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A066919
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a(n) = least number of applications of f to n to reach 1, where f is defined by f(n) = phi(n) if n is even; = sigma(n) if n is odd.
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0
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0, 1, 3, 2, 3, 2, 4, 3, 5, 3, 4, 3, 4, 3, 5, 4, 4, 3, 5, 4, 6, 4, 5, 4, 7, 4, 6, 4, 5, 4, 6, 5, 6, 5, 6, 4, 5, 4, 6, 5, 5, 4, 6, 5, 6, 5, 6, 5, 8, 5, 6, 5, 5, 4, 6, 5, 7, 5, 6, 5, 6, 5, 7, 6, 6, 5, 7, 6, 7, 5, 6, 5, 6, 5, 7, 5, 7, 5, 7, 6, 10, 6, 6, 5, 6, 5, 7, 6, 6, 5, 7, 6, 8, 6, 7, 6, 6, 5, 7, 6, 7, 6, 7, 6
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OFFSET
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1,3
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COMMENTS
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a(n) is in [0,19] for n < 10^5. Conjecture: a(n) exists for all n, i.e. repeated application of f to n eventually yields 1, for any n. The only way this could fail is if n, f(n), f(f(n)), ... are all odd squares.
The conjecture is true, since sequence A055021 (smallest x such that n iterations of sigma() are required for the result to be >= 2x) is complete. - Vim Wenders, Apr 07 2008
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LINKS
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Table of n, a(n) for n=1..104.
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EXAMPLE
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f(f(f(f(7)))) = f(f(f(8))) = f(f(4)) = f(2) = 1 and 4 applications of f are required to achieve this. Therefore a(7) = 4.
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MATHEMATICA
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f[n_] := If[EvenQ[n], EulerPhi[n], DivisorSigma[1, n]]; a[n_] := Module[{b=n, k=0}, While[b>1, b=f[b]; k++ ]; k]; Table[a[i], {i, 1, 105}]
Table[Length[NestWhileList[If[EvenQ[#], EulerPhi[#], DivisorSigma[1, #]]&, n, #!=1&]], {n, 110}]-1 (* Harvey P. Dale, Jun 16 2018 *)
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CROSSREFS
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Sequence in context: A039641 A198729 A093489 * A268714 A084117 A116895
Adjacent sequences: A066916 A066917 A066918 * A066920 A066921 A066922
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KEYWORD
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nonn
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AUTHOR
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Joseph L. Pe, Jan 23 2002
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EXTENSIONS
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Edited by Dean Hickerson, Oct 26 2002
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STATUS
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approved
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