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

Cf. also A326042, A347391, A347392, A347393, A347394, A348736, A348748, A348749, A348941, A348943, A349169, A349174.

Except for Apart from missing 2, this sequence also 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.

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 are a subsequence of A348943.

Cf. also A326042, A347391, A347392, A347393, A347394, A348736, A348941, A348943, A349169, A349174.

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.

(End)

Cf. also A326042, A347391, A347392, A347393, A347394, A348736, A349169, A349174.

approved

editing

proposed

approved

editing

proposed

From Antti Karttunen, Sep 11 2021: (Start)

Above proof more formally: For any n > 2, A003961(n) > sigma(n), and any child x under A003961(n) [the left child of n] certainly is also greater than sigma(n), because sigma(A003961(n)) > sigma(n), and sigma(2^k * n) > sigma(n). Also sigma(A003961(2^k * n)) = 3^k * sigma(A003961(n)) > sigma(n). Thus the only way sigma(n) could be a descendant of n in the Doudna tree is for sigma(n) to be located in the rightmost branch under n, where all the terms are obtained by doubling their parent.

Any odd perfect number x (if such numbers exist) should be included in this sequence, and 2*x should be a term of A347392.

(End)

From Antti Karttunen, Sep 11 2021: (Start)

Any odd perfect number Above proof more formally: For any n > 2, A003961(n) > sigma(n), and any child x under A003961(n) [the left child of n] certainly is also greater than sigma(n), because sigma(A003961(n)) > sigma(if such numbers existn) should be included in this sequence, , and sigma(2^k * n) > sigma(n). Also sigma(A003961(2^k * n)) = 3^k *x should sigma(A003961(n)) > sigma(n). Thus the only way sigma(n) could be a term descendant of A347392n in the Doudna tree is for sigma(n) to be located in the rightmost branch under n, where all the terms are obtained by doubling their parent. - _Antti Karttunen_, Sep 04 2021

Any odd perfect number x (if such numbers exist) should be included in this sequence, and 2*x should be a term of A347392.

(End)

Union with {2} gives the positions of zeros in A347381. Cf. also A347391, A347392, A347393, A347394.

Cf. also A347391, A347392, A347393, A347394.