[number theory] sieving with a_i mod p_i
#1
Consider all the primes between 1 and 100.
Call them p_i.

If we want to count the primes between 100 and 10000 then we sieve that interval with 0 mod p_i.

But what happens if say we sieve 7 mod p_i ?
( 7 mod 2 => 1 mod 2 , 7 mod 3 => 1 mod 3 , 7 mod 5 => 2 mod 5 , 7 mod 7 => 0 mod 7 , 7 mod 11 , ... )

In general what happens if we sieve a_i mod p_i with 0 < a_i < p_i ?


How do we choose the a_i such that we sieve as many numbers as possible ?
Or how do we choose the a_i such that there are as many numbers left as possible ?

regards

tommy1729
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#2
(09/09/2014, 01:10 AM)tommy1729 Wrote: Consider all the primes between 1 and 100.
Call them p_i.

If we want to count the primes between 100 and 10000 then we sieve that interval with 0 mod p_i.

But what happens if say we sieve 7 mod p_i ?
( 7 mod 2 => 1 mod 2 , 7 mod 3 => 1 mod 3 , 7 mod 5 => 2 mod 5 , 7 mod 7 => 0 mod 7 , 7 mod 11 , ... )

In general what happens if we sieve a_i mod p_i with 0 < a_i < p_i ?


How do we choose the a_i such that we sieve as many numbers as possible ?
Or how do we choose the a_i such that there are as many numbers left as possible ?

regards

tommy1729

It took me a couple minutes to really figure out what you were even asking here. Once I did, I threw together some tests in Excel. I was expecting it to be very erratic, but counter-intuitively, it's really stable. I haven't had a chance to validate these numbers or write some gp code, so these numbers are very preliminary, but...

Taking primes between 2 and 97 ("100"), and sieving the numbers from 101 to 10000, I get:

Code:
a_i     Count("Primes")
0..27   1204
28..33  1203
34..51  1202
52..59  1201
60..69  1200
70..71  1199
72..77  1198
78..93  1197
94..96  1196

I suppose the numbers might be more stable if I had sieved from 98 to 97^2? Or perhaps from 98 to 101^2-1? Anyway, this single data point suggests that low numbers, close to 0, maximize the number of "primes".
~ Jay Daniel Fox
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#3
Thank you for your reply and intrest Jay !

However Im not sure what you computed.

There are 25 primes below 100.
Or 24 odd primes.

So i must be 24 or 25 depending on details in the definition.

But then you write

a_i
..
28..33

I do not know what that means ?

28,29,30,31,32,33 ?

Does that mean that for any of 28,29,... 33 as a fixed value of a_i we get the same result ?

Many interpretations are possible I think.

Im betting you are considering a fixed a_i and

0..27 1204 means that for any fixed a_i between 0 and 27 we get the same value : 1204.

Is my guess correct ?

Thanks for the reply.

Btw as for " intuition " I note that for prime twins we sieve
0 mod 2 0 mod 3 0 mod 5 ...
and
(3-2) mod 3 (5-2) mod 5 ...

Then for this variant , the intuition of " stable " at least for simple "patterns" in the a_i , leads to the conjecture of infinitude of the prime twins !

Note that in my OP - which appears confusing apparantly - the a_i are not neccessarily fixed.

So we could consider
1 mod 2 ,1 mod 3, 2 mod 5, 6 mod 7, 5 mod 11 etc

I assume fixed a_i are easier to study perhaps.

My question is thus more general.
I gave an example - between the brackets - of a fixed a_i ...
Probably that confused matters. ( sorry )

**
Probably this reminds some people of the " Lucky numbers ".
**

Thanks for your reply.
Hope this results in something constructive.


regards

tommy1729
Reply
#4
(09/10/2014, 08:27 PM)tommy1729 Wrote: Thank you for your reply and intrest Jay !

However Im not sure what you computed.
(...)
But then you write

a_i
..
28..33

(...)

Does that mean that for any of 28,29,... 33 as a fixed value of a_i we get the same result ?

Many interpretations are possible I think.

Im betting you are considering a fixed a_i and

0..27 1204 means that for any fixed a_i between 0 and 27 we get the same value : 1204.

Is my guess correct ?

Exactly what I meant! For a fixed a_i of 0 (or 1, or 2, or 27, etc.), I found 1204 "primes" in the range of 101 to 10000 inclusive.

Quote:Note that in my OP - which appears confusing apparantly - the a_i are not neccessarily fixed.

So we could consider
1 mod 2 ,1 mod 3, 2 mod 5, 6 mod 7, 5 mod 11 etc

I assume fixed a_i are easier to study perhaps.
Well, my modular arithmetic is rusty, but it seems to me that {1 mod 2, 1 mod 3, 2 mod 5, 6 mod 7, 5 mod 11}, as a set, defines a unique number mod 2*3*5*7*11. Assuming my educated guess is right, you basically are fixing a_i, though not limited to the same range as initially defined. So I think it still makes sense to speak in terms of a fixed a_i, if perhaps we relax the restriction on the size of a_i (up to primorial(p_i)-1 or something like that).
~ Jay Daniel Fox
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#5
{0 mod 2, 0 mod 3 , 0 mod 5 , ... 0 mod prime 't' } does not define a certain a mod b.

