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...

Re: how to find number of solutions of (1/x)+(1/y)=(1/n!)

Quote:

Originally Posted by

**spk** 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.

Re: how to find number of solutions of (1/x)+(1/y)=(1/n!)

Quote:

Originally Posted by

**spk** 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...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$