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A048105
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Number of non-unitary divisors of n.
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56
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0, 0, 0, 1, 0, 0, 0, 2, 1, 0, 0, 2, 0, 0, 0, 3, 0, 2, 0, 2, 0, 0, 0, 4, 1, 0, 2, 2, 0, 0, 0, 4, 0, 0, 0, 5, 0, 0, 0, 4, 0, 0, 0, 2, 2, 0, 0, 6, 1, 2, 0, 2, 0, 4, 0, 4, 0, 0, 0, 4, 0, 0, 2, 5, 0, 0, 0, 2, 0, 0, 0, 8, 0, 0, 2, 2, 0, 0, 0, 6, 3, 0, 0, 4, 0, 0, 0, 4, 0, 4, 0, 2, 0, 0, 0, 8, 0, 2, 2, 5, 0, 0, 0, 4, 0
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OFFSET
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1,8
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COMMENTS
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LINKS
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FORMULA
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a(n) = Sigma(0, n) - 2^r(n), where r() = A001221, the number of distinct primes dividing n.
Dirichlet g.f.: zeta(s)^2 - zeta(s)^2/zeta(2*s). - Geoffrey Critzer, Dec 10 2014
G.f.: Sum_{k>=1} (1 - mu(k)^2)*x^k/(1 - x^k). - Ilya Gutkovskiy, Apr 21 2017
Sum_{k=1..n} a(k) ~ (1-6/Pi^2)*n*log(n) + ((1-6/Pi^2)*(2*gamma-1)+(72*zeta'(2)/Pi^4))*n , where gamma is Euler's constant (A001620). - Amiram Eldar, Nov 27 2022
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EXAMPLE
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Example 1: If n is squarefree (A005117) then a(n)=0 since all divisors are unitary.
Example 2: n=12, d(n)=6, ud(n)=4, nud(12)=d(12)-ud(12)=2; from {1,2,3,4,6,12} {1,3,4,12} are unitary while {2,6} are not unitary divisors.
Example 3: n=p^k, a true prime power, d(n)=k+1, u(d)=2^r(x)=2, so nud(n)=d(p^k)-2=k+1 i.e., it can be arbitrarily large.
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MAPLE
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with(NumberTheory):
seq(SumOfDivisors(n, 0) - 2^NumberOfPrimeFactors(n, 'distinct'), n = 1..105);
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MATHEMATICA
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Table[DivisorSigma[0, n] - 2^PrimeNu[n], {n, 1, 50}] (* Geoffrey Critzer, Dec 10 2014 *)
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PROG
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(Haskell)
a048105 n = length [d | d <- [1..n], mod n d == 0, gcd d (n `div` d) > 1]
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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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