A183985 1/4 the number of (n+1) X 9 binary arrays with all 2 X 2 subblock sums the same.
153, 155, 158, 164, 174, 194, 230, 302, 438, 710, 1238, 2294, 4374, 8534, 16790, 33302, 66198, 131990, 263318, 525974, 1050774, 2100374, 4198550, 8394902, 16785558, 33566870, 67125398, 134242454, 268468374, 536920214, 1073807510, 2147582102
Offset: 1
Keywords
Examples
Some solutions for 5 X 9: ..0..0..1..1..1..0..0..1..1....0..1..0..0..1..1..1..1..1 ..1..1..0..0..0..1..1..0..0....1..0..1..1..0..0..0..0..0 ..0..0..1..1..1..0..0..1..1....0..1..0..0..1..1..1..1..1 ..1..1..0..0..0..1..1..0..0....1..0..1..1..0..0..0..0..0 ..0..0..1..1..1..0..0..1..1....0..1..0..0..1..1..1..1..1
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
Empirical: a(n) = 3*a(n-1) - 6*a(n-3) + 4*a(n-4).
Conjectures from Colin Barker, Apr 09 2018: (Start)
G.f.: x*(153 - 304*x - 307*x^2 + 608*x^3) / ((1 - x)*(1 - 2*x)*(1 - 2*x^2)).
a(n) = 3*2^(n/2-1) + 2^(n-1) + 150 for n even.
a(n) = 2^(n-1) + 2^((n+1)/2) + 150 for n odd.
(End)
The above empirical formula is correct. See note from Andrew Howroyd in A183986.
Comments