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A096335
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Number of iterations of n -> n + tau(n) needed for the trajectory of n to join the trajectory of A064491.
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3
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0, 0, 2, 0, 1, 3, 0, 1, 0, 2, 8, 0, 7, 1, 6, 5, 6, 0, 5, 3, 4, 3, 4, 0, 3, 2, 13, 2, 13, 1, 12, 0, 11, 1, 10, 8, 10, 0, 9, 7, 9, 0, 8, 1, 7, 1, 8, 6, 7, 0, 6, 6, 6, 5, 5, 0, 4, 5, 4, 26, 3, 4, 2, 0, 2, 3, 2, 3, 1, 2, 0, 25, 0, 2, 0, 2, 1, 1, 1, 1, 0, 1, 39, 24, 38
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OFFSET
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1,3
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COMMENTS
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Conjecture: For any positive integer starting value n, iterations of n -> n + tau(n) will eventually join A064491 (verified for all n up to 50000).
The graph looks like a forest of stalks. The tops of the stalks form A036434. - N. J. A. Sloane, Jan 17 2013
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REFERENCES
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Claudia Spiro, Problem proposed at West Coast Number Theory Meeting, 1977. - From N. J. A. Sloane, Jan 11 2013
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LINKS
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T. D. Noe, Table of n, a(n) for n = 1..11000
T. D. Noe, Logarithmic plot of 10^6 terms
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EXAMPLE
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a(6)=3 because the trajectory for 1 (sequence A064491) starts
1->2->4->7->9->12->18->24->32->38->42...
and the trajectory for 6 starts
6->10->14->18->24->32->38->42->50->56...
so the sequence beginning with 6 joins A064491 after 3 steps.
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MATHEMATICA
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s = 1; t = Join[{s}, Table[s = s + DivisorSigma[0, s], {n, 2, 1000}]]; mx = Max[t]; Table[r = n; gen = 0; While[r < mx && ! MemberQ[t, r], gen++; r = r + DivisorSigma[0, r]]; If[r >= mx, gen = -1]; gen, {n, 100}] (* T. D. Noe, Jan 13 2013 *)
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CROSSREFS
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Cf. A000005, A036434, A064491.
Sequence in context: A119900 A141097 A278045 * A191910 A129503 A225682
Adjacent sequences: A096332 A096333 A096334 * A096336 A096337 A096338
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KEYWORD
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nonn
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AUTHOR
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Jason Earls, Jun 28 2004
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STATUS
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approved
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