A379233 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).

-5, 6, -7, -161, 1045, -2525, 2795, 4825, 9725, -159115, 307993, -359315, -18377525, 25484825

Offset: 1

Author

Antti Karttunen, Dec 23 2024

Keywords

• sign
• hard
• more

Comments

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.

If it exists, abs(a(15)) > 2^32.

Links

• Index entries for sequences related to prime indices in the factorization of n.

Crossrefs

Cf. A003961, A104210, A252748, A319630.

Cf. also A048674, A348514, A378980, A379230, A379231, A379235, A379237.

Formula

{sign(A252748(k)) * k, for k such that abs(A252748(k)) = 3}.