%I #26 Jun 07 2024 14:50:02

%S 1,8,14,20,26,35,38,44,50,62,64,65,68,74,77,86,92,95,110,112,116,119,

%T 122,125,134,143,146,155,158,160,161,164,170,185,188,194,196,203,206,

%U 208,209,212,215,218,221,230,236,242,254,275,278,280,284,287,290,299,302,304,305,314,323,326,329,332,335,341,343

%N Numbers k such that A001414(k) and A083345(k) are both multiples of 3, where A001414 is fully additive with a(p) = p, and A083345 is the numerator of the fully additive function with a(p) = 1/p.

%C If k is a term, then 3^9 * k is also a term. See A373476.

%C A369659 is a subsequence of this sequence, giving the terms that are not multiples of 3. This follows because A083345(n) = n' / gcd(n',n) and from the following lemma: When k is not a multiple of 3, then either sopfr(k) [= A001414(k)] and k' [= A003415(k)] are both multiples of 3, or both are non-multiples of 3.

%C Proof of the lemma: As k is not a multiple of 3, all its prime factors p, q, r, s, t, u, v, w, ... (not necessarily all distinct) are either of the form 3m+1 or 3m-1. Let's first eliminate from k all triplets of primes that are of the same type modulo 3, either -1 or +1, (marked now as p, q, r) as they do not affect the divisibility by 3 of either the sopfr(k) or k'. In the case of the arithmetic derivative this is because we have k' = (pqr)' * (k/pqr) + (k/pqr)' * pqr, and as we know that the first summand is a multiple of 3 (because (pqr)' is), therefore the divisibility of the whole expression by 3 depends only on whether (k/pqr)' is a multiple of 3, as certainly pqr is not a multiple of 3.

%C What will remain after such elimination process has been completed as far as possible, must be either 1, or of the form p*q (p and q of different types), or p*q*r*s (with two primes of one type, and two primes of the other type), in which cases both sopfr(k) and k' are multiples of 3, or then alternatively, what remains must be of the form p*q (p and q of the same type), or p*q*r (with two primes of one type and the third of the other type), both cases which indicate that both sopfr(k) and k' are non-multiples of 3.

%H Antti Karttunen, <a href="/A373475/b373475.txt">Table of n, a(n) for n = 1..33911</a>

%H <a href="/index/Se#sequences_which_agree_for_a_long_time">Index entries for sequences which agree for a long time but are different</a>

%F a(n) = A373476(n) / 3^9.

%e 110 = 2*5*11 is a term of this sequence because 2+5+11 = 18 is a multiple of 3, and also 2*5 + 2*11 + 5*11 = 87 is a multiple of 3.

%e 54 (= A369644(10)) is NOT a term of this sequence, because A001414(54) = 11 is not a multiple of 3, although A083345(54) = 3 is.

%e 19683 = 3^9 is a term of this sequence, because both A001414(19683) = 9*3 = 27 and A083345(19683) = A003415(3^9)/gcd(3^9, A003415(3^9)) = 3, are multiples of 3.

%o (PARI) isA373475 = A373474;

%Y Cf. A001414, A003415, A083345, A373474 (characteristic function).

%Y Positions of multiples of 3 in A373363.

%Y Intersection of A289142 and A369644.

%Y Subsequence of A373478.

%Y Disjoint union of A369659 and A373476.

%Y Differs from A369659 for the first time at n=4186, where a(4186) = A373476(1) = 19683, a term not present in A369659, as it is the first multiple of 3 in this sequence.

%K nonn

%O 1,2

%A _Antti Karttunen_, Jun 06 2024