Monday Math 60

Let ω(n) be the number of distinct prime factors of the integer n, with us defining ω(1)=0 (see, for example, here). What, then, is the value of the series ?
How about ?

Note first that if a and b are mutually prime, then , and thus .
Now the fundamental theorem of arithmetic tells us that every integer n>1 has a unique prime factorization , where the p1, p2, …, pk are distinct primes, and the m1, m2, …, mk are positive integers (note that k=ω(n)). Then, for n>1, and .
We see then that the prime product
, when expanded into a sum, will contain terms whose numerator is of the form , with corresponding denominator , with each such combination appearing once. But these are just and , with each positive integer n appearing once and only once. Thus

and as , and for k>0, we have

Now, via the geometric series,


and thus

Now, we showed previously that , and so
To find this last product, note that . Thus
, and again using , we see that
, and so

For s=2, this then answers our original question:
, and using and (see here), we find
(see here), and so


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2 Responses to “Monday Math 60”

  1. Monday Math 86 « Twisted One 151’s Weblog Says:

    […] factors; namely, zero. In terms of the distinct prime factor counting function ω(n) (see here), we can write the Möbius function as: One can easily confirm that μ(n) is a multiplicative […]

  2. Monday Math 91 « Twisted One 151’s Weblog Says:

    […] of the Möbius function μ(n), and the multiplicative function 2ω(n) appearing in this post. Since both are multiplicative, μ*2ω will be multiplicative, so we need only examine its […]

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