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.
> 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
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.
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