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%I A163755 #15 Apr 25 2022 08:10:43
%S A163755 1,2,6,4,12,30,18,8,24,90,210,60,36,150,54,16,48,270,1050,180,420,
%T A163755 2310,630,120,72,450,1470,300,108,750,162,32,96,810,5250,540,2100,
%U A163755 16170,3150,360,840,6930,30030,4620,1260,11550,1890,240,144,1350,7350,900,2940,25410
%N 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.
%C A163755 This sequence is a permutation of the terms of sequence A055932.
%C A163755 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.
%H A163755 Andrew Weimholt, <a href="/A163755/b163755.txt">Table of n, a(n) for n = 0..1000</a>
%H A163755 Christian Krause, Simon Strandgaard, et al, <a href="https://github.com/loda-lang/loda-programs/blob/main/oeis/163/A163755.asm">A mined LODA assembly source for this sequence</a>
%F A163755 a(n) = A057335(A341915(n)). [Found by LODA miner, should be easy to prove] - _Antti Karttunen_, Apr 22 2022
%e A163755 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.
%Y A163755 Cf. A055932, A057335, A341915.
%Y A163755 Cf. also A005940, A163511.
%K A163755 base,nonn
%O A163755 0,2
%A A163755 _Leroy Quet_, Aug 03 2009

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