A257066 Number of length 4 1..(n+1) arrays with every leading partial sum divisible by 2 or 3

2, 11, 45, 81, 256, 364, 738, 1149, 1905, 2401, 4096, 4912, 7172, 9297, 12685, 14641, 20736, 23436, 30344, 36455, 45633, 50625, 65536, 71872, 87438, 100767, 120141, 130321, 160000, 172300, 201782, 226521, 261745, 279841, 331776, 352944, 402848

Offset

1,1

Comments

Row 4 of A257062

Links

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

Formula

Empirical: a(n) = a(n-1) +4*a(n-6) -4*a(n-7) -6*a(n-12) +6*a(n-13) +4*a(n-18) -4*a(n-19) -a(n-24) +a(n-25)

Empirical for n mod 6 = 0: a(n) = (16/81)*n^4 + (4/9)*n^3 + (1/3)*n^2

Empirical for n mod 6 = 1: a(n) = (16/81)*n^4 + (50/81)*n^3 + (107/108)*n^2 + (44/81)*n - (113/324)

Empirical for n mod 6 = 2: a(n) = (16/81)*n^4 + (46/81)*n^3 + (83/108)*n^2 - (7/162)*n + (25/81)

Empirical for n mod 6 = 3: a(n) = (16/81)*n^4 + (20/27)*n^3 + (7/9)*n^2 + (2/3)*n

Empirical for n mod 6 = 4: a(n) = (16/81)*n^4 + (32/81)*n^3 + (8/27)*n^2 + (8/81)*n + (1/81)

Empirical for n mod 6 = 5: a(n) = (16/81)*n^4 + (64/81)*n^3 + (32/27)*n^2 + (64/81)*n + (16/81)

Example

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

Crossrefs

Cf. A257062

Sequence in context: A110679 A127109 A054208 * A209604 A120279 A037751

Adjacent sequences: A257063 A257064 A257065 * A257067 A257068 A257069

Keyword

nonn

Author

R. H. Hardin, Apr 15 2015

Status

approved