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• Hi I m not sure if this is a new idea but I have not come across it . Its based on the fact that apart from 2 and 5 all primes have a last digit of 1,3,7 or 9.
Message 1 of 15 , Dec 2 4:52 PM
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Hi

I'm not sure if this is a new idea but I have not come across it .
Its based on the fact that apart from 2 and 5 all primes have a last digit of 1,3,7 or 9.

and

only numbers that end with 1,3,7 or 9 can be multiplied together to produce numbers that end with a 1,3,7 or 9

thus

the set of prime numbers = 2 and 5 and not including 1

PLUS

the set of all numbers ending with 1,3,7,or 9

MINUS

the set of all numbers created by multiplying all
numbers ending with 1,3,7 or 9 by all numbers
ending with 1,3,7 or 9.

I'm looking for comments such as this has all been done before, so forget about or should it be further investigated.

Jim

---------------------------------
Find local movie times and trailers on Yahoo! Movies.

[Non-text portions of this message have been removed]
• It seems correct, but we can say it in an equivalent way: the set of prime numbers = 2, 5 PLUS all odd numbers 5, not ending by 5 and not composite.
Message 1 of 15 , Dec 6 3:55 AM
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It seems correct, but we can say it in an equivalent way:

the set of prime numbers = 2, 5 PLUS all odd numbers > 5, not ending by
5 and not composite.

Following your way we could perform primality test on a number N
searching odd numbers ending by 1, 3, 7, 9 and less than sqrt(N).

Moreover, considering the least significant digit:

...1 => 1 x 1 or 3 x 7 or 9 x 9
...3 => 1 x 3 or 7 x 9
...7 => 1 x 7 or 3 x 9
...9 => 1 x 9 or 7 x 7 or 3 x 3

we can limit our search to a set of the order of sqrt(N) x 2/10 or
sqrt(N) x 3/10 elements.
That is: numbers ending by 1, 3, 9 (or 1, 7, 9) if N ends by 1 and so
on.

I was wondering this: for a fixed M, are there more primes < M ending
with 3 and 7 than primes < M ending with 1 and 9?

Gianni

>-----Original Message-----
>From: Jim Doyle [mailto:ozyjim2004@...]
>Sent: Friday, December 03, 2004 1:53 AM
>Subject: [PrimeNumbers] another way to calculate primes
>
>
>Hi
>
>I'm not sure if this is a new idea but I have not come across
>it . Its based on the fact that apart from 2 and 5 all primes
>have a last digit of 1,3,7 or 9.
>
>...
>...
>
>Jim
• Of course I meant to say: the set of prime numbers = 2, 3, 5 PLUS all odd numbers 5, not ending by 5 and not composite. ... From: Mazzarello Gianni Sent:
Message 1 of 15 , Dec 6 4:05 AM
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Of course I meant to say:

the set of prime numbers = 2, 3, 5 PLUS all odd numbers > 5, not ending
by
5 and not composite.

-----Original Message-----
From: Mazzarello Gianni
Sent: Monday, December 06, 2004 12:55 PM
Subject: RE: [PrimeNumbers] another way to calculate primes

> It seems correct, but we can say it in an equivalent way:

> the set of prime numbers = 2, 5 PLUS all odd numbers > 5, not ending
by
> 5 and not composite.
• In a message dated 06/12/2004 11:57:15 GMT Standard Time, ... Everyone believes that asymptotically (as M - oo) the numbers are equal, but no-one can prove
Message 1 of 15 , Dec 6 4:17 AM
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In a message dated 06/12/2004 11:57:15 GMT Standard Time,
g.mazzarello@... writes:

>I was wondering this: for a fixed M, are there more primes < M ending
>with 3 and 7 than primes < M ending with 1 and 9?

Everyone believes that asymptotically (as M -> oo) the numbers are equal,
but no-one can prove this.

