OFFSET
1,1
COMMENTS
Column 1 ofA262488.
LINKS
R. H. Hardin,Table of n, a(n) for n = 1..210
Index entries for linear recurrences with constant coefficients,signature (3,-2,0,0,0,-1,3,-2).
FORMULA
Empirical: a(n) = 3*a(n-1) -2*a(n-2) -a(n-6) +3*a(n-7) -2*a(n-8).
FromRobert Israel,Dec 15 2016: (Start)
All rows are either 0,0,0,0 or 1,1,0,1; first column is base-2 expansion of any multiple of 13 less than 2^(n+3).
a(n) = 1+floor((2^(n+3)/13).
G.f.: (2*x-3*x^2+x^4+x^7-2*x^8)/(1-3*x+2*x^2+x^6-3*x^7+2*x^8).
Since 2^12 == 1 (mod 13), a(n+12) - 2^12*a(n) has period 12, and from this we can derive the g.f. and recursion. (End)
EXAMPLE
Some solutions for n=4
..0..0..0..0....1..1..0..1....1..1..0..1....1..1..0..1....0..0..0..0
..1..1..0..1....0..0..0..0....0..0..0..0....0..0..0..0....0..0..0..0
..0..0..0..0....0..0..0..0....1..1..0..1....0..0..0..0....1..1..0..1
..0..0..0..0....1..1..0..1....1..1..0..1....0..0..0..0....1..1..0..1
..1..1..0..1....1..1..0..1....0..0..0..0....0..0..0..0....0..0..0..0
..1..1..0..1....1..1..0..1....1..1..0..1....0..0..0..0....1..1..0..1
..1..1..0..1....0..0..0..0....1..1..0..1....1..1..0..1....0..0..0..0
MAPLE
seq(1+floor(2^(n+3)/13), n=1..60); #Robert Israel,Dec 15 2016
CROSSREFS
KEYWORD
nonn,base
AUTHOR
R. H. Hardin,Sep 24 2015
STATUS
approved