A379237 Numbers k such that A003961(k) = 2k +- 7, multiplied by the sign of difference A003961(k)-2k, where A003961 is fully multiplicative with a(prime(i)) = prime(i+1).

9, 35, -38, 39, -51, 69, -374, -4521, 7869, 10426, 12639, -16094, -29354, 102579, -103881, 1295206, -3298514, 4267318, 478642449, -2120241621

Offset: 1

Author

Antti Karttunen, Dec 23 2024

Keywords

• sign
• hard
• more

Comments

35 is the only term that is in A104210, the absolute values of all other terms residing in its complement, A319630, thus 7 occurs only once in A379231. Proof: If k is not a multiple of 7 and k is in A104210, then there are primes p (either p=2, p=3 or p > 7 and q = nextprime(p) that both divide k and q also divides A003961(k). However, q does not divide 2k +- 7, therefore the equation 2k +- 7 = A003961(k) is unsolvable in these cases. So let's assume that k is a multiple of 7, which immediately entails that k must be also a multiple of 5, for A003961(k) to be a multiple of 7. Let x = k/35; then the equation can be rewritten as 2*35*x +- 7 = A003961(35)*A003961(x) <=> 70x +- 7 = 77*A003961(x) <=> 7*(10x +- 1) = 7*11*A003961(x). The only value of x that satisfies the equation is x=1 (as A003961(n)>n for all n>1), hence k=35.

If it exists, abs(a(21)) > 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, A379235.

Formula

{sign(A252748(k)) * k, for k such that abs(A252748(k)) = 7}.