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%I #2 Sep 24 2013 00:41:43
%S 1,1,1,6,16,5,2,23,2,4,56,17,56,2,29,22,8,8,2,29,15,4,5,72,4,11,55,56,
%T 42,83,17,43,71,14,2,4,2,5,5,49,17,40,71,48,17,68,13,16,76,115,83,22,
%U 6,125,7,19,10,9,70,1,31,36,72,3,13,28,9,18,57,63,31,34,16,225,26,53,116
%N Length of the cycles in which finish the sequences a(k+1)=sopfr(n a(k)+1), with sopfr=A001414.
%t Lcycle = Function[ x, z = 1; pz = Position[x[[Range[z]]], x[[z + 1]]]; While[Length[pz] == 0, z++; pz = Position[ x[[Range[z]]], x[[z + 1]]]]; z - pz[[1, 1]] + 1]; sopfr = Function[x, Plus @@ Map[Times @@ # &, FactorInteger[x]]]; Table[tb = NestList[sopfr[n# + 1] &, 1, 350]; Lcycle[tb], {n, 1, 100}]
%Y Cf. A136136, A136137, A136138, A136139.
%K nonn
%O 1,4
%A _Carlos Alves_, Dec 16 2007
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