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Restricted growth sequence transform of A366258.

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

The original name was "Bit-reverse of n, including as many leading as trailing zeros." - Antti Karttunen, Dec 25 2024

A permutation of integers consisting only of fixed points and pairs. a(n)=n when n is a binary palindrome (including as many leading as trailing zeros), otherwise a(n)=A003010(n) (i.e. n has no axis of symmetry). A057890 gives the palindromes (fixed points, akin to A006995) while A057891 gives the "antidromes" (pairs). See also A280505.

This is multiplicative in domain GF(2)[X], i.e. with carryless binary arithmetic. A193231 is another such permutation of natural numbers. - Antti Karttunen, Dec 25 2024

Let's define "property S" for sequences as follows: If s is any sequence of positive natural numbers, normalized to begin with offset 1, then it satisfies the S-property if LCM-transform(s) is equal to the sequence obtained by applying A014963 to sequence s, or in other words, when for all n >= 1, lcm {s(1)..s(n)} / lcm {s(1)..s(n-1)} = A014963(s(n)). This holds if and only if, for all n >= 1, when, either (case A): s(n) is of the form p^k, p prime, then gcd(s(n), lcm {s(1)..s(n-1)}) must be equal to p^(k-1), or (case B): when s(n) is not a prime power, then gcd(s(n), lcm {s(1)..s(n-1)}) must be equal to s(n). Together the cases (A) and (B) reduce to the condition that each prime power should appear in s before any of its multiples do.

Clearly the Doudna-sequence satisfies the property by the way of its construction, as do many of its variants like A356867 (see A369060).

Also, for any base-2 related permutation b that keeps all the numbers of range [2^k, 2^(1+k)[ in the same range, i.e., if for all n >= 1, A000523(b(n)) = A000523(n), then the above property is automatically satisfied.

Furthermore, because in Doudna-sequence no multiple of any term is located on the same row as the term itself (see the tree-illustration in A005940), it follows that any composition of A005940 with any such base-2 related permutation as mentioned above also automatically satisfies the S-property, for example, the permutations A163511, A243353, A253563, A253565, A366260, A366263 and A366275.

Note: Like A005940 itself, also this sequence might be more logical with the starting offset 0 instead of 1, to better align with the underlying mapping from the binary expansion of n to the prime factorization. - Antti Karttunen, Jan 24 2024

The viabin number of an integer partition is defined in the following way. Consider the southeast border of the Ferrers board of the integer partition and consider the binary number obtained by replacing each east step with 1 and each north step, except the last one, with 0. The corresponding decimal form is, by definition, the viabin number of the given integer partition. "Viabin" is coined from "via binary". For example, consider the integer partition [2,2,2,1]. The southeast border of its Ferrers board yields 10100, leading to the viabin number 20.

From Omar E. Pol, Jul 24 2017: (Start)

Consider that this is also an irregular triangle read by rows T(n,k), n>=0, k>=1, in which the row lengths are the elements of A011782 (see example).

Conjectures:

1) It appears that if k > 1 and k is a power of 2 then column k lists the positive integers (A000027).

2) It appears that column k lists the nonnegative integers (A001477) starting from the first elements of the column k.

3) It appears that if n > 0 then row n lists the first 2^(n-1) elements of A063787 in reverse order. (End)

Restricted growth sequence transform of A366279.

For all i, j >= 0, a(i) = a(j) => A290251(i) = A290251(j).