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This is a permutation of the positive integers.

From Antti Karttunen, Jun 20 2014: (Start)

Note the indexing: the domain starts from 0, while the range excludes zero, thus this is neither a bijection on the set of nonnegative integers nor on the set of positive natural numbers, but a bijection from the former set to the latter.

Apart from that discrepancy, this could be viewed as yet another "entanglement permutation" where the two complementary pairs to be interwoven together are even and odd numbers (A005843/A005408) which are entangled with the complementary pair even numbers (taken straight) and odd numbers in the order they appear in A003961: (A005843/A003961). See also A246375 which has almost the same recurrence.

Note how the even bisection halved gives the same sequence back. (For a(0)=1, take ceiling of 1/2).

(End)

From Antti Karttunen, Dec 30 2017: (Start)

This irregular table can be represented as a binary tree. Each child to the left is obtained by doubling the parent, and each child to the right is obtained by applying A003961 to the parent:

1

|

...................2...................

4 3

8......../ \........9 6......../ \........5

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

16 27 18 25 12 15 10 7

32 81 54 125 36 75 50 49 24 45 30 35 20 21 14 11

etc.

Sequence A005940 is obtained by scanning the same tree level by level in mirror image fashion. Also in binary trees A253563 and A253565 the terms on level of the tree are some permutation of the terms present on the level n of this tree. A252464(n) gives the distance of n from 1 in all these trees, and A252463 gives the parent of the node containing n.

A252737(n) gives the sum and A252738(n) the product of terms on row n (where 1 is on row 0, 1 on row 1, 3 and 4 on row 2, etc.). A252745(n) gives the number of nodes on level n whose left child is smaller than the right child, and A252744(n) is an indicator function for those nodes.

(End)

Note that the idea behind maps like this (and the mirror image A005940) admits also using alternative orderings of primes, not just standard magnitude-wise ordering (A000040). For example, A332214 is a similar sequence but with primes rearranged as in A332211, and A332817 is obtained when primes are rearranged as in A108546. - Antti Karttunen, Mar 11 2020

From Lorenzo Sauras Altuzarra, Nov 28 2020: (Start)

This sequence is generated from A228351 by applying the following procedure: 1) eliminate the compositions that end in one unless the first one, 2) subtract one unit from every component, 3) replace every tuple [t_1, ..., t_r] by Product_{k=1..r} A000040(k)^(t_k) (see the examples).

Is it true that a(n) = A337909(n+1) if and only if a(n+1) is not a term of A161992?

Does this permutation have any other cycle apart from (1), (2) and (6, 9, 16, 7)? (End)

From Antti Karttunen, Jul 25 2023: (Start)

(In the above question, it is assumed that the starting offset would be 1 instead of 0).

Questions:

Does a(n) = 1+A054429(n) hold only when n is of the form 2^k times 1, 3 or 7, i.e., one of the terms of A029748?

It seems that A007283 gives all fixed points of map n -> a(n), like A335431 seems to give all fixed points of map n -> A332214(n). Is there a general rule for mappings like these that the fixed points (if they exist) must be of the form 2^k times a certain kind of prime, i.e., that any odd composite (times 2^k) can certainly be excluded? See also note in A029747.

(End)

If the conjecture given in A364297 holds, then it implies the above conjecture about A007283. See also A364963. - Antti Karttunen, Sep 06 2023

Conjecture: a(n^k) is never of the form x^k, for any integers n > 0, k > 1, x >= 1. This holds at least for squares, cubes, seventh and eleventh powers (see A365808, A365801, A366287 and A366391). - Antti Karttunen, Sep 24 2023, Oct 10 2023.

See A365805 for why the above holds for any n^k, with k > 1. - Antti Karttunen, Nov 23 2023

Theorem (J.-P. Allouche, J. Shallit, G. Skordev): This sequence = A052499 - 1.

Ahnentafel numbers of ancestors contributing the X-chromosome to a female. A280873 gives the male inheritance. - Floris Strijbos, Jan 09 2017 [Equivalence with this sequence pointed out by John Blythe Dobson, May 09 2018]

The k-th composition in standard order (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions. This sequence lists all numbers k such that the k-th composition in standard order has no parts greater than two. See the corresponding example below. - Gus Wiseman, Apr 04 2020

The binary representation of a(n+1) has the same string of digits as the lazy Fibonacci (also known as dual Zeckendorf) representation of n that uses 0s and 1s. (The "+1" is essentially an adjustment for the offset of this sequence.) - Peter Munn, Sep 06 2022

The motivation for this sequence is the observation that one of the necessary conditions for inclusion in A293430 is almost the same as for what is required from k that A163511(k) were squarefree. Namely, all terms of A293430 can be found in A003754 (the former is a subsequence of the latter), while A163511(k) yields a squarefree number if and only if k is in A280873, which is a subsequence of A003754 (actually its intersection with A004760). Thus this sequence must be in the intersection of A293430 and A004760, which implies that the binary expansion of all terms is free of adjacent 0's and furthermore, none begins with bits "10". Indeed, in base-2 the terms look as: 1, 11, 110, 111, 1101, 1110, 1111, 11010, 11101, 11110, 11111, 111010, 111110, 11101111, 11110111, 111011110, 111011111, 111101110, 1110111110, 111011111111111101, 1110111111111111010.

A163511 applied to the first 21 terms yields 2, 3, 6, 5, 15, 10, 7, 30, 21, 14, 11, 42, 22, 187, 119, 374, 247, 238, 494, 6837, 13674, that in binary look like: 10, 11, 110, 101, 1111, 1010, 111, 11110, 10101, 1110, 1011, 101010, 10110, 10111011, 1110111, 101110110, 11110111, 11101110, 111101110, 1101010110101, 11010101101010. These are of course numbers such that both n and A243071(n) are in A293430, but not listed in ascending order.

Question: Is this sequence finite?

See also the binary tree illustration in A293230.

No runs of 1-bits of odd length allowed in the binary expansion of n (A007088), except that when n is an odd number, then the rightmost run may have an odd length. Subsequence A277335 does not allow that exception.

A005940(1+a(n)) yields a permutation of A028982, squares and twice squares.

Running maximum without repetition of the decimal equivalent of Gray code for n (A003188). - Frédéric Nouvier, Aug 14 2020

Related to a model of X-chromosome inheritance:

The two X chromosomes of a female are inherited one from each parent, while the X chromosome of a male is always inherited from his mother. Thus, the probability distribution of inheritance from the parents (mother, father) is (0.5, 0.5) for a female and (1, 0) for a male. For the inheritance of any X-chromosome of a female from the 2^i ancestors of the i-th generation before (right to left on an ahnentafel), the distribution is given by the first 2^i terms of the sequence, divided by 2^i. For example, the X-chromosome of a man, which was inherited from his mother, was inherited from his mother's 16 great-great-grandparents with probabilities 1/16, 1/16, 1/8, 0, 1/8, 1/8, 0, 0, 1/8, 1/8, 1/4, 0, 0, 0, 0, 0.