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A330786
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Number of steps to reach 1 by iterating the absolute alternating sum-of-divisors function (A206369).
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3
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0, 1, 2, 3, 4, 2, 3, 5, 4, 4, 5, 3, 4, 3, 6, 6, 7, 4, 5, 4, 4, 5, 6, 5, 5, 4, 5, 5, 6, 6, 7, 5, 5, 7, 6, 5, 6, 5, 6, 5, 6, 4, 5, 7, 6, 6, 7, 6, 6, 5, 6, 6, 7, 5, 6, 7, 6, 6, 7, 6, 7, 7, 5, 6, 7, 5, 6, 7, 8, 6, 7, 7, 8, 6, 5, 6, 7, 6, 7, 8, 8, 6, 7, 6, 7, 5, 8, 6, 7, 6
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OFFSET
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1,3
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COMMENTS
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A preliminary for investigating iterates of A206369.
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REFERENCES
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LINKS
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FORMULA
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a(1) = 0; for n > 1, a(n) = 1 + a(A206369(n)).
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EXAMPLE
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Iterating A206369 with a starting value of 27 gives 20, 12, 6, 2, 1, taking 5 steps to reach 1. So a(27) = 5.
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MATHEMATICA
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f[p_, e_] := Sum[(-1)^(e-k)*p^k, {k, 0, e}]; s[1] = 1; s[n_] := Times @@ (f @@@ FactorInteger[n]); a[n_] := Length @ FixedPointList[s, n] - 2; Array[a, 90] (* Amiram Eldar, Jan 01 2020 *)
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PROG
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(PARI) f(n) = sumdiv(n, d, eulerphi(n/d) * issquare(d)); \\ A206369
a(n) = {if (n==1, return (0)); my(nb = 1); while ((n = f(n)) != 1, nb++); nb; } \\ Michel Marcus, Jan 01 2020
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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