cp's OEIS Frontend

Note how 6, 720 and 612360000 occur in A244743 as its 0th, 4th and 8th term, from which my bold conjecture that A244743(12) or A244743(16) = 1697781042840960000000000.

According to preliminary results from Janis Iraids, the value of A005245(a(5)) = ||1697781042840960000000000|| = 160, while ||1697781042840960000000000 - 1|| = 169, which lays to rest my naive conjecture above, as 169 - 160 is neither 12 nor 16. Note also how 5, 719 and 612359999 are all primes, while a(5)-1 factorizes as 1697781042840959999999999 = 13 * 89443 * 908669 * 1606890407869. - Antti Karttunen, Dec 20 2015

Meyers (see Guy reference) conjectures that for all r >= 1, the least odd number not in the set {a(i): i < prime(r)} is prime(r+1). - N. J. A. Sloane, Jan 08 2021

Meyers' conjecture would be refuted if and only if for some r there were such a large gap between prime(r) and prime(r+1) that there existed a composite c for which prime(r) < c < a(c) < prime(r+1), in which case (by Bertrand's postulate) c would necessarily be a term of A246281. - Antti Karttunen, Mar 29 2021

a(n) is odd for all n and for each odd m there exists a k with a(k) = m (see A064216). a(n) > n for n > 1: bijection between the odd and all numbers. - Reinhard Zumkeller, Sep 26 2001

a(n) and n have the same number of distinct primes with (A001222) and without multiplicity (A001221). - Michel Marcus, Jun 13 2014

From Antti Karttunen, Nov 01 2019: (Start)

More generally, a(n) has the same prime signature as n, A046523(a(n)) = A046523(n). Also A246277(a(n)) = A246277(n) and A287170(a(n)) = A287170(n).

Many permutations and other sequences that employ prime factorization of n to encode either polynomials, partitions (via Heinz numbers) or multisets in general can be easily defined by using this sequence as one of their constituent functions. See the last line in the Crossrefs section for examples.

(End)

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

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

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

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

1

|

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

4 3

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

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

16 12 18 10 27 15 25 7

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

etc.

Sequence A253565 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) tells distance of n from 1 in all these trees. Of these four trees, this is the one where the left child is always larger than the right child.

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

a(n) (n>=1) can be obtained by the composition of a bijection between {1,2,3,4,...} and the set of integer partitions and a bijection between the set of integer partitions and {2,3,4,...}. Explanation on the example n=10. Write 2*n = 20 as a binary number: 10100. Consider a Ferrers board whose southeast border is obtained by replacing each 1 by an east step and each 0 by a north step. We obtain the Ferrers board of the partition p = (2,2,1). Finally, a(10) = 2'*2'*1', where m' = m-th prime. Thus, a(10)= 3*3*2 = 18. - Emeric Deutsch, Sep 17 2016

x^(2^n) - a(n) is the minimal polynomial over Q for the algebraic number sqrt(p(1)*sqrt(p(2)*...*sqrt(p(n-1)*sqrt(p(n)))...)), where p(k) is the k-th prime. Each such monic polynomial is irreducible by Eisenstein's Criterion (using p = p(n)).

A prime version of Somos's quadratic recurrence sequence A052129(n) = A052129(n-1)^2 * n = Product_{k=1..n} k^(2^(n-k)). - Jonathan Sondow, Mar 29 2014

All positive integers have unique factorizations into powers of distinct primes, and into powers of squarefree numbers with distinct exponents that are powers of 2. (See A329332 for a description of the relationship between the two.) a(n) is the least number such that both factorizations have n factors. - Peter Munn, Dec 15 2019

From Peter Munn, Jan 24 2020 to Feb 06 2020: (Start)

For n >= 0, a(n+1) is the n-th power of 12 in the monoid defined by A306697.

a(n) is the least positive integer that cannot be expressed as the product of fewer than n terms of A072774 (powers of squarefree numbers).

All terms that are less than the order of the Monster simple group (A003131) are divisors of the group's order, with a(6) exceeding its square root.

(End)

It is remarkable that 4 of the first 5 terms are factorials. - Hal M. Switkay, Jan 21 2025