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In analytic number theory, Dirichlet theorem says that all of the sequence(exactly progress) $p_n=an+b$ of which $a$ and $b$ are relatively prime for each other contains infinitely many primes . For example, there are infinitely many primes of the form $4n+1$ as explicit formula. (This theorem does not mean that all of the terms for sequence are primes)

So then I though the sequence that is given as recurrence relation, like that $p_{n+1}$=$2p_n+1$. If a initial term of sequence is $p_1=2$, then it would be listed as $2 ,5 , 11 ,23 , 47 ,95 , 191 ... $

Now, here are two questions in general case. For given positive integers $a$, $b$, let's consider the progress(or sequence) $ap_n+b=p_{n+1}$ as recurrence relation

$1.$ How many primes are there in this sequence under some condition ?

$2.$ What is a condition?

$3.$ Is there any sequence {p_n} that contains only primes?

I did try to solve the previous two questions with Dirichlet theorem. If a sequence is given as recurrence relation has an explicit formula, then Dirichlet theorem would solve it. And we can consider some condition.

The 3rd question, If yes, both $a$ and $b$ are even or odd in the same time for $ap_n+b=p_{n+1}$ If both $a$ and $b$ are even, {$p_n$} would contain at least 2 even number. If both $a$ and $b$ are odd, then there would be term $p_n$ as even for $n \geq2$ These are contradictions that $p_n$ is odd for $n \geq2$

Next, How should i consider? Any opinions?

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  • $\begingroup$ The term you were trying to use is, probably, progression. $\endgroup$ – DonAntonio Oct 24 '16 at 0:53
  • $\begingroup$ Your question is boring $\endgroup$ – Planche Oct 31 '16 at 2:17
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Well for one thing, a sequence satisfying $x_n = ax_{n-1} + b$ have closed forms of the form $c_1a^n + c_2$ for some constants $c_1$ and $c_2$. For example, your sequence $2,5,11, \dots$ has closed form $3\cdot2^n-1$.

One thing that is clear from this is that modulo any prime the sequence is periodic, in particular this implies the answer to question 3 is no (just pick some prime dividing some entry of the sequence and look modulo that).

Now there are some obvious obstructions to a sequence of this form taking a prime value: If $c_2$ has a common factor with $c_1$ or $a$, or if $c_1 = -c_2 = 1$ then $a^n-1$ factors as a polynomial in $a$ for all $n$, and it will never (other than maybe a few small values of $n$) be prime.

Now if we throw out sequences with these obvious obstructions it is expected (based on probabilistic hueristics) that for general values of $c_1, c_2$ and $a$ there should be infinitely many prime entries of such a sequence. However to my knowledge there is no specific such sequence that has been proven to exhibit this generic behavior.

Pinning down this "generic" behavior tricky business though, for example the your sequence $3\cdot 2^n-1$ looks like it's probably "generic" but the sequence given by $2^n+1$ almost certainly has only finitely many primes (such primes are called Fermat primes, and proving there are only finitely many is a famous open problem).

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