OFFSET
1,1
COMMENTS
Each row must be either 01 or 10. The two columns are therefore binary complements with sum 2^k-1, where k = n + 1 is the number of rows. If k is even then 2^k-1 is divisible by 3 and the number of solutions is 2*(2^k-1)/3. If k is odd then 2^k-1 == 1 (mod 3) and the number of solutions is (2^k-2)/3. - Andrew Howroyd, Feb 03 2022
LINKS
R. H. Hardin, Table of n, a(n) for n = 1..210
FORMULA
a(n) = a(n-1) + 4*a(n-2) - 4*a(n-3).
From Colin Barker, Jan 01 2019: (Start)
G.f.: 2*x / ((1 - x)*(1 - 2*x)*(1 + 2*x)).
a(n) = 2^n - 2/3 - (-2)^n/3.
(End)
a(n) = 2*A052992(n). - Pascal Bisson, Feb 03 2022
EXAMPLE
All solutions for n=4:
0 1 0 1 1 0 1 0 1 0 0 1 1 0 1 0 0 1 0 1
0 1 0 1 0 1 1 0 0 1 1 0 1 0 0 1 1 0 1 0
1 0 0 1 0 1 1 0 1 0 1 0 0 1 1 0 0 1 0 1
0 1 1 0 0 1 0 1 1 0 1 0 1 0 0 1 1 0 0 1
1 0 0 1 1 0 1 0 1 0 0 1 0 1 0 1 1 0 0 1
PROG
(Python) [int(2**n - 2/3 -((-2)**n)/3) for n in range(1, 40)] # Pascal Bisson, Feb 03 2022
CROSSREFS
KEYWORD
nonn
AUTHOR
R. H. Hardin, Oct 08 2015
STATUS
approved