_Leroy Quet, _, Aug 03 2009

Andrew Weimholt, <a href="/A163755/b163755.txt">Table of n, a(n) for n = 0..1000</a>

base,nonn,new

base,nonn,new

Leroy Quet (http://www.prism-of-spirals.net/), , Aug 03 2009

base,nonn,new

Leroy Quet (q1qq2qqq3qqqq(AT)yahoohttp://www.prism-of-spirals.comnet/), Aug 03 2009

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

0,2

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.

Andrew Weimholt, <a href="b163755.txt">Table of n, a(n) for n = 0..1000</a>

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.

Cf. A055932

base,nonn

Leroy Quet (q1qq2qqq3qqqq(AT)yahoo.com), Aug 03 2009

approved