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A136138
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a(n+1)=sopfr(4a(n)+1), with sopfr=A001414. Finishes with the cycle (34, 137, 67, 269, 362, 36).
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3
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1, 5, 10, 41, 19, 18, 73, 293, 43, 173, 24, 97, 389, 179, 242, 39, 157, 54, 38, 23, 34, 137, 67, 269, 362, 36, 34
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OFFSET
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0,2
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COMMENTS
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In the class of recurrence sequences a(n+1)=sopfr(C*a(n)+D), with C=4, D=1. It is a simple example where it finishes with a nontrivial cycle.
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LINKS
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FORMULA
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MATHEMATICA
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sopfr = Function[x, Plus @@ Map[Times @@ # &, FactorInteger[x]]]; NestList[sopfr[4# + 1] &, 1, 40]
NestList[Total[Times@@@FactorInteger[4*#+1]]&, 1, 30] (* Harvey P. Dale, Sep 23 2019 *)
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CROSSREFS
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KEYWORD
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fini,nonn
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AUTHOR
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STATUS
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approved
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