for $x \ge 2863:$

$$\ln\left(\left\lfloor\frac{x}{6}\right\rfloor!\right) < \sum_{k=5}^{\infty}-\mu(k)\ln\left(\left\lfloor\frac{x}{k}\right\rfloor!\right)$$

I've written a java application which checked from 2,863 to 24,600 which is how I came up with $2,863$.

I'm looking for tips on how to proceed to prove or disprove this comparison.


$$\ln\left(\left\lfloor\frac{x}{5}\right\rfloor!\right) > \ln\left(\left\lfloor\frac{x}{6}\right\rfloor!\right)$$

The problems comes down to showing that for $x \ge 2863$:

$$\sum_{k=6}^{\infty}\mu(k)\ln\left(\left\lfloor\frac{x}{k}\right\rfloor!\right) \le 0$$

or showing that:

$$\sum_{k=6}^{\infty}\mu(k)\ln\left(\left\lfloor\frac{x}{k}\right\rfloor!\right) \le \ln\left(\left\lfloor\frac{x}{5}\right\rfloor!\right) - \ln\left(\left\lfloor\frac{x}{6}\right\rfloor!\right)$$

Does anyone have any suggestions?

Thanks very much!


  • 1
    $\begingroup$ You are essentially trying to prove bounds on chebyshevs prime counting function, I don't think this representation for $\psi(x)$ will prove useful. $\endgroup$ – Ethan May 11 '13 at 21:19
  • $\begingroup$ Thanks, Ethan! I was trying to understand if this representation was useful or not. $\endgroup$ – Larry Freeman May 11 '13 at 21:59

Did you mean $$ \sum_{k=6}^\infty \mu(k) \ln\left(\left\lfloor \frac{x}{k} \right\rfloor !\right) \geq 0 $$

If not, I can't reproduce your verification for this inequality using the following Matlab code:

function s = larry( x )

    % Compute \sum_{k=6}^\infty \mu(k) \ln\left(\lfloor \frac{x}{k} \rfloor !\right)
    s = 0;
    for k = 6:ceil(x/2)
        s = s + moebiusmu(k)*logfactorial( floor(x/k) );

    % Compute \mu(k)
    function mu = moebiusmu(n)

        if n == 1, mu = 1; return; end

        p = factor(n); 
        r = histc(p, unique(p));

        if all(r < 2) 
            mu = (-1)^length(unique(p)); 
            mu = 0; 

    % Compute \ln\left(\lfloor \frac{x}{k} \rfloor !\right)
    function p = logfactorial(n)        
        if n < 2
            p = 0;
            p = sum(log( 2:n ));

  • $\begingroup$ Thanks. I didn't know how to check the expression. I'll take a look at purchasing a Student License for MatLab for the future. :-) $\endgroup$ – Larry Freeman May 11 '13 at 22:27
  • $\begingroup$ :) I'm not sure I could come up with a Java version given my poor skills in this language.. C++ maybe if you're interested? $\endgroup$ – Sheljohn May 11 '13 at 22:29
  • 1
    $\begingroup$ I think that your Matlab code is very clear. :-) I just meant that I should probably pick up a MatLab license and start using that instead of Java. :-) $\endgroup$ – Larry Freeman May 11 '13 at 22:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.