## Re: [PrimeNumbers] another attempt at Twin prime conjecture...

Expand Messages
• ... No, I mean that you can t use the PNT to count primes in an interval in an exact fashion; there ll always be an error bound. According to the book by
Message 1 of 7 , Nov 5, 2004
On Friday 05 November 2004 15:42, you wrote:
> Do you mean that one cannot compare 1,p+3 and
> p^2,(p+2)^2 bounds or ranges?

No, I mean that you can't use the PNT to count primes in an interval in an
exact fashion; there'll always be an error bound. According to the book by
Crandall & Pomerance, the PNT as currently proved could be stated as

pi(x) = li(x) + O(x e^(-C sqrt(log x))),

where pi(x) is the prime counting function and li(x) is the logarithmic
integral int(dt/log t) from 2 to x, and is asymptotic to x/log x. The problem
is that you're neglecting the error term O(...). Even if one were to assume
the Riemann Hypothesis, through which the error term would be decreased to
O(x^(1/2 + eps)) for eps -> 0, the bounds are not tight enough for your
method of proof to work.

While you replied to me in private, I'm going to get this back into the list,
as it might help others pursuing similar lines of proof, whether for the twin
prime conjecture or something else.

Décio

[Non-text portions of this message have been removed]
• Also, here s another argument regarding your method of proof, which I ll recall involves showing that if p, p + 2 is a twin prime pair, then there exists
Message 2 of 7 , Nov 5, 2004
Also, here's another argument regarding your method of proof, which I'll
recall involves showing that if p, p + 2 is a twin prime pair, then there
exists another twin prime pair between p^2 and (p + 2)^2.

The book by Hardy & Wright claims that it is an unsolved problem whether there
exists a prime between n^2 and (n+1)^2. In a sense this is a weaker form of
your statement, as it requires the existence of a single prime, not a prime
pair. On the other hand it is a little stronger, as your interval is (p +
2)^2 - p^2 = (p + 2 - p)(p + 2 + p) = 2(2p + 2) = 4p + 4, while (n+1)^2 - n^2
= (n + 1 - n)(n + 1 + n) = 2n + 1. Also, your argument is only required to
hold for primes p, p + 2, whereas no restrictions are placed on n. Still, I
don't see why your argument would be easier to prove.

Décio

[Non-text portions of this message have been removed]
• OK. If the range isn t tight, why not consider the bounds p^2 and p*(p+2)? The range would be just 2p. Then, the difference would be.. p^2 + 2p / log (p)*(p+2)
Message 3 of 7 , Nov 5, 2004
OK. If the range isn't tight, why not consider the bounds p^2 and
p*(p+2)? The range would be just 2p.

Then, the difference would be..

p^2 + 2p / log (p)*(p+2) - p^2 / log p^2

p^2 + 2p / log p * log (p+2) - p^2 / 2 log p

Reducing log (p+2) to log p when p--> Infinity, we get,

p^2 + 2p / 2 log p - p^2 / 2 log p

2p / 2 log p

p / log p ----------- 1 (PNT)

Compare this to the original range, p+3 / log p.
This reduces to p / log p for large p.

So, both of the ranges show same order.

Can we now not say that, since both the ranges have comparable prime
quantities, if a P+3 has a twin prime, then squared range
((p+2)^2, p^2) should have too have a twin prime?

My effort stems from the encouragement of a recent observation i
made. I counted how may twin primes are there between the squares of
a twin prime.
It's something like:
5-7 2
11-13 2
17-19 2
29-31 2
41-43 3
59-61 5
71-73 3
101-103 7
107-109 6
137-139 6
149-151 10
179-181 13
191-193 7
197-199 8

When I graphed this and looked at the regression equation, it looked
like the same rate of growth as p/log p. This made me believe that
may be Squaring a number or infact any exponential order does not
change any thing. pi(x) stays the same. Hence this attempt.

Suresh
• ... This is a gain of a constant factor. I believe you need asymptotically tighter bounds to achieve a proof. That is, instead of p times a constant, you would
Message 4 of 7 , Nov 5, 2004
On Saturday 06 November 2004 01:17, you wrote:
> OK. If the range isn't tight, why not consider the bounds p^2 and
> p*(p+2)? The range would be just 2p.

This is a gain of a constant factor. I believe you need asymptotically tighter
bounds to achieve a proof. That is, instead of p times a constant, you would
need sqrt(p) or log(p) or whatever. I'm not an expert in analytical number
theory, I'm just sure the bounds are not tight enough or this proof would
have surfaced much earlier.

