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See comments and examples in A302032 to see how Ludic factorization proceeds.

See comments and examples in A302032 to see how Ludic factorization proceeds.

Permutation of natural numbers obtained from the Ludic sieve. Note the indexing: Domain starts from 0, range from 1.

This sequence can be represented as a binary tree. Each left hand child is obtained by applying A269379 to the parent's contents, and each right hand child is obtained by doubling the parent's contents:

1

|

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

3 4

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

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

7 10 15 12 19 18 21 16

11 14 27 20 35 30 33 24 31 38 51 36 49 42 45 32

etc.

Sequence A269385 is obtained from the mirror image of the same tree.

Permutation of natural numbers obtained from the Ludic sieve. Note the indexing: Domain starts from 0, range from 1.

This sequence can be represented as a binary tree. Each left hand child is obtained by doubling the parent's contents, and each right hand child is obtained by applying A269379 to the parent's contents:

1

|

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

4 3

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

/ \ / \ / \ / \

/ \ / \ / \ / \

/ \ / \ / \ / \

16 21 18 19 12 15 10 7

32 45 42 49 36 51 38 31 24 33 30 35 20 27 14 11

etc.

Sequence A269387 is obtained from the mirror image of the same tree.

This is a variant of A269356, from which it differs for the first time at n=18.

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)

The starting offset of the sequence giving the terms of square array is 2. However, we can tacitly assume that a(1) = 1 when the sequence is used as a permutation of natural numbers. However, term 1 itself is out of the array.

The choice of offset = 2 for the terms starting in rows >= 1 is motivated by the desire to have a permutation of the integers n -> a(n) with a(n) = A(A002260(n-1), A004736(n-1)) for n > 1 and a(1) := 1. However, since this sequence is declared as a "table", offset = 2 would mean that the first *row* (not element) has index 2. I think the sequence should have offset = 1 and the permutation of the integers would be n -> a(n-1) with a(0) := 1 (if a(1) = A(1,1) = 2). Or, the sequence could have offset 0, with an additional row 0 of length 1 with the only element a(0) = A(0,1) = 1, the permutation still being n -> a(n-1) if a(n=0, 1, 2, ...) = (1, 2, 4, ...). This would be in line with considering 1 as the first ludic number, and A(n, 1) = A003309(n+1) for n >= 0. - M. F. Hasler, Nov 12 2024