$a_1=k,a_{n}=2a_{n-1}+1(n\geq 2).$ Does there exist $k\in\mathbb N$ such that $a_n,n=1,2,3,\cdots$ are all composite numbers? Let $a_1=k,a_{n}=2a_{n-1}+1(n\geq 2).$
If $k=1$ then $a_n=1,3,7,15,31,63,\cdots$ here $3,7,31$ are prime numbers. I'm interested in this problem:

Does there exist $k\in\mathbb N$ such that $a_n,n=1,2,3,\cdots$ are all composite numbers?

If $k=147$ then $a_n,n=1,2,\cdots 2551$ are all composite, but $a_{2552}$ is prime. So I doubt the existence of such number.
 A: The numbers you mention are Riesel numbers http://en.wikipedia.org/wiki/Riesel_number
and there is the similar Sierpinski numbers where it is $2a_{n-1}-1$ instead.
http://en.wikipedia.org/wiki/Sierpinski_number
A: I'll just add the proof that for $a_1=1018405=2\cdot509203-1=2\cdot c-1$ all $a_n$ are composite. 
First note that 
$a_1=2\cdot c-1, \ a_2=2a_1+1=2^2\cdot c-1$
and inductively 
$a_n=2a_{n-1}+1=2\cdot(2^{n-1}\cdot c-1)+1=2^n\cdot c-1$ for all $n\geq 1$.

Now if $p$ is a prime such that $a_m\equiv a_{m+l}\equiv0\pmod p$ with $m\in\mathbb N$ minimum then it's not so hard to see that $a_{n}\equiv0\pmod p$ for all $n\equiv m\pmod l$.

By a direct computation we see that
$$
a_2\equiv a_4\equiv 0\pmod 3\\
a_1\equiv a_5\equiv 0\pmod 5\\
a_2\equiv a_5\equiv 0\pmod 7\\
a_7\equiv a_{19}\equiv 0\pmod {13}\\
a_7\equiv a_{15}\equiv 0\pmod {17}\\
a_3\equiv a_{27}\equiv 0\pmod {241}.
$$
Since the solution of the system of the consequences
$$
n\equiv 0\pmod 2\\
n\equiv 1\pmod 4\\
n\equiv 2\pmod 3\\
n\equiv 7\pmod {12}\\
n\equiv 7\pmod 8\\
n\equiv 3\pmod {24}
$$
is $n\in\mathbb N$ it follows that $a_n\equiv0\pmod {5592405}$ for all $n\in\mathbb N$(note that ${5592405=3\cdot5\cdot7\cdot13\cdot17\cdot241}$).
