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Number of composite divisors of n.
10

%I #60 Dec 07 2023 01:43:55

%S 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,

%T 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,

%U 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

%N Number of composite divisors of n.

%C 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

%H Reinhard Zumkeller, <a href="/A055212/b055212.txt">Table of n, a(n) for n = 1..10000</a>

%F a(n) = A033273(n) - 1.

%F a(n) = tau(n)-omega(n)-1, where tau=A000005 and omega=A001221. - _Reinhard Zumkeller_, Jun 13 2003

%F 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

%F 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

%e a[20] = 3 because the composite divisors of 20 are 4, 10, 20.

%t Table[ Count[ PrimeQ[ Divisors[n] ], False] - 1, {n, 1, 105} ]

%t Table[Count[Divisors[n],_?CompositeQ],{n,120}] (* Requires Mathematica version 10 or later *) (* _Harvey P. Dale_, Jul 09 2018 *)

%t a[n_] := DivisorSigma[0, n] - PrimeNu[n] - 1; Array[a, 100] (* _Amiram Eldar_, Jun 18 2022 *)

%o (Haskell)

%o a055212 = subtract 1 . a033273 -- _Reinhard Zumkeller_, Sep 15 2015

%o (PARI) a(n) = numdiv(n) - omega(n) - 1; \\ _Michel Marcus_, Oct 17 2015

%Y Complement of A083399.

%Y Cf. A000005, A001221, A008578 (indices of 0's), A033273, A056809, A086005, A115393, A137944, A137945, A344713.

%Y Cf. A001620, A077761.

%K easy,nonn

%O 1,8

%A _Leroy Quet_, Jun 23 2000