Rather it sieves out possibilities of a mod b.

Its like 0 mod 2 , 0 mod 3 , 0 mod 5 gives :

1,7,11,13,17,19,23,29 mod 30.

A similar thing happens with a_i mod p_i.

But maybe that is what you actually meant (to say).

regards

tommy1729
Reply
#6
If we sieve an interval [a,b] with 0 mod 2 , 0 mod 3 , 0 mod 5 etc
we get close to the primes in that interval.

If we sieve an interval [a,b] with 5 mod 2, 5 mod 3, 0 mod 5, 5 mod 7 , 5 mod 11 etc , we get close to the primes in the interval [a-5,b-5].

Therefore it is not so surprising the results for fixed a_i are so similar.

This is also why I say fixed a_i are less intresting.

regards

tommy1729
Reply
#7
(09/11/2014, 08:24 AM)tommy1729 Wrote: {0 mod 2, 0 mod 3 , 0 mod 5 , ... 0 mod prime 't' } does not define a certain a mod b.

Rather it sieves out possibilities of a mod b.

Its like 0 mod 2 , 0 mod 3 , 0 mod 5 gives :

1,7,11,13,17,19,23,29 mod 30.

A similar thing happens with a_i mod p_i.

But maybe that is what you actually meant (to say).

regards

tommy1729

Hmm, maybe we're talking about different things. I'm talking about an a_i that gives the respective remainders mod p_i, all at the same time, not individually. I agree that for sieving, we care about a single match, so we get back a set of sieved (or unsieved) numbers.

The only numbers that are 0 mod 2, 0 mod 3, 0 mod 5 (all at the same time, not individually) are the numbers that are 0 mod 30. The only numbers that are 1 mod 2, 2 mod 3, 1 mod 5 are the numbers that are 11 mod 30. So by giving me a set of a_i's to go with the p_i's, like a_i={1, 2, 1} and p_i={2, 3, 5}, I can give you back a new "a" to go with product(p_i): a=11 and primorial(5)=30.

In this sense, giving me a set of a_i's in the range 0 <= a_i < p_i is just equivalent to giving me a fixed a in the range 0 <= a < primorial(max(p_i))

In light of this, then yes, having unfixed a_i's is an interesting problem in its own right, as long as we can agree that it's just another way of looking at a fixed value of a.

Edit: So for example, sieving numbers against 1 mod 2, 2 mod 3, and 1 mod 5, is like sieving them against 11 mod 2, 11 mod 3, and 11 mod 5. Hopefully that made sense...End Edit

And I reserve the right to be wrong. Tongue
~ Jay Daniel Fox
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#8
(09/11/2014, 08:00 PM)jaydfox Wrote:
(09/11/2014, 08:24 AM)tommy1729 Wrote: {0 mod 2, 0 mod 3 , 0 mod 5 , ... 0 mod prime 't' } does not define a certain a mod b.

Rather it sieves out possibilities of a mod b.

Its like 0 mod 2 , 0 mod 3 , 0 mod 5 gives :

1,7,11,13,17,19,23,29 mod 30.

A similar thing happens with a_i mod p_i.

But maybe that is what you actually meant (to say).

regards

tommy1729

Hmm, maybe we're talking about different things. I'm talking about an a_i that gives the respective remainders mod p_i, all at the same time, not individually. I agree that for sieving, we care about a single match, so we get back a set of sieved (or unsieved) numbers.

The only numbers that are 0 mod 2, 0 mod 3, 0 mod 5 (all at the same time, not individually) are the numbers that are 0 mod 30. The only numbers that are 1 mod 2, 2 mod 3, 1 mod 5 are the numbers that are 11 mod 30. So by giving me a set of a_i's to go with the p_i's, like a_i={1, 2, 1} and p_i={2, 3, 5}, I can give you back a new "a" to go with product(p_i): a=11 and primorial(5)=30.

In this sense, giving me a set of a_i's in the range 0 <= a_i < p_i is just equivalent to giving me a fixed a in the range 0 <= a < primorial(max(p_i))

In light of this, then yes, having unfixed a_i's is an interesting problem in its own right, as long as we can agree that it's just another way of looking at a fixed value of a.

Edit: So for example, sieving numbers against 1 mod 2, 2 mod 3, and 1 mod 5, is like sieving them against 11 mod 2, 11 mod 3, and 11 mod 5. Hopefully that made sense...End Edit

And I reserve the right to be wrong. Tongue

Unfortunately the primorial is a quickly growing function.

If the primorial is much larger than the interval we sieve , I wonder if that has practical use ?

Notice your table does not contain such large numbers.

regards

tommy1729
Reply


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