS
Row 3 of A202784.
LINKS
R. H. Hardin, Table of n, a(n) for n = 1..210
Robert Israel, Proof of empirical formula for A202785
Wikipedia, Ehrhart polynomial
FORMULA
Empirical: a(n) = (3/10)*n^5 + (3/2)*n^4 + (7/2)*n^3 + (9/2)*n^2 + (16/5)*n + 1.
Conjectures from Colin Barker, Jun 01 2018: (Start)
G.f.: x*(7 - 2*x + x^2)*(2 + x + 4*x^2 - x^3) / (1 - x)^6.
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6) for n>6.
(End)
Empirical formula verified (see link): Robert Israel, May 02 2019
EXAMPLE
Some solutions for n=7:
..3..2..1....3..5..5....0..6..2....0..7..5....4..2..1....5..6..0....1..6..1
..2..0..4....5..6..2....2..1..5....6..1..5....3..2..2....0..4..7....5..2..1
..1..4..1....5..2..6....6..1..1....6..4..2....0..3..4....6..1..4....2..0..6
MAPLE
seq((3/10)*n^5 + (3/2)*n^4 + (7/2)*n^3 + (9/2)*n^2 + (16/5)*n + 1, n=1..30); # Robert Israel, May 02 2019
CROSSREFS
KEYWORD
nonn
AUTHOR
R. H. Hardin, Dec 24 2011
STATUS
approved