cp's OEIS Frontend

A natural number n occurs here if and only if it is either a power of 2, or satisfies A001511(n) = A071178(n) [the exponent of highest power of 2 dividing n is one less than the exponent of the largest prime factor of n], and all the intermediate exponents form a palindrome. [Please see the definition of A241916.]

Numbers for which the corresponding rows in A112798 and A241918 are the conjugate partitions of each other.

Terms that occur in 2-cycles of permutation A241916. (E.g., A241916(6)=9, A241916(9)=6.)

Apart from its initial terms, 1 and 2, all the terms of A088902 occur here because A241909 has no other fixed points than 1 and 2.

See comments in A368900.

A permutation of the natural numbers. - Robert G. Wilson v, Feb 22 2005

Fixed points: A029747. - Reinhard Zumkeller, Aug 23 2006

The even bisection, when halved, gives the sequence back. - Antti Karttunen, Jun 28 2014

From Antti Karttunen, Dec 21 2014: (Start)

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

1

|

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

3 4

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

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

7 10 15 12 25 18 27 16

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

etc.

Sequence A163511 is obtained by scanning the same tree level by level, from right to left. 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.

A252737(n) gives the sum and A252738(n) the product of terms on row n (where 1 is on row 0, 2 on row 1, 3 and 4 on row 2, etc.). A252745(n) gives the number of nodes on level n whose left child is larger than the right child, A252750 the difference between left and right child for each node from node 2 onward.

(End)

-A008836(a(1+n)) gives the corresponding numerator for A323505(n). - Antti Karttunen, Jan 19 2019

(a(2n+1)-1)/2 [= A244154(n)-1, for n >= 0] is a permutation of the natural numbers. - George Beck and Antti Karttunen, Dec 08 2019

From Peter Munn, Oct 04 2020: (Start)

Each term has the same even part (equivalently, the same 2-adic valuation) as its index.

Using the tree depicted in Antti Karttunen's 2014 comment:

Numbers are on the right branch (4 and descendants) if and only if divisible by the square of their largest prime factor (cf. A070003).

Numbers on the left branch, together with 2, are listed in A102750.

(End)

According to Kutz (1981), he learned of this sequence from American mathematician Byron Leon McAllister (1929-2017) who attributed the invention of the sequence to a graduate student by the name of Doudna (first name Paul?) in the mid-1950's at the University of Wisconsin. - Amiram Eldar, Jun 17 2021

From David James Sycamore, Sep 23 2022: (Start)

Alternative (recursive) definition: If n is a power of 2 then a(n)=n. Otherwise, if 2^j is the greatest power of 2 not exceeding n, and if k = n - 2^j, then a(n) is the least m*a(k) that has not occurred previously, where m is an odd prime.

Example: Use recursion with n = 77 = 2^6 + 13. a(13) = 25 and since 11 is the smallest odd prime m such that m*a(13) has not already occurred (see a(27), a(29),a(45)), then a(77) = 11*25 = 275. (End)

The odd bisection, when transformed by replacing all prime(k)^e in a(2*n - 1) with prime(k-1)^e, returns a(n), and thus gives the sequence back. - David James Sycamore, Sep 28 2022

Factor n; replace each prime(i) with i, take the conjugate partition, replace parts i with prime(i) and multiply out.

From Antti Karttunen, May 12-19 2014: (Start)

For all n >= 1, A001222(a(n)) = A061395(n), and vice versa, A061395(a(n)) = A001222(n).

Because the partition conjugation doesn't change the partition's total sum, this permutation preserves A056239, i.e., A056239(a(n)) = A056239(n) for all n.

(Similarly, for all n, A001221(a(n)) = A001221(n), because the number of steps in the Ferrers/Young-diagram stays invariant under the conjugation. - Note added Apr 29 2022).

Because this permutation commutes with A241909, in other words, as a(A241909(n)) = A241909(a(n)) for all n, from which follows, because both permutations are self-inverse, that a(n) = A241909(a(A241909(n))), it means that this is also induced when partitions are conjugated in the partition enumeration system A241918. (Not only in A112798.)

(End)

From Antti Karttunen, Jul 31 2014: (Start)

Rows in arrays A243060 and A243070 converge towards this sequence, and also, assuming no surprises at the rate of that convergence, this sequence occurs also as the central diagonal of both.

Each even number is mapped to a unique term of A102750 and vice versa.

Conversely, each odd number (larger than 1) is mapped to a unique term of A070003, and vice versa. The permutation pair A243287-A243288 has the same property. This is also used to induce the permutations A244981-A244984.

Taking the odd bisection and dividing out the largest prime factor results in the permutation A243505.

Shares with A245613 the property that each term of A028260 is mapped to a unique term of A244990 and each term of A026424 is mapped to a unique term of A244991.

Conversely, with A245614 (the inverse of above), shares the property that each term of A244990 is mapped to a unique term of A028260 and each term of A244991 is mapped to a unique term of A026424.

(End)

The Maple program follows the steps described in the first comment. The subprogram C yields the conjugate partition of a given partition. - Emeric Deutsch, May 09 2015

The Heinz number of the partition that is conjugate to the partition with Heinz number n. The Heinz number of a partition p = [p_1, p_2, ..., p_r] is defined as Product(p_j-th prime, j=1...r). Example: a(3) = 4. Indeed, the partition with Heinz number 3 is [2]; its conjugate is [1,1] having Heinz number 4. - Emeric Deutsch, May 19 2015

If n >= 2, n occurs in column a(n) of A246278.

By convention, a(1) = 0 because 1 does not occur in A246278.

This permutation maps between the partitions as ordered in A112798 and A241918 (the original motivation for this sequence).

For all n > 2, A007814(a(n)) = A055396(n)-1, which implies that this self-inverse permutation maps between primes (A000040) and the powers of two larger than one (A000079(n>=1)), and apart from a(1) & a(2), this also maps each even number to some odd number, and vice versa, which means there are no fixed points after 2.

A122111 commutes with this one, that is, a(n) = A122111(a(A122111(n))).

Conjugates between A243051 and A242424 and other rows of A243060 and A243070.

This sequence can be represented as a binary tree. Each child to the left is obtained by applying A253550 to the parent, and each child to the right is obtained by applying A253560 to the parent:

1

|

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

3 4

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

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

7 25 15 27 10 18 12 16

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

etc.

Sequence A253563 is the mirror image of the same tree. Also in binary trees A005940 and A163511 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. Of these four trees, this is the one where the left child is always smaller than the right child.

Note that the indexing of sequence starts from 0, although its range starts from one.

The term a(n) is the Heinz number of the adjusted partial sums of the n-th composition in standard order, where (1) the k-th composition in standard order (graded reverse-lexicographic, 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, (2) the Heinz number of a partition (y_1,...,y_k) is prime(y_1)*...*prime(y_k), and (3) we define the adjusted partial sums of a composition to be obtained by subtracting one from all parts, taking partial sums, and adding one back to all parts. See formula for a simplification. A triangular form is A242628. The inverse is A253566. The non-adjusted version is A358170. - Gus Wiseman, Dec 17 2022

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