For evidence at M=10^13, see my post
_http://listserv.nodak.edu/scripts/wa.exe?A2=ind0403&L=nmbrthry&P=2486_
(http://listserv.nodak.edu/scripts/wa.exe?A2=ind0403&L=nmbrthry&P=2486)

-Mike Oakes

[Non-text portions of this message have been removed]
• ... I thought that they were proven to be asymptotically equal by an extension of Dirichlet s Theorem. http://www.utm.edu/research/primes/notes/Dirichlet.html
Message 1 of 15 , Dec 6 6:19 AM
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--- Mike Oakes wrote:
>
> In a message dated 06/12/2004 11:57:15 GMT Standard Time,
> g.mazzarello@t... writes:
>
> >I was wondering this: for a fixed M, are there more primes < M
> >ending with 3 and 7 than primes < M ending with 1 and 9?
>
> Everyone believes that asymptotically (as M -> oo) the numbers
> are equal, but no-one can prove this.

I thought that they were proven to be asymptotically equal by
an extension of Dirichlet's Theorem.

http://www.utm.edu/research/primes/notes/Dirichlet.html

See the last equation on that page, which gives bounds for an
error term.
• It s likely that they are asymptotically equal. In this moment I m running a routine (I m over 8.5 x 10^6) that counts separately the number of primes ending
Message 1 of 15 , Dec 6 7:01 AM
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It's likely that they are asymptotically equal.
In this moment I'm running a routine (I'm over 8.5 x 10^6) that counts
separately the number of primes ending with the digit 1, 3, 7, 9.
The fraction of the total number of primes found (that is: p1/PN, p3/PN,
p7/PN, p9/PN, where pk is a prime with k as last digit) is < 0.2500 for
the primes ending with 1 and 9, is > 0.2500 for the primes ending with 3
and 7.
They are all approaching 0.25: always from below p1 and p9 (0.2499),
always from above p3 and p7 (0.2501).

Gianni

>-----Original Message-----
>From: jbrennen [mailto:jack@...]
>Sent: Monday, December 06, 2004 3:19 PM
>Subject: [PrimeNumbers] Re: another way to calculate primes
>
>
>
>--- Mike Oakes wrote:
>>
>> In a message dated 06/12/2004 11:57:15 GMT Standard Time,
>> g.mazzarello@t... writes:
>>
>> >I was wondering this: for a fixed M, are there more primes
>< M ending
>> >with 3 and 7 than primes < M ending with 1 and 9?
>>
>> Everyone believes that asymptotically (as M -> oo) the numbers are
>> equal, but no-one can prove this.
>
>I thought that they were proven to be asymptotically equal by
>an extension of Dirichlet's Theorem.
>
> http://www.utm.edu/research/primes/notes/Dirichlet.html
>
>See the last equation on that page, which gives bounds for an
>error term.
>
>
>
• In a message dated 06/12/2004 14:22:21 GMT Standard Time, jack@brennen.net writes: I thought that they were proven to be asymptotically equal by an extension
Message 1 of 15 , Dec 6 8:25 AM
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In a message dated 06/12/2004 14:22:21 GMT Standard Time, jack@...
writes:

I thought that they were proven to be asymptotically equal by
an extension of Dirichlet's Theorem.

http://www.utm.edu/research/primes/notes/Dirichlet.html

See the last equation on that page, which gives bounds for an
error term.

Yes, you are right of course.
My mistake :-(

-Mike Oakes

[Non-text portions of this message have been removed]
• I noticed I forgot to include 3, it was only distraction. However I think this statement is enaugh: the set of prime numbers = 2, 3, 5 PLUS all odd numbers
Message 1 of 15 , Dec 6 4:45 PM
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I noticed I forgot to include 3, it was only distraction.

However I think this statement is enaugh:

the set of prime numbers = 2, 3, 5 PLUS all odd numbers > 5, not ending by 5 and not composite.

7 is included: 7>5, 7 is odd not ending by 5 AND not composite.
9 is not included: 9 is composite (3^2)
11 is included: 11>5, 11 is odd not ending by 5 AND not composite.

By "composite number" I meant an integer that is the product of two or more integers. I hope I used
the right translation of what I meant.

Of course we can use a more concise statement:
pimes = all not composite numbers > 1.
We will find that, with the exception of 2 and 5, all prime numbers have as final digit one of
these: 1, 3, 7, 9.

