Estimating an infinite product over primes

CRGreathouse · 2010-07-22, 01:19

I'm trying to estimate the product
$\prod_{p<q<r<s}1-\frac{24}{(pqrs)^2}$
where p,q,r,s are primes.

This can also be calculated as a sum, if desired: for each quadruple, add $\frac{24-24S}{(pqrs)^2}$ to the running sum S.

This is for the purpose of calculating the density of Sloane's A070284. For some reason my naive calculations so far have diverged strongly from evidence obtained from examining the first few hundred members of this sequence, so I may be mistaken about the whole thing -- or my calculations may simply be far from adequate.

Any insight to either the problem or the product would be appreciated.

R.D. Silverman · 2010-07-22, 11:12

Quote:

Originally Posted by CRGreathouse

I'm trying to estimate the product
$\prod_{p<q<r<s}1-\frac{24}{(pqrs)^2}$
where p,q,r,s are primes.

This can also be calculated as a sum, if desired: for each quadruple, add $\frac{24-24S}{(pqrs)^2}$ to the running sum S.

This is for the purpose of calculating the density of Sloane's A070284. For some reason my naive calculations so far have diverged strongly from evidence obtained from examining the first few hundred members of this sequence, so I may be mistaken about the whole thing -- or my calculations may simply be far from adequate.

Any insight to either the problem or the product would be appreciated.

Products of this sort can be quite 'delicate' to estimate. I would ask at

http://listserv.nodak.edu/archives/nmbrthry.html

CRGreathouse · 2010-07-22, 12:48

I'll try that. I think there might be a way to express the product with the primezeta function, which (if true) would not save any effort but would give better precision (since calculating up to a given limit would let s be arbitrarily large rather than just as large as the limit). It's not hard to get that with the (naive) sum version of the problem, but that appears to give a sum greater than 1... clearly there's a lot of cancellation going on.

science_man_88 · 2010-07-22, 13:20

Quote:

Originally Posted by CRGreathouse

I'm trying to estimate the product
$\prod_{p<q<r<s}1-\frac{24}{(pqrs)^2}$
where p,q,r,s are primes.

This can also be calculated as a sum, if desired: for each quadruple, add $\frac{24-24S}{(pqrs)^2}$ to the running sum S.

This is for the purpose of calculating the density of Sloane's A070284. For some reason my naive calculations so far have diverged strongly from evidence obtained from examining the first few hundred members of this sequence, so I may be mistaken about the whole thing -- or my calculations may simply be far from adequate.

Any insight to either the problem or the product would be appreciated.

well if I got the equation right (probably not) 1-(24/(pqrs)^2) would be .90625 for them all equaling 2.
if we add ((24-24Ans)/(24^2)) to this we get .91015625 I'm probably wrong on this one as I only changed one to 3 so technically I should use 81^2 I think. this would make the second one .906592935528 or 90659293553 depending on if my pocket calculator wants to help me lol.

CRGreathouse · 2010-07-23, 16:19

I posted it to MathOverflow; maybe it'll get some eyes there.

I'm calculating the product with s < 7000 now; I expect it will take ~16 hours to finish (around 6pm my time). That should give me some kind of estimate, though I really want 6+ decimal places and that's going to be hard to get.

Quote:

Originally Posted by science_man_88

well if I got the equation right (probably not) 1-(24/(pqrs)^2) would be .90625 for them all equaling 2.

Yes, except I need $p<q<r<s$ , not $p\le q\le r\le s$ , so all 2s isn't an option. (In my problem, this corresponds to the fact that there are no four consecutive numbers that are each multiples of 4.) The smallest multiplicand is $1-\frac{24}{(2\cdot3\cdot5\cdot7)^2}=3673/3675=0.99945\ldots$ .

R. Gerbicz · 2010-07-23, 19:26

Quote:

Originally Posted by CRGreathouse

I'm trying to estimate the product
$\prod_{p<q<r<s}1-\frac{24}{(pqrs)^2}$
where p,q,r,s are primes.

This can also be calculated as a sum, if desired: for each quadruple, add $\frac{24-24S}{(pqrs)^2}$ to the running sum S.

This is for the purpose of calculating the density of Sloane's A070284. For some reason my naive calculations so far have diverged strongly from evidence obtained from examining the first few hundred members of this sequence, so I may be mistaken about the whole thing -- or my calculations may simply be far from adequate.

Any insight to either the problem or the product would be appreciated.

That product is wrong, you are calculating multiple times the same cases, say for example:

Code:

n==0 mod p1^2*p2^2
n+1==0 mod p3^2
n+2==0 mod p4^2
n+3==0 mod p5^2

You are calculating this residue class two times.
Another argument suppose you are using only 5 primes p1,..,p5 in this case the product using these primes multipled by p1^2...*p5^2 should be an integer, but that is not true here.

It would be possible to use inclusion-exclusion principle to get a correct sieve/product for the density. (It is not trivial).

CRGreathouse · 2010-07-23, 19:30

Good catch! That would explain the very slight divergence (0.0001) I was seeing between the first 100,000 terms of the sequence and the constant as calculated.

You're right, that is nontrivial. Ick.

