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%I #11 Dec 23 2024 09:42:14
%S -5,6,-7,-161,1045,-2525,2795,4825,9725,-159115,307993,-359315,
%T -18377525,25484825
%N Numbers k such that A003961(k) = 2k +- 3, multiplied by the sign of difference A003961(k)-2k, where A003961 is fully multiplicative with a(prime(i)) = prime(i+1).
%C 6 is the only term that is in A104210, the absolute values of all other terms residing in its complement, A319630, thus 3 occurs only once in A379231. Proof: If k is not a multiple of 3 and k is in A104210, then there are primes p > 3 and q = nextprime(p) that both divide k and q also divides A003961(k). However, q does not divide 2k +- 3, therefore the equation 2k +- 3 = A003961(k) is unsolvable in these cases. So let's assume that k is a multiple of 3, which immediately entails that k must be also even, for A003961(k) to be a multiple of 3. Let x = k/6; then the equation can be rewritten as 2*6*x +- 3 = A003961(6)*A003961(x) <=> 12x +- 3 = 15*A003961(x) <=> 3*(4x +- 1) = 3*5*A003961(x). The only value of x that satisfies the equation is x=1 (as A003961(n)>n for all n>1), hence k=6.
%C If it exists, abs(a(15)) > 2^32.
%H <a href="/index/Pri#prime_indices">Index entries for sequences related to prime indices in the factorization of n</a>.
%F {sign(A252748(k)) * k, for k such that abs(A252748(k)) = 3}.
%Y Cf. A003961, A104210, A252748, A319630.
%Y Cf. also A048674, A348514, A378980, A379230, A379231, A379235, A379237.
%K sign,hard,more,new
%O 1,1
%A _Antti Karttunen_, Dec 23 2024