cp's OEIS Frontend

Numbers k > 1 such that A332224(A156552(k)) = A087808(sigma(A156552(k))) is equal to 2*A048675(k) = A048675(k^2).

Notably, of the first 150 terms (4 .. 9996), 156 = 2^2 * 3 * 13 is the only even term that does not map to a prime, as A156552(156) = 267 = 3*89 (and sigma(267) = 360 = 4*90).

Although sigma(A156552(k)) = A323243(k) is a multiple of 4 for most of the terms k present in this sequence, there are exceptions, for example 840350 = A005940(1+A332445(1)) = 2^1 * 5^2 * 7^5 is one, as A048675(A332223(840350)) = 98 = 2*A048675(840350) and A323243(840350) = 2394 == 2 (mod 4).

Numbers k such that A332223(k) is in A353308.

If there existed any 3-perfect number (A005820) of the form x = 4u+2 and not divisible by 3, then x would be also included in A353365 and A005940(1+x) would be included in this sequence. See comments in A353365.

Numbers k such that A348717(A332223(k)) = A348717(k) form a subsequence of this sequence. As its subsequence, we further have sequences A005940(1+A336702(n)) and A005940(1+A027687(n)), computed for n >= 1, and sorted into ascending order.

For any other number n than those in A326182 we have a(n) < A003961(n).

Fixed points k (for which a(k) = k) satisfy A003973(k) = 2^e * A003961(k) for some exponent e >= 0. Applying A003961 to such numbers gives the odd terms in A336702, of which there are likely to be just a single instance, its initial 1. (Clarified Nov 07 2021).

Conjecture: There are no other fixed points than a(1) = 1. If true, then there are no odd perfect numbers. This condition is equivalent to the condition that if A161942 has no fixed points larger than one, then there are no odd perfect numbers. This follows as whenever k is a fixed point, that is, a(k) = k, then we should also have A003961(a(k)) = A003961(A064989(sigma(A003961(k)))) = A161942(A003961(k)) = A003961(k). Note that A003961 is an injective and surjective mapping from natural numbers to odd numbers, A064989 is its (left) inverse, and composition A003961(A064989(n)) is equivalent to A000265(n).

From Antti Karttunen, Aug 05 2020: (Start)

For any hypothetical odd perfect number x, we would have A003973(k) = 2 * A003961(k), with k = A064989(x) and x = A003961(k). Thus we would have a(k) = A064989(sigma(A003961(k))) = A064989(sigma(x)) = A064989(2*x) = A064989(x) = k. On the other hand, A003973(k) = sigma(A003961(k)) < A003961(A003961(k)) [see A286385 for the reason why], so a necessary condition for this is that x should be one of the terms of A246282. (Clarified Dec 01 2020).

(End)

Numbers k such that A336698(k) [= A000265(1+A161942(k))] is equal to A000265(1+k).

Numbers k such that A337194(k) = 2^e * A000265(1+k), for some e >= 1, where that e = A337195(k).

Any odd perfect number would trivially satisfy this condition.

Also, all hypothetical quasiperfect numbers, numbers k that satisfy sigma(k) = 2k+1, would be members.

Question: Is A066175 a subsequence of this sequence?

From Antti Karttunen, Aug 23 2020: (Start)

Numbers k such that (1+k) = 2^e * A336698(k), for some e >= 0.

Thus numbers k such that for some e >= 0, (1+k) = 2^(e-A337195(k)) * A337194(k), or equally, that A337194(k) = 2^(A337195(k)-e) * (1+k).

Conjecture: There are no even terms. This is equivalent to claim that there are no k such that A336698(k) = 1+k: If we assume that k is even, then in above equations we set e=0, and the requirement will then become that A337194(k) = 2^A337195(k)*(1+k), thus 1+k = A336698(k) = A000265(1+A000265(sigma(k))).

(End)

All odd squares (A016754) are present, but not all terms are squares. A348743 gives the nonsquare terms.

Odd terms of A336702 form a subsequence. Also all odd terms of A005820 would be present here, as well as any hypothetical quasi-perfect numbers (see comments and references in A332223, A336700), both in A016754. - Antti Karttunen, Nov 28 2024

Positions of 0's in this sequence is given by such numbers n that sigma(n) = 2^k * n + r, for some n >= 1, k >= 0, 0 <= r < 2^k. These would include also quasi-perfect numbers and their generalizations, numbers n such that sigma(n) = 2^k * n + 2^k - 1, for some n > 1, k > 0 (see comments in A332223), if such numbers exist. However, it is conjectured that there are no other zeros than those given by A336702.