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%I #42 Aug 29 2020 02:46:25
%S 1,10,584,3824,23008,5033216
%N Numbers k such that A048673(k) divides 1+A003973(k).
%C Numbers k such that A048673(k) = A337335(k). Equivalently, numbers k such that (A003961(k)+1)/2 divides 1+A003973(k).
%C 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.
%C 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.
%C a(7) > 2^28.
%C 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).
%o (PARI)
%o A003961(n) = { my(f = factor(n)); for(i=1, #f~, f[i, 1] = nextprime(f[i, 1]+1)); factorback(f); };
%o isA337342(n) = { my(s=A003961(n)); !((1+sigma(s))%((1+s)/2)); };
%Y Cf. A003961, A003973, A007593, A048673, A064989, A209922, A324899, A325311, A336700, A337335, A337339.
%K nonn,hard,more
%O 1,2
%A _Antti Karttunen_, Aug 24 2020
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