## Twin primes

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• Hello all, I ve been working on the twin prime conjecture for about five years and would like to share some of my ideas (no proofs as yet though, sorry!) The
Message 1 of 9 , Mar 18 5:24 PM
Hello all,

I've been working on the twin prime conjecture for about five years
and would like to share some of my ideas (no proofs as yet though,
sorry!)

The moebius function: M(n)

M(1)=1
M(n)=0 if n is divisible by m^2 for some m>1
M(n)=(-1)r if n is the product of r distinct primes,

The Mangoldt function: A(n)

A(n)= log p if n=p^k, k>0
0 otherwise

a function that defines the primes: F(n)
F(n)= M(n)*A(n) gives 0 if n is square-divisible or divides a product
of primes, or log(1/n)=-log n if and only if n is prime

therefore, F(n)*F(n+2) is equal to log n * log (n+2) if and only if n
is the first prime in a twin pair. (The second prime n+2 would not
produce a nonzero result since n+4 is not prime, and n>3) So then,

sum n<=x(F(n) * F(n+2))/(n * log n * log (n+2))

would sum the reciprocals of the first prime in a prime pair (by
using the error term (2*c)/log n (c=.6601618158 the twin prime
constant) I've supposed this limits to about 1.05893... (*)

I'm using (2*c)/log n instead of Prof Brent's term (4*c)/ log n
because I've summed this using one term in the pair instead of the
usual two.

The constant I marked (*) appears to be constructed out of a ratio of
prime power products:

product n=1, infinity, 1-M(n)*n^-2 divided by
product odd primes 1-(p-1)^-2

(The denominator being the constant c=.66016...) This appears to be
the same as the sum of reciprocals of the twins defined above, in
terms of a limit (I dont think this assumes infinitely many twins
since Brun's constant is the limit of 1/p+1/(p+2) regardless of the
number of terms in the sum). Any one like to prove this? Or, if I am
moebius function in a product...!) If this is proven, what will it
say about the total number of twin pairs?

Thanks,
from Guy H Bearman
• ... Has this been checked numerically to Nicely-level (Intel destruction level:-) accuracy? As a sum, it looks like a subset of the Brun reciprocals, which
Message 2 of 9 , Mar 18 7:38 PM
"Guy" <anastasis_1999@y...> wrote:

> appears to be constructed out of a ratio

Has this been checked numerically to Nicely-level
(Intel destruction level:-) accuracy?

As a sum, it looks like a subset of the Brun reciprocals,
as I recall.

As a product it looks like the sort of thing
that Pieter Moree or Gerhard Niklasch might eat for breakfast.

So, if you use best practice for each form,
how good is the match, please?

If it's a few parts per billion,
then it's a capital empirical Ansatz, and makes
me wonder whether you can guess a product form for Brun?
If not, why for one and not for the other?

Intrigued..

David
• Prod_n (1-mu(n)/n^2) looks easy to get to (say) 18 sf. 1) Take product up to N=10^6. 2) For the rest take logs and retain only sum_{n N} mu(n)/n^2 = 6/pi^2 -
Message 3 of 9 , Mar 18 9:28 PM
Prod_n (1-mu(n)/n^2) looks easy to get to (say) 18 sf.
1) Take product up to N=10^6.
2) For the rest take logs and retain only
sum_{n>N} mu(n)/n^2 = 6/pi^2 - sum_{n<=N} mu(n)/n^2.
3) Then combine with twin-prime constant.
4) Mail 18-digit prediction for "semi-Brun" sum
to Thomas Nicely, and pray hard
that the first 9 digits hold up.
David
• Finally: you do not actually need Nicely help, since 2*semi-Brun - Brun is much better convergent and you can test it deep yourself, when you have fixed the
Message 4 of 9 , Mar 18 11:18 PM
Finally: you do not actually need Nicely help, since
2*semi-Brun - Brun is much better convergent
and you can test it deep yourself, when you have fixed
the glitch that prod_{n>=1} (1-mu(n)/n^2) is zero!
• Pm=prod_{n 1} (1-mu(n)/n^2) [cannot include n=1, else vanishes!] C2=twin-prime, as per usual SB=semi-Brun, only take first reciprocal for each twin At speed, I
Message 5 of 9 , Mar 19 10:42 AM
Pm=prod_{n>1} (1-mu(n)/n^2) [cannot include n=1, else vanishes!]
C2=twin-prime, as per usual
SB=semi-Brun, only take first reciprocal for each twin
At speed, I make those
Pm=1.430380788
C2=0.660161816
SB=1.059064265
You said SB*C2=Pm, which is clearly no good.
SB*C2*Pm=1 is better, but not good enough.
I make it 1.0000561 [needs checking]
Back to the drawing board?
• ... Guy Bearman has lost his Yahoo connectivity. Offlist, he confirmed that I had guessed what he meant and said ... It would be in the spirit of this list if
Message 6 of 9 , Mar 19 3:30 PM
I wrote:

