OFFSET
1,1
LINKS
Chai Wah Wu, Table of n, a(n) for n = 1..10000
FORMULA
From Chai Wah Wu, Jul 05 2021: (Start)
Theorem: 2^i for i >= 3 are terms.
Proof: This can be shown by induction on i. For the inductive step, A345055(1)=1, A345055(2)=-3, A345055(3)=2, and A011772(2^i)=2^(i+1)-1.
So for the divisors 1,2,4 for 2^i, A011772(2^i)*A345055(1)+A011772(2^(i-1))*A345055(2)+A011772(2^(i-2))*A345055(4)=0.
A345055(d)=0 for the other proper divisors d of 2^i by the inductive hypothesis as d is a power of 2 and this implies A345033(2^i)=0 for i>=3.
(End)
Conjecture: all terms are of the form 2^i, 2^i*p, 2^i*p*q or 7^2*p for some primes p and q. - Chai Wah Wu, Jul 05 2021
PROG
(PARI) isA345053(n) = (0==A345055(n));
CROSSREFS
KEYWORD
nonn
AUTHOR
Antti Karttunen, Jul 01 2021
STATUS
approved