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Math Help - how to find number of solutions of (1/x)+(1/y)=(1/n!)

  1. #1
    spk
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    how to find number of solutions of (1/x)+(1/y)=(1/n!)

    How to find number of positive unique solutions of 1/x + 1/y = 1/n!

    Well, the answer is [ factors(n!) ^ 2 - 1 ] / 2

    As it is n!, finding factors for it seem to be hard when they turn out to be big numbers. In such case how should I do..

    can you tell me the algorithms for that...
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  2. #2
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    Re: how to find number of solutions of (1/x)+(1/y)=(1/n!)

    Quote Originally Posted by spk View Post
    How to find number of positive unique solutions of 1/x + 1/y = 1/n!

    Well, the answer is [ factors(n!) ^ 2 - 1 ] / 2

    As it is n!, finding factors for it seem to be hard when they turn out to be big numbers. In such case how should I do..

    can you tell me the algorithms for that...
    The number of proper divisors of n! (which is what you must mean here) is listed at A153823 - OEIS for n≤35. There does not appear to be an explicit formula for this number.
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    MHF Contributor chisigma's Avatar
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    Re: how to find number of solutions of (1/x)+(1/y)=(1/n!)

    Quote Originally Posted by spk View Post
    How to find number of positive unique solutions of 1/x + 1/y = 1/n!

    Well, the answer is [ factors(n!) ^ 2 - 1 ] / 2

    As it is n!, finding factors for it seem to be hard when they turn out to be big numbers. In such case how should I do..

    can you tell me the algorithms for that...
    The problem to find an approximation of the number of divisors of n! has been 'attacked' in...

    http://www.math.dartmouth.edu/~carlp/factorial.pdf

    The autors found the following result...

     \ln d(n!) \sim c_{0}\ \frac{\ln n!}{(\ln \ln n!)^{2}},\ c_{0}= \sum_{k=2}^{\infty} \frac{\ln k}{k\ (k-1)} \sim 1.25775 (1)

    Kind regards

    \chi \sigma
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