First prime of the form $x_i$ for $x_0=658$ and $x_i=1+2x_{i-1}$

Question

Given an initial integer $x_0>0$, one can consider the first prime of the recursive sequence $x_i=1+2x_{i-1}$.

Naïvely such a prime should exist for $x_0$ arbitrary since the sequence $\log(x_i)$ is asymptotically an arithmetic progression. Sometimes it takes however some time: for $x_0=147$ my Maple algorithm stops when hitting a 771-digit number labelled as prime by Maple (Maple does not use a primality proof but some strong primality tests if I am not mistaken).

Starting with $x_0=658$ I lost patience: No prime among the first $56000$ iterates leading to numbers with almost $17000$ digits.

I tried to find an easy reason: If $x\longmapsto 2x+1$ is $k$-periodic for $k$ prime numbers $p_0,\dotsc,p_{k-1}$ such that $x_i\equiv 0\pmod{p_i}$ and $x_i>p_i$ then there is obviously no prime in this sequence. Such an easy argument fails for all small values of $k$ for the sequence $x_0=658,\dotsc$.

Perhaps my patience ran out a bit early and the sequence will hit eventually a prime:

Does an integral sequence given by $x_0>0$, $x_i=1+2x_{i-1}$ necessarily contain a prime number?

Answer 1

The $n$th term of your sequence is $x_{n} = 659 \cdot 2^{n} - 1$. People have long searched for prime values of numbers $k \cdot 2^{n} - 1$ for small $k$, with the goal of proving that $k = 509203$ is the smallest positive integer for which $k \cdot 2^{n} - 1$ is composite for all $n$. (This is called the Riesel problem.)

According to the MathWorld entry on Riesel numbers, in 2004 Dave Linton discovered that $659 \cdot 2^{800516} - 1$ is prime, and that $800516$ is the smallest $n$ so that $659 \cdot 2^{n} - 1$ is prime. A lot of patience would have been required indeed.

Currently, the smallest $k$ for which it is unknown if $k \cdot 2^{n} - 1$ is ever prime is $k = 23669$ and a distributed computing project related to this is run by PrimeGrid. For $k = 23669$ all $n < 13968972$ have been tested.