A183983 1/4 the number of (n+1) X 7 binary arrays with all 2 X 2 subblock sums the same.
45, 47, 50, 56, 66, 86, 122, 194, 330, 602, 1130, 2186, 4266, 8426, 16682, 33194, 66090, 131882, 263210, 525866, 1050666, 2100266, 4198442, 8394794, 16785450, 33566762, 67125290, 134242346, 268468266, 536920106, 1073807402, 2147581994, 4295098410
Offset: 1
Keywords
Examples
Some solutions for 5 X 7. ..0..0..1..0..1..0..1....1..0..1..0..1..0..1....0..1..0..1..0..1..0 ..1..1..0..1..0..1..0....0..0..0..0..0..0..0....0..1..0..1..0..1..0 ..0..0..1..0..1..0..1....1..0..1..0..1..0..1....0..1..0..1..0..1..0 ..1..1..0..1..0..1..0....0..0..0..0..0..0..0....0..1..0..1..0..1..0 ..0..0..1..0..1..0..1....1..0..1..0..1..0..1....0..1..0..1..0..1..0
Links
- R. H. Hardin, Table of n, a(n) for n = 1..200
- Index entries for linear recurrences with constant coefficients, signature (3, 0, -6, 4).
Crossrefs
Cf. A183986.
Formula
Conjectures from Colin Barker, Apr 09 2018: (Start)
G.f.: x*(45 - 88*x - 91*x^2 + 176*x^3) / ((1 - x)*(1 - 2*x)*(1 - 2*x^2)).
a(n) = 3*2^(n/2-1) + 2^(n-1) + 42 for n even.
a(n) = 2^(n-1) + 2^((n+1)/2) + 42 for n odd.
a(n) = 3*a(n-1) - 6*a(n-3) + 4*a(n-4) for n>4.
(End)
The above empirical formula is correct. See note from Andrew Howroyd in A183986.
Comments