cp's OEIS Frontend

Note: 1 might be a more natural starting offset for this sequence, although the identities concerning A053645 and A161511 would have to be changed. - Antti Karttunen, Oct 12 2009.

This can be regarded as an arrangement of the partitions, indexed by position in A125106. The partitions in a given row all have the same remaining partition when the largest part is removed; specifically, the partition indexed by the row number in A125106 (with row 0 having the empty partition remaining).

The first value on row n is A004760(n+1). The second value on each row is A004760(n+1) plus A062383(n); subsequent values increase by ever enlarging powers of two. Or equivalently, each subsequent value on the row after the first nonzero value is given by A004754(previous value on the same row).

A055941(r) tells how many terms the row r (>= 0) has been shifted rightward from its "natural position", i.e. with how many zeros that row has been prepended.

The number of (nonzero) entries in column k is A000041(k).

a(n) gives the one-based position of the first nonzero term on the row n-1 of A126441.

Sequence A016116 can be used to identify the extracted subsequence by computing the number of terms to alternately extract and skip. [This comment is from the original submitter. I don't understand it. - Antti Karttunen, Oct 12 2009]

Note that the values of this triangle identify the number partitioned described in A125105 and A161924.

A pseudo-logarithmic function in the sense that a(b*c) = a(b)+a(c) and so a(b^c) = c*a(b) and f(n) = k^a(n) is a multiplicative function. [Cf. A248692 for example.] Essentially a function from the positive integers onto the partitions of the nonnegative integers (1->0, 2->1, 3->2, 4->1+1, 5->3, 6->1+2, etc.) so each value a(n) appears A000041(a(n)) times, first with the a(n)-th prime and last with the a(n)-th power of 2. Produces triangular numbers from primorials. - Henry Bottomley, Nov 22 2001

Michael Nyvang writes (May 08 2006) that the Danish composer Karl Aage Rasmussen discovered this sequence in the 1990's: it has excellent musical properties.

All A000041(a(n)) different n's with the same value a(n) are listed in row a(n) of triangle A215366. - Alois P. Heinz, Aug 09 2012

a(n) is the sum of the parts of the partition having Heinz number n. We define the Heinz number of a partition p = [p_1, p_2, ..., p_r] as Product_{j=1..r} (p_j-th prime) (concept used by Alois P. Heinz in A215366 as an "encoding" of a partition). For example, for the partition [1, 1, 2, 4, 10] we get 2*2*3*7*29 = 2436. Example: a(33) = 7 because the partition with Heinz number 33 = 3 * 11 is [2,5]. - Emeric Deutsch, May 19 2015

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

The primes become the powers of 2 (2 -> 1, 3 -> 2, 5 -> 4, 7 -> 8); the composite numbers are formed by taking the values for the factors in the increasing order, multiplying them by the consecutive powers of 2, and summing. See the Example section.

From Antti Karttunen, Jun 27 2014: (Start)

The odd bisection (containing even terms) halved gives A244153.

The even bisection (containing odd terms), when one is subtracted from each and halved, gives this sequence back.

(End)

Question: Are there any other solutions that would satisfy the recurrence r(1) = 0; and for n > 1, r(n) = Sum_{d|n, d>1} 2^A033265(r(d)), apart from simple variants 2^k * A156552(n)? See also A297112, A297113. - Antti Karttunen, Dec 30 2017

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

Length of n-th row given by A000120(n);

Min of n-th row given by A001511(n);

Sum of n-th row given by A029931(n);

Product of n-th row given by A096111(n);

Max of n-th row given by A113473(n);

Numerator of sum of reciprocals of n-th row given by A116416(n);

Denominator of sum of reciprocals of n-th row given by A116417(n);

LCM of n-th row given by A271410(n).

The first appearance of n is at A001787(n - 1).

n-th row begins at index A000788(n - 1) for n > 0.

Also the reversed positions of 1's in the reversed binary expansion of n. Also the reversed partial sums of the n-th composition in standard order (row n of A066099). Reversing rows gives A048793. - Gus Wiseman, Jan 17 2023

For other numbers than the powers of 2 (that are fixed), this permutation reverses the sequence of exponents in the prime factorization of n from the exponent of 2 to that of the largest prime factor, except that the exponents of 2 and the greatest prime factor present are adjusted by one. Note that some of the exponents might be zeros.

Self-inverse permutation of natural numbers, composition of A122111 & A241909 in either order: a(n) = A122111(A241909(n)) = A241909(A122111(n)).

This permutation preserves both bigomega and the (index of) largest prime factor: for all n it holds that A001222(a(n)) = A001222(n) and A006530(a(n)) = A006530(n) [equally: A061395(a(n)) = A061395(n)].

From the above it follows, that this fixes both primes (A000040) and powers of two (A000079), among other numbers.

Even positions from n=4 onward contain only terms of A070003, and the odd positions only the terms of A102750, apart from 1 which is at a(1), and 2 which is at a(2).

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