CRGreathouse · 2010-07-23, 19:37

Wait, that might not be as bad as I had thought. What if I multiplied in
$\left(1-\frac{4}{p^2}\right)^{-1}\left(1-\frac{4}{q^2}\right)^{-1}\left(1-\frac{4}{r^2}\right)^{-1}\left(1-\frac{4}{s^2}\right)^{-1}\prod_t1-\frac{4}{t^2}$
to ensure that none of the terms are divisible by the squares of any other primes? Then I'd have to multiply in the case of 5 prime squares, the case of 6 prime squares, etc. -- but these would be small enough that hopefully the first 500 primes would suffice for their calculations.

Edit: Of course I'd need to special-case the prime 2 out... but that actually makes the primary calculation more tractable.

CRGreathouse · 2010-07-23, 20:20

OK, maybe
$\prod_{k=3}^\infty\;\;\;\prod_{2<p_1<\cdots<p_k}1-\frac{12(3^{k-1}-2^k+1)}{p_1p_2\cdots p_k}$
?

R. Gerbicz · 2010-07-23, 20:27

This is my "sum" version for the density, up to 6 primes in the sieve (using inclusion-exclusion principle):

Code:

F(L)=c=0.0;M=[0,0,0,24,-240,1560,0,0,0,0,0,0,0,0,0];\
return(sum(n=2,L,1.0*abs(moebius(n))*M[omega(n)]/n/n))

F(10^7)=0.002148844425826054164408309480
And this should be an upper bound for the density if the M array is correct up to M[6], obiously one could fill up for higher values also.

CRGreathouse · 2010-07-23, 20:47

Quote:

Originally Posted by R. Gerbicz

F(10^7)=0.002148844425826054164408309480

That compares favorably with my empirical finding of 0.0022 in A070284.

2010-07-22, 01:19	#1
CRGreathouse Aug 2006 3·1,993 Posts	Estimating an infinite product over primes I'm trying to estimate the product $\prod_{p<q<r<s}1-\frac{24}{(pqrs)^2}$ where p,q,r,s are primes. This can also be calculated as a sum, if desired: for each quadruple, add $\frac{24-24S}{(pqrs)^2}$ to the running sum S. This is for the purpose of calculating the density of Sloane's A070284. For some reason my naive calculations so far have diverged strongly from evidence obtained from examining the first few hundred members of this sequence, so I may be mistaken about the whole thing -- or my calculations may simply be far from adequate. Any insight to either the problem or the product would be appreciated.

2010-07-23, 19:30	#7
CRGreathouse Aug 2006 3×1,993 Posts	Good catch! That would explain the very slight divergence (0.0001) I was seeing between the first 100,000 terms of the sequence and the constant as calculated. You're right, that is nontrivial. Ick. Last fiddled with by CRGreathouse on 2010-07-23 at 19:32

2010-07-23, 19:37	#8
CRGreathouse Aug 2006 3×1,993 Posts	Wait, that might not be as bad as I had thought. What if I multiplied in $\left(1-\frac{4}{p^2}\right)^{-1}\left(1-\frac{4}{q^2}\right)^{-1}\left(1-\frac{4}{r^2}\right)^{-1}\left(1-\frac{4}{s^2}\right)^{-1}\prod_t1-\frac{4}{t^2}$ to ensure that none of the terms are divisible by the squares of any other primes? Then I'd have to multiply in the case of 5 prime squares, the case of 6 prime squares, etc. -- but these would be small enough that hopefully the first 500 primes would suffice for their calculations. Edit: Of course I'd need to special-case the prime 2 out... but that actually makes the primary calculation more tractable. Last fiddled with by CRGreathouse on 2010-07-23 at 19:39

2010-07-23, 20:20	#9
CRGreathouse Aug 2006 3×1,993 Posts	OK, maybe $\prod_{k=3}^\infty\;\;\;\prod_{2<p_1<\cdots<p_k}1-\frac{12(3^{k-1}-2^k+1)}{p_1p_2\cdots p_k}$ ? Last fiddled with by CRGreathouse on 2010-07-23 at 20:20

2010-07-23, 20:27	#10
R. Gerbicz "Robert Gerbicz" Oct 2005 Hungary 2²×7×53 Posts	This is my "sum" version for the density, up to 6 primes in the sieve (using inclusion-exclusion principle): Code: F(L)=c=0.0;M=[0,0,0,24,-240,1560,0,0,0,0,0,0,0,0,0];\ return(sum(n=2,L,1.0abs(moebius(n))M[omega(n)]/n/n)) F(10^7)=0.002148844425826054164408309480 And this should be an upper bound for the density if the M array is correct up to M[6], obiously one could fill up for higher values also.

2010-07-22, 12:48	#3
CRGreathouse Aug 2006 3·1,993 Posts	I'll try that. I think there might be a way to express the product with the primezeta function, which (if true) would not save any effort but would give better precision (since calculating up to a given limit would let s be arbitrarily large rather than just as large as the limit). It's not hard to get that with the (naive) sum version of the problem, but that appears to give a sum greater than 1... clearly there's a lot of cancellation going on.

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