Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Wikipedia has a page about the prime zeta function which is defined as follows:

$$P(s)=\sum_{p\;\text{prime}} \frac1{p^s}$$

I entered this additional definition:

Define a sequence: $$a_n=\prod_{d\mid n} \frac{\Lambda(d)}{\log(d)}$$

where zeros are not included in the multiplication and $a_1=1$ then:

$$P(s)=\log\sum_{n=1}^\infty \frac{a_n}{n^s}$$

Is it a problem that this later definition doesn't include the zeros in the multiplication?

share|cite|improve this question
What is $\Lambda(n/d)$? Is it the Von Mangoldt function ( Then that should be explained both here and at Wikipedia. –  joriki Apr 14 '11 at 12:19
I might have made a mistake but what I had in mind was the Mangoldt function applied to the divisors of n. In oeis notation which I understand better I would have said: For row>1: a(n)^-1 = row products of A100995(A126988). –  Mats Granvik Apr 14 '11 at 12:23
@joriki Ok, nothing new then. But I don't think this is mentioned in the oeis. –  Mats Granvik Apr 14 '11 at 12:27
What do you mean by "nothing new"? I only linked to the Von Mangoldt function to make what you wrote intelligible to someone who doesn't know this symbol (like I didn't), not to imply that there is nothing new in what you wrote. As you may be aware, it's Wikipedia policy that you shouldn't put anything new in there ( I personally don't think this should apply too rigorously to mathematics, since proofs can be checked without secondary sources. But what you wrote doesn't form a natural part of the article. I think you should either flesh it out or delete it. –  joriki Apr 14 '11 at 12:37
It's an interesting relationship, by the way. But why don't you write $d$ instead of $n/d$? –  joriki Apr 14 '11 at 12:40

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.