A337342 Numbers k such that A048673(k) divides 1+A003973(k).

1, 10, 584, 3824, 23008, 5033216

Offset: 1

Author

Antti Karttunen, Aug 24 2020

Keywords

• nonn
• hard
• more

Comments

Numbers k such that A048673(k) = A337335(k). Equivalently, numbers k such that (A003961(k)+1)/2 divides 1+A003973(k).

No squares larger than one in this sequence => No quasiperfect numbers. See also A337339. For any x corresponding to a quasiperfect number qp = A003961(x), the quotient (1+A003973(x)) / A048673(x) should be 4. Thus that A003961(x) should also be a member of A325311.

At least for the terms x = a(2) .. a(6) here, the quotient (1+A003973(x)) / A048673(x) = 3. The terms for which the quotient is 3 are precisely those which by prime shifting become the terms of A007593 (that are all odd), thus the terms y = A064989(A007593(n)), for n >= 1, form a subsequence of this sequence.

a(7) > 2^28.

Terms 65810851904356352, 30943274395471606363637940224, 40102483616531202199118491418624 are also in the sequence, but their positions are unknown. (Adapted from Jud McCranie's Dec 16 1999 comment in A007593).

Crossrefs

Cf. A003961, A003973, A007593, A048673, A064989, A209922, A324899, A325311, A336700, A337335, A337339.

Programs

PARI

A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
isA337342(n) = { my(s=A003961(n)); !((1+sigma(s))%((1+s)/2)); };