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 1 of 7 , Nov 5, 2004View SourceOK. 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:
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.
... 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 wouldMessage 1 of 7 , Nov 5, 2004View SourceOn Saturday 06 November 2004 01:17, you wrote:
> OK. If the range isn't tight, why not consider the bounds p^2 andThis 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.
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.
[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 expectationMessage 1 of 7 , Nov 6, 2004View SourceSuresh Batta wrote:
> Can we now not say that, since both the ranges have comparable primeDefinitely 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
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
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