> Pm=prod_{n>1} (1-mu(n)/n^2) [cannot include n=1, else vanishes!]
> C2=twin-prime, as per usual
> SB=semi-Brun, only take first reciprocal for each twin
> SB*C2*Pm=1 is better, but not good enough.
> I make it 1.0000561 [needs checking]

Guy Bearman has lost his Yahoo connectivity.
Offlist, he confirmed that I had guessed what he meant and said

> Thank you for making the time to refute this idea

I replied:

> It was fun. When I worked out what you meant and got 1.00005,
> at low accuracy, the adrenaline started to flow.
> I hope that someone will check my numerics,
> lest I have missed a gem.

It would be in the spirit of this list if someone
(Andrey Kulsha where are you?)
checks that I am not mistaken.

Had it worked, it would have been sensational!

Goodness knows how Guy was led to guess something that got so close.
It's a salutory reminder of how much dedicated private study
surfaces in our amateur (fine word!) pages.

David
• Yes, it s me again :) (all logs are to base e.) Brent (1975) uses the formula, sum(k=1, +inf, (k+1)/zeta(k+1) - k/zeta(k+2))*(log x)^(k-1)/(k+1)! in his work
Message 7 of 9 , Oct 5, 2001
Yes, it's me again :)

(all logs are to base e.)
Brent (1975) uses the formula,

sum(k=1, +inf, (k+1)/zeta(k+1) - k/zeta(k+2))*(log x)^(k-1)/(k+1)!

in his work on Brun's sum.

Just now I was thinking (again), how do I do a product equalling
Brun's sum? And came up with this:

4*c*prod(k=1, +inf, (k+1)/zeta(k+1) - k/zeta(k+2)), where c is the
twin prime constant, 0.6601618158...

this comes to the value (approx.) 0.40888 (each term in the product
tends to 1 as k->+inf) - yes I know this doesn't look like Brun's
constant!

call this constant M. Then calculate -1/(log(1-M)). With only twenty
terms of the product, and using 9 dp for each zeta function, I get
1.90204 already. Would anyone like to do this with say, 100 dp of
each zeta function, for several hundred terms, and see if it gets any
closer to 1.90216?

Thanks, from Guy
• I ve got a bit of a fascination at the moment for composites (c) of the form c = p^2 One of the things I seem to be observing is that there seems to be twin
Message 8 of 9 , Sep 9, 2003
I've got a bit of a fascination at the moment for composites (c) of the form

c = p^2

One of the things I seem to be observing is that there seems to be twin
primes in close
proximity to composites of the above form.

eg.

5^2 = 25 17,19 and 29,31

7^2 = 49 41,43 and 59,61

317^2 = 100489 100391,100393 and 100517,100519

Is this significant?

Navid

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• ... Not particularly. The first two examples are far too small to draw any reasonable conclusions from. It seems strange that you suddenly have to jump from 7
Message 9 of 9 , Sep 9, 2003
> I've got a bit of a fascination at the moment for composites (c) of the form
>
> c = p^2
>
> One of the things I seem to be observing is that there seems to be twin
> primes in close
> proximity to composites of the above form.
>
> eg.
>
> 5^2 = 25 17,19 and 29,31
>
> 7^2 = 49 41,43 and 59,61
>
> 317^2 = 100489 100391,100393 and 100517,100519
>
>
> Is this significant?

Not particularly. The first two examples are far too small to draw any
reasonable conclusions from. It seems strange that you suddenly have to
jump from 7 to 317, but maybe there are other examples in between, I
can't be bothered to check.

According to the prime number theorem, the average distance from a prime
p to the next prime is about log p. Suppose now that p is part of a twin
prime pair, and that q is the first of the next twin prime pair. Then
the expected distance between p and q is roughly log^2 p instead. So the
twin primes are expected to be fairly common, and certainly a lot more
common than numbers of the form p^2. So it's not surprising that you may
well find twin primes "fairly close" to some prime squares.

Andy
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