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A337342
Numbers k such that A048673(k) divides 1+A003973(k).
4
1, 10, 584, 3824, 23008, 5033216
OFFSET
1,2
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).
PROG
(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)); };
KEYWORD
nonn,hard,more
AUTHOR
Antti Karttunen, Aug 24 2020
STATUS
approved