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

0,2

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.

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

Formula

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

Example

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.

Crossrefs

Cf. A055932, A057335, A341915.

Cf. also A005940, A163511.

Sequence in context: A127730 A118416 A046204 * A100851 A351500 A365902

Adjacent sequences: A163752 A163753 A163754 * A163756 A163757 A163758

Keyword

base,nonn

Author

Leroy Quet, Aug 03 2009

Status

approved