# Can someone give me a counterexample to disprove this statement?

Claim :

For any even number $n$ there is at least one prime number of the form :

$$p=k\cdot2^{n}-1$$

with following properties : $k=2^{a-n}+1 , n\leq a < 2n ,$ and $a,n\in \mathbf{Z^{+}}$

Can someone give me a counterexample ?

I have checked statement for each $n$ up to $n=100$ , but I am also aware of strong law of small numbers so this statement could be false.

EDIT:

One necessary condition is that there is at least one prime number on interval :

$[2^{n+1}-1,2^{2n-1}+2^{n}-1]$

One can easily show that this is true by using Bertrand's theorem.

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Just to unfold the slightly convoluted definition, the condition is equivalent to $p=2^a+2^n-1$, or $p=2^{n+b}+2^n-1$ with $0\le b<n$. –  Henning Makholm Nov 5 '11 at 13:10
If instead of primes we look for a random property distributed "like the primes" (according to the prime number theorem), the expected number of hits for any $n$ tends towards $1$ for $n\to\infty$. The limiting distribution of number-of-hits would be somewhat Poisson-like, with uniform positive probability for each $n$ to have no hits. Of course, however, the primes are not randomly distributed ... –  Henning Makholm Nov 5 '11 at 13:26
Where does this problem come from? What is the motivation? –  lhf Nov 5 '11 at 13:36
@lhf,no specific source,my experimentation...motivation is to prove that there is infinitely many primes of the $p=k\cdot2^{n}-1$ form –  pedja Nov 5 '11 at 13:42
Without the restriction on $k$, this is a special case of Dirichlet's theorem –  Henning Makholm Nov 5 '11 at 14:38

## 2 Answers

In the n=100 to 200 range, I think that 114, 134, 146 and 158 are all counterexamples.

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you are right,I have checked them by Maple...thanks –  pedja Nov 5 '11 at 17:22

Mathematica code:

mymax = 200;

mycount = Range[1, mymax];

Do[{n = 2*k; mya = Range[0, n - 1]; mycount[[k]] = Count[PrimeQ[2^n + 2^(mya + n) - 1], True]}, {k, 1, mymax}];

Count[mycount, 0]

The output is 9. This tell me that there are 9 counterexamples between n = 2 and n = 400. Peter listed four between 100 and 200, and there are 5 more between 200 and 400.

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you are right,I have checked Peter's counterexamples by Maple..thanks –  pedja Nov 5 '11 at 17:49