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For all i, j: a(i) = a(j) => A292261(i) = A292261(j).

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

Inverse of sequence A064216 considered as a permutation of the positive integers. - Howard A. Landman, Sep 25 2001

From Antti Karttunen, Dec 20 2014: (Start)

Permutation of natural numbers obtained by replacing each prime divisor of n with the next prime and mapping the generated odd numbers back to all natural numbers by adding one and then halving.

Note: there is a 7-cycle almost right in the beginning: (6 8 14 17 10 11 7). (See also comments at A249821. This 7-cycle is endlessly copied in permutations like A250249/A250250.)

The only 3-cycle in range 1 .. 402653184 is (2821 3460 5639).

For 1- and 2-cycles, see A245449.

(End)

The first 5-cycle is (1410, 2783, 2451, 2703, 2803). - Robert Israel, Jan 15 2015

From Michel Marcus, Aug 09 2020: (Start)

(5194, 5356, 6149, 8186, 10709), (46048, 51339, 87915, 102673, 137205) and (175811, 200924, 226175, 246397, 267838) are other 5-cycles.

(10242, 20479, 21413, 29245, 30275, 40354, 48241) is another 7-cycle. (End)

From Antti Karttunen, Feb 10 2021: (Start)

Somewhat artificially, also this permutation can be represented as a binary tree. Each child to the left is obtained by multiplying the parent by 3 and subtracting one, while each child to the right is obtained by applying A253888 to the parent:

1

|

................../ \..................

2 3

5......../ \........4 8......../ \........6

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

14 13 11 7 23 9 17 18

41 10 38 12 32 28 20 15 68 25 26 63 50 16 53 19

etc.

Each node's (> 1) parent can be obtained with A253889. Sequences A292243, A292244, A292245 and A292246 are constructed from the residues (mod 3) of the vertices encountered on the path from n to the root (1).

(End)

Note the indexing: the domain starts from 1, while the range includes also zero.

See also the comments at A163511, which is the inverse permutation to this one.

Note the indexing: the domain starts from 0, while the range excludes zero.

From Antti Karttunen, May 30 2017: (Start)

This sequence can be represented as a binary tree. Each left hand child is obtained by applying A254049(n) when the parent contains n, and each right hand child is obtained by applying A016789(n-1) (i.e., multiply by 3, subtract one) to the parent's contents:

1

|

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

3 5

4......../ \........8 13......../ \........14

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

6 11 18 23 25 38 63 41

7 17 28 32 39 53 88 68 61 74 123 113 172 188 313 122

etc.

This is a mirror image of the tree depicted in A245612.

(End)

Note the indexing: the domain starts from 1, while the range includes also zero.

The odd bisection of A243071 decremented by one and halved. (For a(1) = 0, take ceiling of -1/2).