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A136140
Length of the cycles in which finish the sequences a(k+1)=sopfr(n a(k)+1), with sopfr=A001414.
3
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, 42, 83, 17, 43, 71, 14, 2, 4, 2, 5, 5, 49, 17, 40, 71, 48, 17, 68, 13, 16, 76, 115, 83, 22, 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
OFFSET
1,4
MATHEMATICA
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}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Carlos Alves, Dec 16 2007
STATUS
approved