----- Original Message -----
From: "Mary & Graeme" <candlesque@...>
To: "Mazzarello Gianni" <g.mazzarello@...>
Sent: Tuesday, December 07, 2004 12:32 AM
Subject: Re: [PrimeNumbers] another way to calculate primes

> May I make an observation here?

>> the set of prime numbers = 2, 5 PLUS all odd numbers > 5, not ending by
>> 5 and not composite.

> Might it not be more correct to state:

> the set of prime numbers = 2,3,5,7,11 PLUS all odd numbers > 5, not ending by
> 5 and not composite.

> ... since the first statement would seem to allow 9 as a prime, and omits 3?
• ... I m missing something. Why single out 2, 3 and 5? The set of prime numbers is all numbers that are not composite, by definition. Paul
Message 1 of 15 , Dec 7 12:23 AM
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On Tue, 2004-12-07 at 00:45, Gianni Mazzarello wrote:
> I noticed I forgot to include 3, it was only distraction.
>
> However I think this statement is enaugh:
>
> the set of prime numbers = 2, 3, 5 PLUS all odd numbers > 5,
> not ending by 5 and not composite.

I'm missing something. Why single out 2, 3 and 5?

The set of prime numbers is all numbers that are not composite, by
definition.

Paul
• ... Whoa there! That leaves a mighty narrow definition of numbers . Units? Rationals? I m guessing that there s some implied context to this remark which I m
Message 1 of 15 , Dec 7 12:20 PM
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--- In primenumbers@yahoogroups.com, Paul Leyland <pcl@w...> wrote:
>
> The set of prime numbers is all numbers that are not composite, by
> definition.
>
Whoa there! That leaves a mighty narrow definition of "numbers".
Units? Rationals?

I'm guessing that there's some implied context to this remark which
I'm missing. Care to supply it?

Rick
• ... Let s not be pedantic here. From the context it s pretty clear that we re talking about integers, so the only gap in Paul s declaration relates to units.
Message 1 of 15 , Dec 7 12:52 PM
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On Tuesday 07 December 2004 18:20, you wrote:
> --- In primenumbers@yahoogroups.com, Paul Leyland <pcl@w...> wrote:
> > The set of prime numbers is all numbers that are not composite, by
> > definition.
>
> Whoa there! That leaves a mighty narrow definition of "numbers".
> Units? Rationals?

Let's not be pedantic here. From the context it's pretty clear that we're
talking about integers, so the only gap in Paul's declaration relates to
units. Furthermore, even if his definition was not 100% mathematically
correct, I think his point was well communicated, and that's what matters
here -- he's not trying to write a book, but rather point out the flaw in the
previous poster's definiton.

Décio

[Non-text portions of this message have been removed]
• ... Thanks Decio. Two people, including Rick and another by personal email, have told me I omitted the unit. Nobody told me I omitted zero. I freely admit
Message 1 of 15 , Dec 8 2:37 AM
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On Tue, 2004-12-07 at 20:52, Décio Luiz Gazzoni Filho wrote:
> On Tuesday 07 December 2004 18:20, you wrote:
> > --- In primenumbers@yahoogroups.com, Paul Leyland <pcl@w...> wrote:
> > > The set of prime numbers is all numbers that are not composite, by
> > > definition.
> >
> > Whoa there! That leaves a mighty narrow definition of "numbers".
> > Units? Rationals?
>
> Let's not be pedantic here. From the context it's pretty clear that we're
> talking about integers, so the only gap in Paul's declaration relates to
> units. Furthermore, even if his definition was not 100% mathematically
> correct, I think his point was well communicated, and that's what matters
> here -- he's not trying to write a book, but rather point out the flaw in the
> previous poster's definiton.

Thanks Decio.

Two people, including Rick and another by personal email, have told me I
omitted the unit. Nobody told me I omitted zero. I freely admit that I
should have included them. As for rationals, irrationals, algebraics,
transcendentals, gaussian integers, complex numbers, quaternions and all
the rest of the zoo, I think it pretty obvious from context that the
area of discourse was N or Z. Otherwise, I may have been tempted to
point out that 3 is prime but that 2 is (1+i)(1-i) and 5 is (2+i)(2-i)
in the Gaussians.

