A379235 Numbers k such that A003961(k) = 2k +- 5, multiplied by the sign of difference A003961(k)-2k, where A003961 is fully multiplicative with a(prime(i)) = prime(i+1).

14, 15, -22, -46, 91, -2782, -269434, -1056574, 14129726, -25652506, 26594126, 34233062, 147087493

Offset: 1

Author

Antti Karttunen, Dec 23 2024

Keywords

• sign
• hard
• more

Comments

15 is the only term that is in A104210, the absolute values of all other terms residing in its complement, A319630, thus 5 occurs only once in A379231. Proof: If k is not a multiple of 5 and k is in A104210, then there are primes p (either p=2 or p > 5 and q = nextprime(p) that both divide k and q also divides A003961(k). However, q does not divide 2k +- 5, therefore the equation 2k +- 5 = A003961(k) is unsolvable in these cases. So let's assume that k is a multiple of 5, which immediately entails that k must be also a multiple of 3, for A003961(k) to be a multiple of 5. Let x = k/15; then the equation can be rewritten as 2*15*x +- 5 = A003961(15)*A003961(x) <=> 30x +- 5 = 35*A003961(x) <=> 5*(6x +- 1) = 5*7*A003961(x). The only value of x that satisfies the equation is x=1 (as A003961(n)>n for all n>1), hence k=15.

If it exists, abs(a(14)) > 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, A379231, A379233, A379237.

Formula

{sign(A252748(k)) * k, for k such that abs(A252748(k)) = 5}.