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Homework Help: Number of squareful integers less than x

  1. Aug 29, 2015 #1
    I'm doing the exercises from Introduction to Analytic Number Theory by A.J. Hildebrand (online pdf lecture notes) from Chapter 2: Arithmetic Functions II - Asymptotic Estimates, and some of them leave me stumped...

    1. The problem statement, all variables and given/known data

    Problem 2.14:
    Obtain an asymptotic estimate with error term [itex]O(x^{\frac{1}{3}})[/itex] for the number of squarefull integers [itex] ≤ x [/itex], i.e., for the quantity [itex]S(x) = \left\{n ≤ x : p | n => p^{2} | n\right\} [/itex].

    2. Relevant equations

    The text describes a method known as the "convolution method" to evaluate sums of arithmetic functions asymptotically. In our case, the arithmetic function would be the characteristic function of the squarefull integers, a(n) = 1 if n is squarefull and 0 otherwise.

    If [itex]a = f * g[/itex] (Dirichlet convolution), then [tex]\sum_{n ≤ x} a(n) = \sum_{d ≤ x} g(d) F\left(\frac{x}{d}\right)[/tex] where [itex]F(x)[/itex] is the summatory function of [itex]f[/itex].

    3. The attempt at a solution
    I am not even sure how to start. I tried expressing [itex]a(n) = 1 * (\mu * a)[/itex] thus using [itex]f = 1[/itex] and [itex]g = \mu * a[/itex]. But I can't find a way to estimate the sum [tex]\sum_{d ≤ x} g(d) \left\lfloor{\frac{x}{d}}\right\rfloor[/tex] Here [itex]\mu[/itex] means the Mobius function.
  2. jcsd
  3. Aug 31, 2015 #2


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    Maybe it would be a good idea to investigate S(x) for different values of x. You could plot the number of squarefull integers for x up to some cutoff, for example. It couldn't hurt! Maybe it will reveal some pattern that will give you an idea of how to solve the problem.
  4. Sep 16, 2015 #3
    So I finally solved the problem, in a totally different (and way simpler) method -_-

    I used an earlier result from another problem, saying that squarefull integers can be written uniquely in the form [itex]a^2b^3[/itex].
    So to count the squarefull integers [itex]\leq x[/itex] you can just do [tex]\sum_{a^2b^3 \leq x} 1 = \sum_{b \leq x^{1/3}} \left\lfloor{\frac{\sqrt x}{b^{3/2}}}\right\rfloor[/tex] which eventually simplifies to [tex]\zeta(3/2)\sqrt x + O(x^{1/3})[/tex]

    Ah, turns out I was wrong... The representation [itex]a^2b^3[/itex] is not unique. For example, [itex]p^8 = (p^4)^2 * 1^3 = p^2 * (p^2)^3[/itex]. I think we need the additional condition that [itex]b[/itex] is squarefree...
    So the sum is actually [tex]\sum_{a^2b^3 \leq x} µ^2(b) = \sum_{b \leq x^{1/3}}µ^2(b) \sum_{a \leq \frac{x^{1/2}}{b^{3/2}}}1 = \sqrt x \sum_{b \leq x^{1/3}}\frac{µ^2(b)}{b^{3/2}} + O(x^{1/3})[/tex]

    which still becomes something of the form [itex] C \sqrt x + O(x^{1/3}) [/itex] but [itex] C = \sum_{b = 1}^{\infty} \frac{µ^2(b)}{b^{3/2}}[/itex] which I can't really simplify...

    After playing with this expression and using Dirichlet series of [itex]µ(n)[/itex] and [itex](µ^2 * µ)(n)[/itex], I found that [itex]C = \frac{\zeta(3/2)}{\zeta(3)} [/itex].
    Last edited: Sep 16, 2015
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