A257065 Number of length 3 1..(n+1) arrays with every leading partial sum divisible by 2 or 3.

2, 6, 18, 27, 64, 81, 141, 200, 293, 343, 512, 578, 776, 954, 1208, 1331, 1728, 1875, 2291, 2652, 3147, 3375, 4096, 4356, 5070, 5678, 6494, 6859, 8000, 8405, 9497, 10416, 11633, 12167, 13824, 14406, 15956, 17250, 18948, 19683, 21952, 22743, 24831, 26564

Offset: 1

Author

R. H. Hardin, Apr 15 2015

Keywords

• nonn

Examples

			Some solutions for n=4:
..3....2....2....4....2....4....2....3....3....4....2....3....4....3....4....2
..5....1....1....2....4....5....4....3....5....5....4....5....4....1....4....1
..4....5....1....2....3....3....2....3....1....1....4....2....1....5....2....3
		

Links

• R. H. Hardin, Table of n, a(n) for n = 1..210

Crossrefs

Row 3 of A257062.

Formula

Empirical: a(n) = a(n-1) + 3*a(n-6) - 3*a(n-7) - 3*a(n-12) + 3*a(n-13) + a(n-18) - a(n-19).

Empirical for n mod 6 = 0: a(n) = (8/27)*n^3 + (4/9)*n^2 + (1/6)*n

Empirical for n mod 6 = 1: a(n) = (8/27)*n^3 + (2/3)*n^2 + (17/18)*n + (5/54)

Empirical for n mod 6 = 2: a(n) = (8/27)*n^3 + (2/3)*n^2 + (7/9)*n - (16/27)

Empirical for n mod 6 = 3: a(n) = (8/27)*n^3 + (8/9)*n^2 + (1/2)*n + (1/2)

Empirical for n mod 6 = 4: a(n) = (8/27)*n^3 + (4/9)*n^2 + (2/9)*n + (1/27)

Empirical for n mod 6 = 5: a(n) = (8/27)*n^3 + (8/9)*n^2 + (8/9)*n + (8/27).

Empirical g.f.: x*(2 + 4*x + 12*x^2 + 9*x^3 + 37*x^4 + 17*x^5 + 54*x^6 + 47*x^7 + 57*x^8 + 23*x^9 + 58*x^10 + 15*x^11 + 24*x^12 + 13*x^13 + 11*x^14 + x^16) / ((1 - x)^4*(1 + x)^3*(1 - x + x^2)^3*(1 + x + x^2)^3). - Colin Barker, Dec 20 2018