OFFSET
0,3
COMMENTS
Conjecture: a(n) = A007306(floor(n/2)+1). - Georg Fischer, Nov 28 2022
LINKS
Georg Fischer, Table of n, a(n) for n = 0..1000
Jeffrey Shallit and Lukas Spiegelhofer, Continuants, run lengths, and Barry's modified Pascal triangle, arXiv:1710.06203 [math.CO], 2017.
FORMULA
a(n) = Sum_{k=0..floor(n/2)} mod(Sum_{j=0..n-2k} C(k, j) C(n-2k, j) (1+(-1)^j)/2, 2). (corrected by Jeffrey Shallit, May 18 2016)
PROG
(PARI) a(n) = sum(k=0, n\2, sum(j=0, n-2*k, binomial(k, j)*binomial(n-2*k, j)*(1+(-1)^j)/2) % 2); \\ Michel Marcus, Jun 06 2021
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Paul Barry, Nov 17 2005
STATUS
approved