OFFSET
1,1
COMMENTS
a(n) exists: write n = r*2^i, where r is odd. Then r divides 2^phi(r) - 1, where phi is the Euler phi function. Choose k such that k phi(r) >= i.
Then n divides (2^{k*phi(r)} - 1)*2^{k*phi(r)}, which is a binary antipalindrome.
LINKS
Rémy Sigrist, Table of n, a(n) for n = 1..10000
Rémy Sigrist, PARI program for A318569
PROG
(PARI) See Links section.
(Python)
def comp(s): z, o = ord('0'), ord('1'); return s.translate({z:o, o:z})
def BCR(n): return int(comp(bin(n)[2:])[::-1], 2)
def a(n):
kn = n
while BCR(kn) != kn: kn += n
return kn//n
print([a(n) for n in range(1, 72)]) # Michael S. Branicky, Mar 20 2022
CROSSREFS
KEYWORD
AUTHOR
Jeffrey Shallit, Oct 12 2018
STATUS
approved