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%I #11 Nov 01 2019 12:52:00
%S 1,1,2,2,2,2,4,2,1,3,5,3,3,3,3,1,4,4,4,4,1,5,5,5,5,5,5,5,5,5,10,5,5,5,
%T 5,14,5,5,5,5,5,5,1,6,5,6,5,6,5,0,5,6,5,6,5,6,1,7,5,7,5,7,5,3,5,7,5,7,
%U 5,7,1,8,5,8,5,8,5,6,5,8,5,8,5,8,1,9,5,9,5,9,5,6,5,9,5,9,5,9,1,10,5,10,5,10
%N a(0)=1. a(n) = the number of earlier terms equal to GCD(a(k),n), where a(k) is the largest term among terms a(0) through a(n-1).
%e The largest term among terms a(0) through a(9) is a(6)=4. GCD(4,10)=2. So a(10) is the number of earlier terms equal to 2. a(2) =a(3) =a(4) =a(5) =a(7) =2. So a(10) = 5.
%t f[l_List] :=Append[l, Count[l, GCD[Max[l], Length[l]]]];Nest[f, {1}, 105] (* _Ray Chandler_, Oct 16 2006 *)
%K easy,nonn
%O 0,3
%A _Leroy Quet_, Oct 15 2006
%E Extended by _Ray Chandler_, Oct 16 2006
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