OFFSET
1,3
COMMENTS
It appears that for n = 2^i*p, i>0, p prime, a(n) = p^(2^i) or a multiple of it.
For which n is a(n) negative?
For prime n and related conjectures, see A133954.
PROG
(PARI) a(n)=sum(i=0, floor(2^n/n), subst(Pol(binary(i*n)), x, 1)%2==0)-sum(i=0, floor(2^n/n), subst(Pol(binary(i*n)), x, 1)%2)
CROSSREFS
KEYWORD
sign,hard,more
AUTHOR
Ralf Stephan, Aug 31 2013
EXTENSIONS
a(31)-a(40) from Amiram Eldar, Apr 23 2022
STATUS
approved