# Prime numbers of the form: $k\cdot 2^n \pm 1$ , where $k<3n$

Is it true that :

For every $n$ there exists a number $k<3n$ such that:

$k\cdot 2^n-1$ or $k\cdot 2^n+1$ is prime,where $k,n\in \mathbf{N}$

Maple code that prints least $k$ such that $k\cdot 2^n-1$ is prime and $k<3n$ :

for $n=624$ we have exception,there is no prime of the form $k\cdot 2^{624}-1$ such that $k<3\cdot 624$ , but number $85\cdot 2^{624}+1$ is prime number and $85<3\cdot 624$

Any idea how to prove or disprove statement above ?

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Such numbers are called Sierpinski and Riesel numbers, respectively: mathworld.wolfram.com/SierpinskiNumberoftheSecondKind.html, mathworld.wolfram.com/RieselNumber.html – JavaMan Nov 9 '11 at 17:16
@JavaMan,What do you think,is there any other way to disprove this statement besides finding an counterexample ? – pedja Nov 9 '11 at 17:33
I have no intuition regarding your question. – JavaMan Nov 9 '11 at 17:42