OFFSET
1,1
COMMENTS
Column 2 of A201353.
LINKS
R. H. Hardin, Table of n, a(n) for n = 1..210
Ângela Mestre, José Agapito, Square Matrices Generated by Sequences of Riordan Arrays, J. Int. Seq., Vol. 22 (2019), Article 19.8.4.
FORMULA
Empirical: a(n) = (1/6)*n^3 - (1/2)*n^2 + (10/3)*n - 2 for n>1.
Conjectures from Colin Barker, May 22 2018: (Start)
G.f.: x*(2 - 4*x + 4*x^2 - 2*x^3 + x^4) / (1 - x)^4.
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) for n>5.
(End)
EXAMPLE
Some solutions for n=10:
..0..0....0..0....0..0....0..0....0..1....0..1....0..0....0..0....0..0....0..0
..0..0....0..0....0..0....0..0....0..1....0..1....0..0....0..1....0..1....0..0
..0..1....0..0....0..0....0..0....1..0....0..1....0..0....0..1....0..1....0..0
..0..1....0..0....1..1....0..1....1..0....0..1....0..0....0..1....0..1....0..0
..0..1....0..0....1..1....0..1....1..0....0..1....0..0....1..0....1..0....0..0
..0..1....0..0....1..1....0..1....1..0....1..0....0..0....1..0....1..0....0..1
..0..1....0..0....1..1....1..1....1..0....1..0....0..1....1..0....1..0....0..1
..0..1....0..0....1..1....1..1....1..0....1..0....0..1....1..1....1..0....0..1
..1..0....0..1....1..1....1..1....1..1....1..0....0..1....1..1....1..0....1..0
..1..0....1..1....1..1....1..1....1..1....1..0....1..1....1..1....1..1....1..0
MATHEMATICA
Rest@ CoefficientList[Series[x (2 - 4 x + 4 x^2 - 2 x^3 + x^4)/(1 - x)^4, {x, 0, 47}], x] (* Michael De Vlieger, Mar 27 2020 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
R. H. Hardin, Nov 30 2011
STATUS
approved