%I
%S 0,1,1,2,1,5,1,3,2,5,1,9,1,5,5,4,1,9,1,9,5,5,1,13,2,5,3,9,1,19,1,5,5,
%T 5,5,16,1,5,5,13,1,19,1,9,9,5,1,17,2,9,5,9,1,13,5,13,5,5,1,33,1,5,9,6,
%U 5,19,1,9,5,19,1,23,1,5,9,9,5,19,1,17,4,5,1,33,5,5,5,13,1,33,5,9,5,5,5
%N Number of nonsquare divisors of n^2.
%C Seems to be equal to the number of unordered pairs of coprime divisors of n. (Checked up to 2*10^14.)  _Charles R Greathouse IV_, May 03 2013
%C Outline of a proof for this observation, _R. J. Mathar_, May 05 2013: (Start)
%C i) To construct the divisors of n, write n=product_i p_i^e_i as the standard prime power decomposition, take any subset of the primes p_i (including the empty set representing the 1) and run with the associated list exponents from 0 up to their individual e_i.
%C To construct the *nonsquare* divisors of n, ensure that one or more of the associated exponents is/are odd. (The empty set is interpreted as 1^0 with even exponent.) To construct the nonsquare divisors of n^2, the principle remains the same, although the exponents may individually range from 0 up to 2*e_i.
%C The nonsquare divisor is therefore a nonempty product of prime powers (at least one) with odd exponents times a (potentially empty) product of prime powers (of different primes) with even exponents.
%C The nonsquare divisors of n^2 have exponents from 0 up to 2*e_i, but the subset of exponents in the "even/square" factor has e_i candidates (range 2, 4, .., 2*e_i) and in the "odd/nonsquare" factor also only e_i candidates (range 1,3,5,2*e_i1).
%C ii) To construct the pairs of coprime divisors of n, take any two nonintersecting subsets of the set of p_i (possibly the empty subset which represents the factor 1), and let the exponents run from 1 up to their individual e_i in each of the two products.
%C iii) The bijection between the sets constructed in i) and ii) is given by mapping the two nonintersection prime sets onto each other, and observing that the numbers of compositions of exponents have the same orders in both cases.
%C (End)
%H T. D. Noe, <a href="/A055205/b055205.txt">Table of n, a(n) for n = 1..1000</a>
%F a(n) = A000005(n^2)A000005(n) because the number of square divisors of n^2 equals the number of divisors of n.
%F a(n) = A056595(A000290(n)).
%F a(n) = A048691(n)  A000005(n). [_Reinhard Zumkeller_, Dec 08 2009]
%e n = 8, d(64) = 7 and from the 7 divisors {1,4,16,64} are square and the remaining 3 = a(8).
%e n = 12, d(144) = 15, from which 6 divisors are squares {1,4,9,16,36,144} so a(12) = d(144)d(12) = 9
%e a(60) = (number of terms of finite A171425) = 33. [_Reinhard Zumkeller_, Dec 08 2009]
%t Table[Count[Divisors[n^2], d_ /; ! IntegerQ[Sqrt[d]]], {n, 1, 95}] (* _JeanFrançois Alcover_, Mar 22 2011 *)
%t Table[DivisorSigma[0,n^2]DivisorSigma[0,n],{n,100}] (* _Harvey P. Dale_, Sep 02 2017 *)
%o (Haskell)
%o a055205 n = length [d  d < [1..n^2], n^2 `mod` d == 0, a010052 d == 0]
%o  _Reinhard Zumkeller_, Aug 15 2011
%o (PARI) a(n)=my(f=factor(n)[,2]);prod(i=1,#f,2*f[i]+1)prod(i=1,#f,f[i]+1) \\ _Charles R Greathouse IV_, May 02 2013
%Y Cf. A000005, A000290, A048691, A056595.
%K nice,nonn
%O 1,4
%A _Labos Elemer_, Jun 19 2000
