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Complement of A323599.
0

%I #54 Apr 20 2024 10:26:37

%S 2,5,29,39,53,59,73,95,119,123,125,129,137,145,147,149,157,159,163,

%T 173,179,191,199,207,209,213,219,221,235,251,257,263,265,269,271,279,

%U 291,293,299,303,305,325,327,329,343,345,347,359,365,367,369,375,385,395,397

%N Complement of A323599.

%C Terms of this sequence are not solutions of Sum_{d|k} A069359(d), k >= 1.

%C Proof that 2 is not a solution of Sum_{d|k} A069359(d), k >= 1: (Start)

%C If 2 is a solution then the only summands of the above are either (0,2) or (0,1,1).

%C (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.

%C (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).

%C (End)

%C 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.

%e 2 is a term because it is not a solution of Sum_{d|k} A069359(d), k >= 1. See proof in Comments.

%Y Cf. A069359, A323599.

%K nonn

%O 1,1

%A _Torlach Rush_, Feb 20 2024