A163755 a(0)=1. For n>=1, write n in binary. Let b(n,m) be the length of the m-th run of 0's or 1's, reading right to left. Then a(n) = product{m=1 to M} p(m)^b(n,m), where p(m) is the m-th prime, and M is the number of runs of 0's and 1's in binary n.

1, 2, 6, 4, 12, 30, 18, 8, 24, 90, 210, 60, 36, 150, 54, 16, 48, 270, 1050, 180, 420, 2310, 630, 120, 72, 450, 1470, 300, 108, 750, 162, 32, 96, 810, 5250, 540, 2100, 16170, 3150, 360, 840, 6930, 30030, 4620, 1260, 11550, 1890, 240, 144, 1350, 7350, 900, 2940, 25410

Offset: 0

Author

Leroy Quet, Aug 03 2009

Keywords

• base
• nonn

Comments

This sequence is a permutation of the terms of sequence A055932.

Clarification: By "run" of 0's or 1's in binary n, it is meant a group either entirely of 0's, and bounded by 1's or the edge of the binary number interpreted as a string, or entirely of 1's, and bounded by 0's or the edge of the string. In other words, the runs of 0's alternate with the runs of 1's.

Examples

			13 in binary is 1101. So reading right to left, there is a run of one 1, followed by a run of one 0, followed by a run of two 1's. So the lengths of the runs are 1,1,2. Therefore a(13) = p(1)^1 * p(2)^1 * p(3)^2 = 2^1 * 3^1 * 5^2 = 150.
		

Links

• Andrew Weimholt, Table of n, a(n) for n = 0..1000
• Christian Krause, Simon Strandgaard, et al, A mined LODA assembly source for this sequence

Crossrefs

Cf. A055932, A057335, A341915.

Cf. also A005940, A163511.

Formula

a(n) = A057335(A341915(n)). [Found by LODA miner, should be easy to prove] - Antti Karttunen, Apr 22 2022