%I #11 Dec 31 2018 20:23:36
%S 2,3,5,9,18,35,69,137,274,547,1093,2185,4370,8739,17477,34953,69906,
%T 139811,279621,559241,1118482,2236963,4473925,8947849,17895698,
%U 35791395,71582789,143165577,286331154,572662307,1145324613,2290649225,4581298450
%N Number of (n+3) X (1+3) 0..1 arrays with each row and column divisible by 15, read as a binary number with top and left being the most significant bits.
%C a(n) is the number of multiples of 15 from 0 to 2^(n+3)-1. - _Robert Israel_, Dec 31 2018
%H R. H. Hardin, <a href="/A262450/b262450.txt">Table of n, a(n) for n = 1..210</a>
%F Empirical: a(n) = 2*a(n-1) + a(n-4) - 2*a(n-5).
%F Empirical g.f.: x*(2 - x - x^2 - x^3 - 2*x^4) / ((1 - x)*(1 + x)*(1 - 2*x)*(1 + x^2)). - _Colin Barker_, Dec 31 2018
%F a(n) = floor((2^(n+3)+14)/15). This satisfies the empirical recursion and g.f. - _Robert Israel_, Dec 31 2018
%e Some solutions for n=4:
%e ..1..1..1..1....1..1..1..1....0..0..0..0....0..0..0..0....1..1..1..1
%e ..1..1..1..1....1..1..1..1....0..0..0..0....0..0..0..0....0..0..0..0
%e ..1..1..1..1....0..0..0..0....1..1..1..1....0..0..0..0....1..1..1..1
%e ..1..1..1..1....1..1..1..1....1..1..1..1....1..1..1..1....1..1..1..1
%e ..0..0..0..0....0..0..0..0....1..1..1..1....1..1..1..1....0..0..0..0
%e ..0..0..0..0....0..0..0..0....1..1..1..1....1..1..1..1....1..1..1..1
%e ..0..0..0..0....1..1..1..1....0..0..0..0....1..1..1..1....0..0..0..0
%p seq(floor((2^(n+3)+14)/15), n=1..100); # _Robert Israel_, Dec 31 2018
%Y Column 1 of A262457.
%K nonn
%O 1,1
%A _R. H. Hardin_, Sep 23 2015