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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)

Apart from missing 2, this sequence gives all numbers k such that the binary expansion of A156552(k) is a prefix of that of A156552(sigma(k)), that is, for k > 1, numbers k for which sigma(k) is a descendant of k in A005940-tree. This follows because of the two transitions x -> A005843(x) (doubling) and x -> A003961(x) (prime shift) used to generate descendants in A005940-tree, using A003961 at any step of the process will ruin the chances of encountering sigma(k) anywhere further down that subtree.

Proof: Any left child in A005940 (i.e., A003961(k) for k) is larger than sigma(k), for any k > 2 [see A286385 for a proof], and A003961(n) > n for all n > 1. Thus, apart from A003961(2) = 3 = sigma(2), A003961^t(k) > sigma(k), where A003961^t means t-fold application of prime shift, here with t >= 1. On the other hand, sigma(2n) > sigma(n) for all n, thus taking first some doubling steps before a run of one or more prime shift steps will not rescue us, as neither will taking further doubling steps after a bout of prime shifts.

The first terms of A325637 not included in this sequence are 154345556085770649600 and 9186050031556349952000, as they have abundancy index 6.

From Antti Karttunen, Nov 29 2021: (Start)

Odd terms of this sequence are given by the intersection of A349169 and A349174.

A064989 applied to the odd terms of this sequence gives the fixed points of A326042, i.e., the positions of zeros in A348736, and a subset of the positions of ones in A348941.

Odd terms of this sequence form a subsequence of A348943, but should occur neither in A348748 nor in A348749.

(End)

Claim: If there is an odd term y of A336702 larger than one, and it is the least one of such terms, then it should satisfy condition that for all nontrivial unitary divisor pairs d and x/d of x = A064989(y) [with gcd(d,x/d) = 1, 1 < d < x], the other divisor should reside in this sequence, and the other divisor in A348739. Proof: Applying A064989 to the odd terms of A336702 gives the fixed points of A326042. Suppose there are other odd terms in A336702 in addition to its initial 1, and let y be the least of these odd terms > 1 and x = A064989(y). Because A326042 (from here on indicated with f) is multiplicative, it follows that if we take any two nontrivial unitary divisors a and b of x, with x = a*b, gcd(a,b) = 1, 1 < a,b < x, then f(a)*f(b) = f(x) = x. Because f(x)/x = 1, we must have f(a)/a * f(b)/b = 1, as also the ratio f(n)/n is multiplicative. But f(a)/a and f(b)/b cannot be equal to 1, because then a and b would also be fixed by f, which contradicts our assumption that x were the least such fixed point larger than one. Therefore f(a) < a and f(b) > b, or vice versa. See also the comments in A348930, A348933.

Moreover, all odd perfect numbers (a subsequence of A336702), if such numbers exist, should also satisfy the same condition, regardless of whether they are the least of such numbers or not, because having a non-deficient proper divisor will push the abundancy index (ratio sigma(n)/n) of any number over 2. That is, for any such pair of nontrivial unitary divisors d and x/d, both A003961(d) and A003961(x/d) should be deficient, i.e., neither one should be in A337386. See also the condition given in A347383.

Terms that occur also in A337386 are: 120, 240, 360, 420, 480, 504, 540, 600, 630, ...

It is conjectured that there are no other zeros after a(1) = 0.

Terms that occur also in A337386 are: 180, 300, 720, 900, 960, 1008, 1200, 1440, 1620, 1800, 2016, 2400, ...

Denominator of ratio A326042(n) / n.

If there are no more 1's in this sequence after the initial one, then there are no odd terms of A336702 (numbers whose abundancy index is a power of 2) larger than one, and neither there are odd terms in A005820 or in A046060. Compare to similar conditions given in A336848, A336849 and A337339.

Multiplicative because A055615 and A326042 are.