One of your key identities EQ2.1

2^w(n)= sum(d^2|n)of mobius(d) * tau(n/d^2)

can be generalized. Let m be any integer with m>=0.

2^(m*w(n)) =

m-fold-sum( d1^2|n, d2^2|n, …, dm^2|n )of

mobius(d1)*mobius(d2)*…*mobius(dm) * tau(n/d1^2) * tau(n/d2^2) *…* tau(n/dm^2).

PROOF SKETCH:

It is easy to see this identity is valid in the following easy cases:

(i) m=0 when it is just 1=1, we agree 0-fold-sum is 1.

(ii) m=1 when it is your old identity.

(iii) n=squarefree when only summand comes from d1=d2=…=dm=1.

(iv) n=p^k=prime power when all dj=1 or p:

If k=0 we get n=1 and 2^(m*w(1))=2^0=1=1.

If k=1 we get 2^m=2^m.

If k>=2 we get 2^m=([k+1]-[k-1])^m=2^m.

Now use coprime-multiplicativity to see the easy cases imply validity for all cases.

QED.

This would enable you to count primes in [a,b] not mod 2, but in fact mod 2^m,

for any desired m>=0. If m>log2(b-a+1) this would count the primes, full stop.

Can this be made efficient? I have not tried to figure that out.

Warren D Smith, warren.wds AT gmail.com

Is there an algorithm which on input , finds a prime larger than , with running time bounded by a polynomial in ?

]]>http://arxiv.org/abs/1009.3956

and submitted to Mathematics of Computation.

]]>1. Page 7, there are two of’s.

2. Page 7, “coefficients O(x)” –> “coefficients of size O(x)”.

3. page 7, “If we restrict to the range… the second term”, I think should be “third term”. And “third term” should be “fourth term”.

4. Page 8, “It suffices to show… in time” –> “… that we can compute in time.”

5. Pag 10, near the bottom: You write that , but all we have is ; however, since , where are all “small”, this isn’t much of an issue.

]]>I totally understand, especially given the ICM conference and all.

[I am unsure about the wording, so please check.]

The wording looks fine to me; and I’m sure David and David would appreciate seeing that they are mentioned (and perhaps when they go to apply for grad school… etc.). It’s looking like we might be able to prove a fast algorithm for , as hoped; I'll know for sure in another two weeks or so (after I return from India, and get to meet with them again).

…

I see that you list the Borodin and Moenk paper, but didn't see that you replaced the use of matrices in the proof of Lemma 3.1 with a reference to it. Now that I think about it, I think I like including the matrix argument instead of just doing a citation. The fact is that everyone is familiar with Strassen's matrix algorithm, while few people know the argument in B-M; and so, the present argument in Lemma 3.1 is more self-contained (given the Strassen matrix alg.), and would be but one or two lines shorter anyways, if we just used the result in B-M (instead of matrices).

…

I would like to look over the paper again carefully before we submit it; but that will be a week or two from now, as I will be leaving for India tomorrow.

]]>-Zomega ]]>

Thanks for your edit. Unfortunately, there is more to a natural number n being squarefree than simply not being a square; one must also show that n is not *divisible* by any square larger than 1, see http://en.wikipedia.org/wiki/Square-free_integer .