I've grossly overstated things in my two posts before the last one. Thank you for providing references that have returned me to reality.

Conjecture For arbitrary integers $\ 0 \le k \le m\ $ there exists integer $\ n\ge m\ $ such that for every natural number $\ s\ $ at least one of the numbers

$$\ p(x+s+1)-p(x+s)\ \ne\ p(x+1)-p(x) $$

where $\ k\le x < n$.

Here, $\ p(0)=2, p(1)=3,\ldots\ $ is the strictly increasing sequence of all prime numbers.

An assembler-like equivalent formulation:


$$ \forall_{m\in\mathbb Z_{\ge 0}}\,\forall_{k\in 0..m}\, \exists_{n\in\mathbb Z_{\ge m}}\,\forall_{s\in\mathbb N}\, \exists_{x\in k..n\!-\!1} $$ $$ p(x+s+1)-p(x+s)\ \ne\ p(x+1)-p(x) $$

Here (Perl notation),

$$\ u..v\ :=\ \{x\in\mathbb Z:\ u\le x\le v\} $$

  • 1
    $\begingroup$ I don't understand the statement, in particular the part "at least one of the numbers" followed by a formula... I would welcome a clarification, perhaps using less natural language and more quantifiers. $\endgroup$ – Yaakov Baruch Sep 23 at 20:44
  • $\begingroup$ @YaakovBaruch or anybody, if you would like too, I'd provide an Image-Processing/Bourbaki-like terminology for my answer below (it was my first choice but I don't mean to write here purely for my own sake).. $\endgroup$ – Wlod AA Sep 24 at 4:21
  • 1
    $\begingroup$ Thank you! It's not that I'm so much of Bourbaki style fan myself, but simpy I didn't understand that "at least one of the numbers <formula>, for $k\le x <n$" meant "for at least one $x$ such that $k\le x <n$, <formula>"... $\endgroup$ – Yaakov Baruch Sep 24 at 10:43
  • 1
    $\begingroup$ Correct me if I'm wrong. Isn't it true that $\forall_{m\in\mathbb Z_{\ge 0}}\,\forall_{k\in 0..m}\, \exists_{n\in\mathbb Z_{\ge m}} :\forall_{s\in\mathbb N}\, \exists_{x\in k..n-1} : P(x,s)$ is equivalent to $\forall_{a\in \mathbb N}\, \exists_{b(a)\in\mathbb Z_{\ge a}} :\forall_{s\in\mathbb N}\, \exists_{x\in a..b(a)} : P(x,s)$? $\endgroup$ – Yaakov Baruch Sep 24 at 12:08
  • $\begingroup$ @YaakovBaruch, u'r right! ### I had the feeling that the initial segment was not important but, how nice! you took a step back and saw it clearly. Indeed, formally, by setting a=k=m we get your equivalence. (There is often a bit of psychology or inertia left beside the straight logic). $\endgroup$ – Wlod AA Sep 24 at 17:44

Like in the previous post, let $q=p(k)$ and let $n\geq m$ be such that primes $p(x),k\leq x<n$ cover all residue classes mod $q$.

Suppose this $n$ doesn't work. This means that $p(x+s+1)-p(x+s)=p(x+1)-p(x)$ for all $k\leq x<n$. Adding up a bunch of such equalities we get $p(x+s)-p(k+s)=p(x)-p(k)$ for all $k\leq x\leq n$. As $s>0$, we have $p(k+s)>q$ so it is indivisible by $q$. There is some $k<x<n$ such that $p(x)\equiv -p(k+s)\pmod{q}$, and we get $p(x+s)=p(k+s)+p(x)-p(k)\equiv 0\pmod q$, which is impossible as $p(x+s)>q$, hence a contradiction.

