A355889 Concatenate the exponents of the powers of 2 in A354169(k) in increasing order, for k = 1, 2, 3, ...

0, 1, 2, 3, 0, 1, 4, 5, 6, 2, 3, 7, 8, 9, 0, 4, 10, 1, 5, 11, 12, 13, 2, 6, 14, 3, 7, 15, 16, 17, 8, 9, 18, 19, 20, 0, 10, 21, 1, 4, 22, 5, 11, 23, 24, 25, 12, 13, 26, 27, 28, 2, 14, 29, 3, 6, 30, 7, 15, 31, 32, 33, 16, 17, 34, 35, 36, 8, 18, 37, 9, 19, 38, 39, 40, 0, 20, 41, 10, 21, 42, 43, 44, 1, 22, 45, 4, 5, 46

Offset: 1

Author

Rémy Sigrist and N. J. A. Sloane, Jul 20 2022

Keywords

• nonn

Comments

It is conjectured that the Hamming weight of A354169(k) is always 0, 1, or 2. This is known to be true for at least the first 2^25 terms. (The present sequence is well-defined even if the conjecture is false.)

So this is a far more efficient way to present A354169 than by listing the decimal expansions.

The terms of A354169 that are pure powers of 2 appear in order, so it is obvious how to recover A354169 from this sequence.

This could be regarded as a table with (presumably) two columns, and could therefore have keyword "tabf", but that is not really appropriate, since basically it consists of the nonnegative integers with some interjections.

Examples

			A354169 begins 0, 1, 2, 4, 8, 3, 16, 32, 64, 12, 128, ... We ignore the initial 0, and then the binary expansions are 2^0, 2^1, 2^2, 2^3, 2^0+2^1, 2^4, 2^5, 2^6, 2^2+2^3, 2^7, ..., so the present sequence begins 0, 1, 2, 3, 0, 1, 4, 5, 6, 2, 3, 7, ...
		

Links

• Rémy Sigrist, Table of n, a(n) for n = 1..20000 (first 6585 terms from N. J. A. Sloane)
• Rémy Sigrist, C++ program

Crossrefs

Cf. A354169, A355150, A354773, A354774, A354767, A354798, A354680.