# $a_1=k,a_{n}=2a_{n-1}+1(n\geq 2).$ 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.

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
• Thank you! But that is the case $k=1$ which I have excluded. – lsr314 Nov 5 '13 at 15:03
• If you rename your number $k-1$, the next term is $2k-1$, then $4k-1,8k-1$, and so on. That is why those pages talk about $k2^n-1$. The first starting point where they are all composite is 509202, where every number in the sequence is a multiple of either 2,3,5,7,13,17 or 241. – Empy2 Nov 5 '13 at 15:17
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 ${5592405=3\cdot5\cdot7\cdot13\cdot17\cdot241}$).