| cite | improve this answer | |
  • $\begingroup$ This is so nice! You've applied a simple-sounding but powerful Dirichlet Theorem twice (in the previous thread too), and this time this application is even more surprising since the profile definition seems to describe complicated objects, not linear. (On a personal note, I am happy that my intuition was justified -- I just got carried away far too far in my -2 and -3 note, by now already deleted). $\endgroup$ – Wlod AA Sep 20 at 15:57
  • $\begingroup$ In effect, Wojowu, you have provided a large class of lonely profiles. $\endgroup$ – Wlod AA Sep 20 at 16:19
  • $\begingroup$ @WlodAA This should be equivalent to "inadmissible" tuples, to my understanding, essentially the "exceptions" to the prime tuples conjecture. $\endgroup$ – user44191 Sep 21 at 4:09
  • $\begingroup$ "This" ? (Let's avoid pronouns -- my general principle for the "Art of Communication" and for poetry). $\endgroup$ – Wlod AA Sep 21 at 4:15
  • $\begingroup$ @WlodAA The profiles that cover all residue classes mod $q$. I should note that they may not actually be fully lonely - $q$ might "slot in" at different points. Wiki gives the example of $(0, 2, 8, 14, 26)$ (the "profile" being the consecutive differences), where $3, 5, 11, 17, 29$ and $5, 7, 13, 19, 31$ both fit. $\endgroup$ – user44191 Sep 21 at 4:28


$u_0\ \ldots\ u_n\,\ $ and $\ v_0\ \ldots\ v_n\,\ $ are arbitrary finite strictly increasing sequences of non-negative integers, of the same sequence's length $\ n+1$.

$p(0)=2, p(1)=3, \ldots\ $ is the strictly increasing sequence of all (consecutive) primes.

$ p(u_0)\ \ldots\ p(u_n)\,\ $ and $\,\ p(v_0)\ \ldots\ p(v_n)\ $ are arbitrary finite strictly increasing sequences of primes (not necessarily consecutive), where the length of the two sequences is the same, namely $\ n+1.$

The goal: We will see that under certain additional assumptions, the above two finite sequences are identical; in particular when one of them is assured to exist as an extension of a different shorter sequence.

Wojowu has posted his answer in the Q&A style. He deserves a more rounded presentation. Let me be the first one to do it. It'll be but a presentation. The results are exclusively due to Wojowu (Wojciech Wawrów).

Theorem 1 Assuming our NOTATION, if there exists a prime $\ q\ $ such that for every integer $\ x\ $ there exists at least one integer (index) $\ a(x)\ $ such that $$ p(u_{a(x)})\ \equiv x\ \mod q $$ and if there exists an integer $\ s\in\mathbb Z\ $ such that

$$\forall_{t=0}^n\quad p(v_t) = p(u_{s+t}) $$ and $$ \forall_{t=1}^n\quad p(v_t)-p(v_{t-1})\ =\ p(u_t)-p(u_{t-1}) $$

then $\ s=0$.

Remark The last assumption can be written equivalently as

$$ \forall_{t=1}^n\quad p(v_t)-p(u_t)\ =\ p(v_{t-1})-p(u_{t-1}) $$

Proof (of Theorem 1) Let the assumptions of the theorem hold. By the above Remark, for every integer $\ x\ $ there exists at least one integer (index) $\ b(x)\ $ such that $$ p(v_{b(x)})\ \equiv x\ \mod q $$

The terms in each prime sequence that are $\ 0 \mod q\ $ are simply equal to $\ q.\ $ If integer $\ s\ $ were different from $\ 0\ $ then the same prime $\ q\ $ would appear twice -- $\ s\ $ places apart -- in the increasing sequence of all primes; that would be a contradiction. End of Proof

More NOTATION: Given integers $\ 0\le m\le n,\ $ prime sequence $ p(u_0)\ \ldots\ p(u_n)\,\ $ is called to be the minimal $n$-extension of a strictly increasing sequence $ p(u_0)\ \ldots\ p(u_m)\,\ \Leftarrow:\Rightarrow$ $$ \forall_{k=m+1}^n\quad u_k=u_m+k-m, $$

Theorem 2 For every strictly increasing finite prime sequence $ p(u_0)\ \ldots\ p(u_m)\,\ $ there exists a minimal extension $ p(u_0)\ \ldots\ p(u_n)\,\ $ which is lonely, meaning that it is the only prime sequence $ p(v_0)\ \ldots\ p(v_n)\,\ $ such that there exists an integer $\ s\in\mathbb Z\ $ such that

$$\forall_{t=0}^n\quad p(v_t) = p(u_{s+t}) $$ and $$ \forall_{t=1}^n\quad p(v_t)-p(v_{t-1})\ =\ p(u_t)-p(u_{t-1}). $$

Proof For $n$ large enough all residue classes \mod p(u_0) appear in $ p(u_0)\ \ldots\ p(u_n)\,\ $ (by Dirichlet Theorem). This, by Theorem 1, makes sequence $ p(u_0)\ \ldots\ p(u_n)\,\ $ lonely. End of Proof

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.