A338059 The Enots Wolley sequence A336957 with the missing prime powers interpolated.

1, 2, 4, 6, 3, 9, 15, 5, 25, 35, 7, 14, 8, 12, 27, 33, 11, 55, 10, 16, 18, 21, 49, 77, 22, 20, 45, 39, 13, 26, 28, 63, 51, 17, 34, 32, 38, 19, 57, 69, 23, 46, 40, 65, 91, 42, 30, 85, 119, 56, 24, 75, 95, 76, 36, 81, 87, 29, 145, 50, 44, 99, 93, 31, 62, 52, 117, 105, 70, 58, 261, 111, 37

Offset: 1

Author

Scott R. Shannon and N. J. A. Sloane, Oct 17 2020

Keywords

• nonn

Comments

There is a strong conjecture that A336957 consists exactly of 1, 2, and all numbers with at least two different prime factors. [The only uncertainty is whether all numbers with at least two prime factors appear.]

The terms in A000961 greater than 2 are definitely missing from A336957, so A336957 is obviously not a permutation of the positive integers.

The present sequence is obtained by inserting the missing prime powers q = p^k, p >= 2, k >= 1, in their natural positions. More precisely, let the terms of A336957 be [W(i), i >= 1].

Between W(i) and W(i+1) we insert, in order, any prime powers q < W(i+1) which are not yet in the new sequence and satisfy gcd(q, W(i)) > 1 and gcd(q, W(i-1)) = 1.

It is conjectured that this is a permutation of the positive integers.

Examples

			Suppose n = 4.
The first 5 terms of A336957 are 1,2,6,15,35. The first 7 terms of the present sequence are 1, 2, 4, 6, 3, 9, 15. To see what comes after a(7) = W(4) = 15, we look at the missing prime powers less than W(5) = 35, which are 5, 7, 8, 11, 13, 16, 17, 19, 23, 25, 27, 29, 31. Just two terms, 5 and 25, have a common factor with 15 and are relatively prime to W(3) = 6, so they are adjoined to the sequence.
In short, we adjoin any missing prime powers which are less than W(n+1), have a common factor with W(n), and are relatively prime to W(n-1). We insert them immediately after W(n).
		

Links

• N. J. A. Sloane, Table of n, a(n) for n = 1..5000

Crossrefs

Cf. A000961, A336957, A338060 (inverse).