- On Friday 05 November 2004 15:42, you wrote:
> Do you mean that one cannot compare 1,p+3 and

No, I mean that you can't use the PNT to count primes in an interval in an

> p^2,(p+2)^2 bounds or ranges?

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 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) - 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 - On Saturday 06 November 2004 01:17, you wrote:
> OK. If the range isn't tight, why not consider the bounds p^2 and

This is a gain of a constant factor. I believe you need asymptotically tighter

> p*(p+2)? The range would be just 2p.

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>

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

>

> 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.

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] - Suresh Batta wrote:

> Can we now not say that, since both the ranges have comparable prime

Definitely no. The only "comparable prime quantities" are about the _expected_

> quantities, if a P+3 has a twin prime, then squared range

> ((p+2)^2, p^2) should have too have a twin prime?

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