OFFSET
1,1
COMMENTS
Terms of this sequence are not solutions of Sum_{d|k} A069359(d), k >= 1.
Proof that 2 is not a solution of Sum_{d|k} A069359(d), k >= 1: (Start)
If 2 is a solution then the only summands of the above are either (0,2) or (0,1,1).
(0,2) cannot be the only summands. If 2 is a summand then it is also a divisor of a(n) and A069359(2) = 1. If 2 is a summand then so must 1 be a summand.
(0,1,1) cannot be the only summands. There must exist an additional summand A069359(p_1*p_2) where p_1 and p_2 (primes) contribute to each 1 in (0,1,1).
(End)
To prove that 5 is not a solution of Sum_{d|k} A069359(d), k >= 1 we need to show that each of the following summands cannot exist: (0,5), (0,1,4), (0,1,2,2), (0,1,1,3), (0,1,1,1,2). (0,1,1,1,1,1). Following from the above proof this is elementary.
EXAMPLE
2 is a term because it is not a solution of Sum_{d|k} A069359(d), k >= 1. See proof in Comments.
CROSSREFS
KEYWORD
nonn
AUTHOR
Torlach Rush, Feb 20 2024
STATUS
approved