Let integers $\ a>1\ $ and $\ b\in\mathbb Z\ $ be relatively prime (hence $\ b\ne 0).\ $ The Dirichlet's prime distribution theorems apply to the arithmetic sequence

$$ (_aG_b(x) : x\in\mathbb Z) $$ where $$ _aG_b(x)\ :=\ a\cdot x + b\ $$

Thus, for arbitrary integer $\ s>0\ $ that is relatively prime to $\ b,\ $ I'd like to see some prime distribution theorems for the nearly geometric sequence

$$ \left(\left(\bigcirc^n\,_aG_b\right)(s): \ s\in\mathbb Z_{_{\ge0}} \right) $$

where $\ \bigcirc^n\ $ is my notation for the composition power (please, let be, do not "correct it"). Can you prove any? It must be difficult but one may try to prove comparison prime distribution theorems where one compares the distribution of primes of two nearly geometric sequences (even without knowing anything specific about any single sequence like this).


$$\ \gamma\ :=\ \frac b{a-1} $$

Then $$ _aG_b(x)\ =\ a\cdot(x+\gamma)-\gamma $$ and $$ \forall_{n\in\mathbb Z_{_{\ge0}}}\quad \left(\bigcirc^n\,_aG_b\right)(s)\ =\ a^n\cdot(x+\gamma)-\gamma $$

Thus, we may rewrite our nearly geometric series as

$$ \left(a^n\!\cdot\!(x+\gamma)\,-\,\gamma\,: \ \ n\in\mathbb Z_{_{\ge0}}\right) $$ (making it more geometric).


  • $\ \left(\bigcirc^n\,_2G_1\right)(1)\ $ -- the $n$-th Mersenne number;

  • $\ \left(\left(\bigcirc^n\,_2G_1\right)(p): \ p\in\mathbb Z_{_{\ge0}} \right)\ $ -- a potential source for Sophie Germain primes (in particular when $\ p\ :=\ _2G_1(s)\ $ is prime, while natural number $\ s\ $ is not prime)

Of course, one would like to know as much as possible about subsequences of consecutive prime terms of nearly geometric sequences.

  • 2
    $\begingroup$ I'm fairly sure there is no sequence of this shape which is known to contains infinitely many primes. $\endgroup$
    – Wojowu
    Jul 26, 2022 at 9:04
  • $\begingroup$ @Wojowu, the key phrase is "no ... known". $\endgroup$
    – Wlod AA
    Jul 26, 2022 at 9:39

1 Answer 1


Modest start. Let integers $\ a>1\ $ and $\ b\in\mathbb Z\ $ be relatively prime (hence $\ b\ne 0),\ $ and let $$ \gamma\ :=\ \frac b{a-1} $$

Let $\ p\in\mathbb P\ $ be a prime, $\ p\not\!|\,a-1,\ $ and let $\ 0<\gamma_p<p\ $ be such that

$$\ (a-1)\cdot\gamma_p \equiv b \mod p $$ (under proper terminology, you could say $\ \gamma_p\equiv \gamma\mod p.)$

Let $\ a\ $ be a primitive root of prime $\ p,\ $ i.e. let

$$ \{a^k: k=0\,...\,(p-2)\} $$ represent all $\ p-1\ $ non-zero residua $\mod p.$

The only (tiny) chance $ \mod p\quad $ for a nearly geometric long finite subsequence of consecutive terms $\ \left(\bigcirc^n\,_aG_b\right)(x)\ $ that are all prime is condition $\ x\equiv -\gamma_p\mod p.\ $ Then $\ p\not\!|\left(\bigcirc^n\,_aG_b\right)(x)\equiv -\gamma_p\mod p.$

Otherwise, let $$ x\not\equiv -\gamma_p\mod p\quad \Rightarrow $$

The above sign $\ \Rightarrow\ $ tells us and let us remember that the above statement is an assumption (please, do not "correct" my notation).

Then, $\ x+\gamma_p\not\equiv 0\mod p,\ $ hence ($a\ $ being a primitive root) there are natural numbers $\ n_p\ $ such that $$ a^{n_p}\cdot(x+\gamma_p)\ \equiv\ \gamma_p\mod p $$ i.e $$ p\ | \left(\bigcirc^{n_p}\,_aG_b\right)(x) \ \equiv a^{n_p}\cdot(x+\gamma_p)-\gamma_p \mod p $$ We have obtained an entire infinite sequence of terms indexed by exponents $\ n_p,\ $ and these terms are not primes, except possible the first one, because they are divisible by $\ p.$ The indices $\ n_p\ $ form an arithmetic series with the difference between the two consecutive of them is $\ n_q-m_p=p-1.$

We may apply the above approach to more general modules besides the primes.

When the same $\ a\ $ is a primitive root for several primes (or similar) then we may combine them, thus, significantly increase the non-prime terms hence severely thinning out the set of potential prime terms.

(All this opens a plethora of further questions, e.g. related to the primitive roots, etc.)

  • 2
    $\begingroup$ I like the originality of your thinking, it reads really promising. $\endgroup$ Jul 28, 2022 at 7:43

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