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A055212
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Number of composite divisors of n.
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10
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0, 0, 0, 1, 0, 1, 0, 2, 1, 1, 0, 3, 0, 1, 1, 3, 0, 3, 0, 3, 1, 1, 0, 5, 1, 1, 2, 3, 0, 4, 0, 4, 1, 1, 1, 6, 0, 1, 1, 5, 0, 4, 0, 3, 3, 1, 0, 7, 1, 3, 1, 3, 0, 5, 1, 5, 1, 1, 0, 8, 0, 1, 3, 5, 1, 4, 0, 3, 1, 4, 0, 9, 0, 1, 3, 3, 1, 4, 0, 7, 3, 1, 0, 8, 1, 1, 1, 5, 0, 8, 1, 3, 1, 1, 1, 9, 0, 3, 3, 6, 0, 4, 0, 5, 4
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OFFSET
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1,8
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COMMENTS
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Trivially, there is only one run of three consecutive 0's. However, there are infinitely many runs of three consecutive 1's and they are at positions A056809(n), A086005(n), and A115393(n) for n >= 1. - Timothy L. Tiffin, Jun 21 2021
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LINKS
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FORMULA
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G.f.: -x/(1 - x) + Sum_{k>=1} (x^k - x^prime(k))/((1 - x^k)*(1 - x^prime(k))). - Ilya Gutkovskiy, Mar 21 2017
Sum_{k=1..n} a(k) ~ n*log(n) - n*log(log(n)) + (2*gamma - 2 - B)*n, where gamma is Euler's constant (A001620) and B is Mertens's constant (A077761). - Amiram Eldar, Dec 07 2023
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EXAMPLE
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a[20] = 3 because the composite divisors of 20 are 4, 10, 20.
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MATHEMATICA
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Table[ Count[ PrimeQ[ Divisors[n] ], False] - 1, {n, 1, 105} ]
Table[Count[Divisors[n], _?CompositeQ], {n, 120}] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jul 09 2018 *)
a[n_] := DivisorSigma[0, n] - PrimeNu[n] - 1; Array[a, 100] (* Amiram Eldar, Jun 18 2022 *)
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PROG
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(Haskell)
(PARI) a(n) = numdiv(n) - omega(n) - 1; \\ Michel Marcus, Oct 17 2015
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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