After that clarification, can someone please tell me why in Z or N that
2, 3 and 5 need singling as special cases?

Paul
• I post this after two days of absence. I still have to read all messages posted, but I d like to answer to this question ... The strange way to define primes
Message 1 of 15 , Dec 10 3:01 AM
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I post this after two days of absence. I still have to read all messages posted, but I'd like to answer to this question

>-----Original Message-----
>From: Paul Leyland [mailto:pcl@...]
>Sent: Wednesday, December 08, 2004 11:37 AM
>To: Décio Luiz Gazzoni Filho
>Subject: Re: [PrimeNumbers] Re: another way to calculate primes
>
>...
>After that clarification, can someone please tell me why in Z or N that
>2, 3 and 5 need singling as special cases?
>
>
>Paul

The strange way to define primes was intended to point out the ending digit of almost all the primes: 1, 3, 7, 9 (all except 2 and 5) and the feature that such numbers, multiplied each other in any way , give as a result another number that ends with one of those digits.

The message posted by Jim Doyle [mailto:ozyjim2004@...] Sent: Friday, December 03, 2004 1:53 AM
>Subject: [PrimeNumbers] another way to calculate primes

made me think in which way the mentioned ending digits are obtained.

...1 => 1 x 1 or 3 x 7 or 9 x 9
...3 => 1 x 3 or 7 x 9
...7 => 1 x 7 or 3 x 9
...9 => 1 x 9 or 7 x 7 or 3 x 3

We notice that "in a period" of 10 Natural numbers (e. g. from 1001 to 1010, or from 1,234,567 to 1,234,576 and so on) we can generate more composite numbers ending with the digit 1 and 9 than numbers with ending digit 3 and 7.
This, at first glance, made me suppose that primes with ending digit 1 or 9 would be less than primes with ending digit 3 or 7.

I heard, though I haven't studied it still, of a theorem that says that primes are equally distributed on the four ending digits as they grow to infinity.
In a easy computer routine that counts the number of primes of the "four ending digit kinds" I found that starting from 2 and going to the order of 10^8, the primes with ending digit 3 and 7 are more (about 200 at the end) than those ending with 1 and 9.

That's all

Gianni
• You say: -- You have to count the reverse cases: ...1 = 1 x 1 or 3 x 7 or 7 x 3 or 9 x 9 ...3 = 1 x 3 or 3 x 1 or 7 x 9 or 9 x 7 ...7 = 1 x 7 or 7 x 1 or
Message 1 of 15 , Dec 12 8:47 AM
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You say:
--
You have to count the "reverse" cases:

...1 => 1 x 1 or 3 x 7 or 7 x 3 or 9 x 9
...3 => 1 x 3 or 3 x 1 or 7 x 9 or 9 x 7
...7 => 1 x 7 or 7 x 1 or 3 x 9 or 9 x 3
...9 => 1 x 9 or 9 x 1 or 7 x 7 or 3 x 3
--

It seems to me you are right (more or less).
But haven't they "reverse" cases also 1x1 and 3x3 and 7 x 7 and 9 x 9?
Then we have to consider:

...1 => 1 x 1 or 3 x 7 or 7 x 3 or 9 x 9 or 1 x 1 or 9 x 9
...3 => 1 x 3 or 3 x 1 or 7 x 9 or 9 x 7
...7 => 1 x 7 or 7 x 1 or 3 x 9 or 9 x 3
...9 => 1 x 9 or 9 x 1 or 7 x 7 or 3 x 3 or 3 x 3 or 7 x 7

Gianni

______________________________________________________
----- Original Message -----
From: "Jens Kruse Andersen" <jens.k.a@...>
To: "Mazzarello Gianni" <g.mazzarello@...>
Sent: Friday, December 10, 2004 4:54 PM
Subject: Re: another way to calculate primes

Re: another way to calculate primes

----- Original Message -----
From: "Mazzarello Gianni" <g.mazzarello@...>
Sent: Friday, December 10, 2004 12:01 PM
Subject: RE: [PrimeNumbers] another way to calculate primes