> <snip>
>
> My effort stems from the encouragement of a recent observation i
> made. I counted how may twin primes are there between the squares of
> a twin prime.
> It's something like:
> 5-7 2
> 11-13 2
> 17-19 2
> 29-31 2
> 41-43 3
> 59-61 5
> 71-73 3
> 101-103 7
> 107-109 6
> 137-139 6
> 149-151 10
> 179-181 13
> 191-193 7
> 197-199 8
>
> When I graphed this and looked at the regression equation, it looked
> like the same rate of growth as p/log p. This made me believe that
> may be Squaring a number or infact any exponential order does not
> change any thing. pi(x) stays the same. Hence this attempt.

In fact there exists a conjecture for the number of twin primes up to a
certain bound

Here's an heuristic for the number of twin primes between p^2 and (p+2)^2. We
start with the approximation pi2(x) ~ 2 C2 x/log^2(x), where pi2(x) is the
twin-prime counting funcion and C2 is the twin-prime constant 0.660161858...
This formula was taken from the trusty book of Crandall & Pomerance. Now

pi2((p+2)^2) - pi2(p^2) ~ 2 C2 ( (p+2)^2/log^2((p+2)^2) - p^2/log^2(p^2) )
~ 2 C2 ( (p+2)^2/log^2(p^2) - p^2/log^2(p^2) )
= 2 C2/log^2(p^2) ( (p+2)^2 - p^2 )
= 2 C2/(2^2 log^2(p)) ( 4p + 4 )
= 2 C2 (4p + 4)/(2^2 log^2(p))
~ 2 C2 4p/(2^2 log^2(p))
= 2 C2 p/log^2(p)

Notice that the first approximation is very good -- for p ~ 1e3 the error is
on the fourth decimal place. The second approximation is good also: the term
I discarded is 2 C2/log^2(p), which is already 0.5 for p as small as 5.

Now for the interpretation of this result: in the interval (p^2,(p+2)^2),
there are as many twin primes as in the interval (0,p). This makes sense:
while there are ~4p primes in the former interval, they are larger than the
primes in the latter interval, and this difference is accounted for by the
factor 1/4 -- which is precisely the square of the ratio between the number
of digits of a typical number in each interval, not coincidently the
dependence embodied by the log^2(p) term in the approximation to pi2(x).
Thus your regression equation is off by a factor 1/log(p).

Please, do not interpret these heuristics as validating your proof: they are
just *unproven heuristics*. Actually if this formula were exact, we wouldn't
need to employ your proof at all; the infinity of twin primes would follow
directly from the formula.

Décio

[Non-text portions of this message have been removed]
• ... Definitely no. The only comparable prime quantities are about the _expected_ number of primes based on unproven heuristics. Even if this expectation
Message 5 of 7 , Nov 6, 2004
Suresh Batta wrote:

> Can we now not say that, since both the ranges have comparable prime
> quantities, if a P+3 has a twin prime, then squared range
> ((p+2)^2, p^2) should have too have a twin prime?

Definitely no. The only "comparable prime quantities" are about the _expected_
number of primes based on unproven heuristics. Even if this expectation could be
proven reasonably accurate, it would not say anything about the number of twin
primes.

p/log p is the approximate number of primes below p.
The prime number theorem says it is asymptotically right. This means the
relative error (NOT the absolute error) tends to 0 when p tends to infinite.

However, as Decio notes, this and better known approximations are too poor to
say anything about the number of primes (let alone twin primes) from p to p+x
when x is much smaller than p.

It can only be used to say things like:
The _average_ number of primes in an interval of x numbers _around_ the size of
p is approximately x/log p.

Little is known about how large the deviation from such averages can be.
To illustrate possible deviations, here are the most extreme values known around
100 digits:

There are 11 primes (0.16 expected) among the 104-digit numbers p to p+36 for
p = 24698258*239# + 28606476153371, found by Norman Luhn and I.

There are 0 primes (30 expected) among the 93-digit numbers c to c+6378 for
c = 5629854038470321802219554908853741163682800524507382035301697914566243\
83980052820124370178769, found by Torbjörn Alm and I.

There probably exists far more extreme cases.

As far as twin primes go, there is not even anything known about averages.

--
Jens Kruse Andersen
Your message has been successfully submitted and would be delivered to recipients shortly.