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a(1)=1. Let m be the number of terms in the longest run of consecutive equal terms from among the first n terms of the sequence. a(n+1) = the number of terms, from among the first n terms of the sequence, that are coprime to m.
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%I #11 Aug 02 2018 17:07:49

%S 1,1,2,2,2,5,6,6,6,6,3,4,4,4,4,4,15,15,15,15,15,15,3,3,3,3,3,3,3,29,

%T 30,31,32,33,34,35,35,35,35,35,35,35,35,28,28,28,28,28,28,28,28,28,32,

%U 33,33,33,33,33,33,33,33,33,33,23,24,24,24,24,24,24,24

%N a(1)=1. Let m be the number of terms in the longest run of consecutive equal terms from among the first n terms of the sequence. a(n+1) = the number of terms, from among the first n terms of the sequence, that are coprime to m.

%H Rémy Sigrist, <a href="/A136347/b136347.txt">Table of n, a(n) for n = 1..10000</a>

%H Rémy Sigrist, <a href="/A136347/a136347.gp.txt">PARI program for A136347</a>

%e Among the first 30 terms of the sequence the run of seven 3's (from a(23) to a(29)) is the longest run of consecutive equal terms. (So m = 7.) Among the first 30 terms of the sequence, all 30 terms are coprime to 7. So a(31) = 30.

%o (PARI) See Links section.

%Y Cf. A136348.

%K nonn,look

%O 1,3

%A _Leroy Quet_, Dec 25 2007

%E More terms from _Rémy Sigrist_, Aug 02 2018