Can someone give me a counterexample to disprove this statement?

Question

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.

Answer 1

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

Answer 2

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.