> ...1 => 1 x 1 or 3 x 7 or 9 x 9
> ...3 => 1 x 3 or 7 x 9
> ...7 => 1 x 7 or 3 x 9
> ...9 => 1 x 9 or 7 x 7 or 3 x 3

You have to count the "reverse" cases:

...1 => 1 x 1 or 3 x 7 or 7 x 3 or 9 x 9
...3 => 1 x 3 or 3 x 1 or 7 x 9 or 9 x 7
...7 => 1 x 7 or 7 x 1 or 3 x 9 or 9 x 3
...9 => 1 x 9 or 9 x 1 or 7 x 7 or 3 x 3

4 cases for each digit.

--
Jens Kruse Andersen (off list)
• Gianni, I think Jens is correct. You have to ask what it is that we are trying to enumerate. We re trying to find the last digits for the factors of some odd
Message 1 of 15 , Dec 13 8:31 AM
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Gianni,
I think Jens is correct. You have to ask what it is that we are
trying to enumerate. We're trying to find the last digits for the
factors of some odd composite, N, call them p and q and we can let
p represent the smaller number, q the larger. The case, for
example where p=3(mod 10) and q=7(mod 10) is different from p=7(mod 10)
and q=3(mod 10) but the case p=1(mod 10) and q=1(mod 10) is not a different
case when the 1 and 1 are swapped.

Another way to think of it: Make a 4x4 multiplication table, mod 10.
With 4 digits there are 16 ways to combine them, with nice symmetry.

1 3 7 9
1 1 3 7 9
3 3 9 1 7
7 7 1 9 3
9 9 7 3 1

-----Original Message-----
From: Gianni Mazzarello [mailto:g.mazzarello@...]
Sent: Sunday, December 12, 2004 10:47 AM
Subject: [PrimeNumbers] Re: another way to calculate primes

You say:
--
You have to count the "reverse" cases:

...1 => 1 x 1 or 3 x 7 or 7 x 3 or 9 x 9
...3 => 1 x 3 or 3 x 1 or 7 x 9 or 9 x 7
...7 => 1 x 7 or 7 x 1 or 3 x 9 or 9 x 3
...9 => 1 x 9 or 9 x 1 or 7 x 7 or 3 x 3
--

It seems to me you are right (more or less).
But haven't they "reverse" cases also 1x1 and 3x3 and 7 x 7 and 9 x 9?
Then we have to consider:

...1 => 1 x 1 or 3 x 7 or 7 x 3 or 9 x 9 or 1 x 1 or 9 x 9
...3 => 1 x 3 or 3 x 1 or 7 x 9 or 9 x 7
...7 => 1 x 7 or 7 x 1 or 3 x 9 or 9 x 3
...9 => 1 x 9 or 9 x 1 or 7 x 7 or 3 x 3 or 3 x 3 or 7 x 7

Gianni

______________________________________________________
----- Original Message -----
From: "Jens Kruse Andersen" <jens.k.a@...>
To: "Mazzarello Gianni" <g.mazzarello@...>
Sent: Friday, December 10, 2004 4:54 PM
Subject: Re: another way to calculate primes

Re: another way to calculate primes

----- Original Message -----
From: "Mazzarello Gianni" <g.mazzarello@...>
Sent: Friday, December 10, 2004 12:01 PM
Subject: RE: [PrimeNumbers] another way to calculate primes

> ...1 => 1 x 1 or 3 x 7 or 9 x 9
> ...3 => 1 x 3 or 7 x 9
> ...7 => 1 x 7 or 3 x 9
> ...9 => 1 x 9 or 7 x 7 or 3 x 3

You have to count the "reverse" cases:

...1 => 1 x 1 or 3 x 7 or 7 x 3 or 9 x 9
...3 => 1 x 3 or 3 x 1 or 7 x 9 or 9 x 7
...7 => 1 x 7 or 7 x 1 or 3 x 9 or 9 x 3
...9 => 1 x 9 or 9 x 1 or 7 x 7 or 3 x 3

4 cases for each digit.

--
Jens Kruse Andersen (off list)

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