A207107 Number of n X 4 0..1 arrays avoiding 0 0 0 and 0 1 1 horizontally and 0 0 1 and 1 1 0 vertically.
9, 81, 271, 643, 1271, 2239, 3641, 5581, 8173, 11541, 15819, 21151, 27691, 35603, 45061, 56249, 69361, 84601, 102183, 122331, 145279, 171271, 200561, 233413, 270101, 310909, 356131, 406071, 461043, 521371, 587389, 659441, 737881, 823073
Offset: 1
Keywords
Examples
Some solutions for n=4: ..1..1..1..1....1..0..0..1....0..0..1..0....1..0..0..1....0..1..0..0 ..1..1..1..1....1..0..0..1....1..0..0..1....0..1..0..0....0..0..1..0 ..1..1..1..1....1..0..0..1....0..0..1..0....0..1..0..1....0..0..1..0 ..1..1..1..1....1..0..0..1....1..0..1..0....0..1..0..0....0..0..1..0
Links
- R. H. Hardin, Table of n, a(n) for n = 1..210
Crossrefs
Cf. A207111.
Formula
Empirical: a(n) = (5/12)*n^4 + (13/2)*n^3 + (115/12)*n^2 - (17/2)*n + 1.
Conjectures from Colin Barker, Feb 20 2018: (Start)
G.f.: x*(9 + 36*x - 44*x^2 + 8*x^3 + x^4) / (1 - x)^5.
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5) for n>5.
(End)
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