# Help on an application of Dirichlet's theorem for primes in progression

Suppose that I have an infinite sequence of positive integers

$$a_1,\ldots,a_m,\ldots$$

with the following recursion

$$a_{m+1} -a_m =b(m+1)$$

So that

$$a_{m+1} =b(m+1) +a_m$$

Suppose further that I know that

$$(b,a_m)=1$$

Would Dirichlet's Theorem tell us that there are infinitely many primes of the form $a_{m+1}$

• Is there a typo? The two displayed equations are not equivalent. – André Nicolas Jul 15 '16 at 2:15

Take $b=1$. Then, (with $a_1=1$), $a_{m+1}=1+2+\cdots +(m+1)$. This is non-prime for $m\gt 1$, since it is $\frac{(m+1)(m+2)}{2}$.
• For general $b$ and $a_1$, we will have that $a_{m+1}$ is a quadratic in $m$. With such quadratics, if there is no obvious reason for the quadratic not to represent infinitely many primes, it probably does. There are even conjectured estimates for the number of such primes $\le n$, which for a number of polynomials that have been examined fit the experimental facts reasonably well. However, proof is another matter! – André Nicolas Jul 15 '16 at 2:52
• How about a modification. If we assume $a_n$ and $b_n$ are strictly monotonically increasing sequences of positive odd integers. Suppose I know there is an even positive number $c$ such that $$b_n-a_n =cn$$ so that $$b_n = cn+a_n$$ Now suppose I also know that $$(c,a_n)=1$$ I want to say that Dirichlet has shown that $b_n$ can assume infinitely many prime numbers. If that is not true then I need help understanding how that subscript $_n$ is throwing off Dirichlet's theorem. – Antonio Hernandez Maquivar Jul 20 '16 at 14:41
• @AnthonyHernandez: One can generate various counterexamples. Or else informally with Dirichlet the $a_0$ is fixed, and even if $cn+a_0$ would like to avoid primes, it turns out that it cannot. However, with an "arbitrary" $a_n$, the AP we are sampling from changes with $n$. – André Nicolas Jul 20 '16 at 15:45
• @AnthonyHernandez: That is one way to think of what the issue is. So we cannot apply Dirichlet's Theorem directly. Counterexamples show that "infinitely many primes" does not always hold. But with suitable additional restrictions on the $b_n$, it very well could be the case that "infinitely many $a_n$ are prime" is forced. The problem is (i) what are reasonable conditions and (ii) can one prove that these conditions suffice. – André Nicolas Jul 20 '16 at 16:40