OFFSET
1,4
COMMENTS
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
Outline of a proof for this observation, R. J. Mathar, May 05 2013: (Start)
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.
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.
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.
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_i-1).
ii) To construct the pairs of coprime divisors of n, take any two non-intersecting 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.
iii) The bijection between the sets constructed in i) and ii) is given by mapping the two non-intersection prime sets onto each other, and observing that the numbers of compositions of exponents have the same orders in both cases.
(End)
LINKS
Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
FORMULA
a(n) = A000005(n^2)-A000005(n) because the number of square divisors of n^2 equals the number of divisors of n.
Sum_{k=1..n} a(k) ~ (n/zeta(2)) * (log(n)^2/2 + c_1 * log(n) + c_2), where c_1 = 3*gamma - 2*zeta'(2)/zeta(2) - zeta(2) - 1 = 0.226634..., c_2 = 3*gamma^2 - (2*gamma - 1)*zeta(2) - 3*gamma_1 + (1 - 3*gamma)*(2*zeta'(2)/zeta(2) + 1) + (2*zeta'(2)/zeta(2))^2 - 2*zeta''(2)/zeta(2) = -0.0529271..., gamma is Euler's constant (A001620), and gamma_1 is the first Stieltjes constant (A082633). - Amiram Eldar, Dec 01 2023
EXAMPLE
n = 8, d(64) = 7 and from the 7 divisors {1,4,16,64} are square and the remaining 3 = a(8).
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
a(60) = (number of terms of finite A171425) = 33. [Reinhard Zumkeller, Dec 08 2009]
MATHEMATICA
Table[Count[Divisors[n^2], d_ /; ! IntegerQ[Sqrt[d]]], {n, 1, 95}] (* Jean-François Alcover, Mar 22 2011 *)
Table[DivisorSigma[0, n^2]-DivisorSigma[0, n], {n, 100}] (* Harvey P. Dale, Sep 02 2017 *)
PROG
(Haskell)
a055205 n = length [d | d <- [1..n^2], n^2 `mod` d == 0, a010052 d == 0]
-- Reinhard Zumkeller, Aug 15 2011
(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
CROSSREFS
KEYWORD
nice,nonn
AUTHOR
Labos Elemer, Jun 19 2000
STATUS
approved