# Is there always a prime between a prime and prime plus the index of that prime?

Is it known is there always a prime strictly between $p_n$ and $p_n+n$, where $p_n$ is the $n$-th prime number and $n\geq5$?

I know about Bertrands postulate which states that for any integer $n>3$ there is always a prime $p$ such that $n<p<2n-2$.

If we would plug $p_n$ instead of $n$ we would get $p_n<p<2p_n-2$ but since I guess $p_n+n<2p_n-2$ will hold for all but finitely many $n$ we have that this problem of mine is stronger than Bertrands postulate and it seems that it is not implied by it.

So, is this known?

• yes it is (until proven wrong), i use this always to find the next prime number in matlab – Abr001am Jun 5 '16 at 21:20
• @Agawa001: I could just as well say "no it isn't (until proven right)". – TonyK Jun 5 '16 at 21:32

jagy@phobeusjunior:~$p p + n n 2 3 1 p/log p 2.885390081777927 p/log^2 p 4.162737962011215 NONONONONO 3 3 5 2 p/log p 2.730717679880512 p/log^2 p 2.485606349070669 NONONONONO 5 5 8 3 p/log p 3.106674672798059 p/log^2 p 1.930285504520985 7 11 4 p/log p 3.597288396588255 p/log^2 p 1.848640544032643 NONONONONO 11 11 16 5 p/log p 4.587356305666709 p/log^2 p 1.913076170467284 13 19 6 p/log p 5.06832618826664 p/log^2 p 1.975994642359189 17 24 7 p/log p 6.00025410570094 p/log^2 p 2.117826431351823 19 27 8 p/log p 6.452842166007064 p/log^2 p 2.191535369442038 23 32 9 p/log p 7.335366744787423 p/log^2 p 2.339461099153619 29 39 10 p/log p 8.612251926827733 p/log^2 p 2.557616663832689 31 42 11 p/log p 9.027406962818835 p/log^2 p 2.628841176527419 37 49 12 p/log p 10.24670205612772 p/log^2 p 2.837700081812219 41 54 13 p/log p 11.04058283064003 p/log^2 p 2.973035835127401 43 57 14 p/log p 11.43252118401864 p/log^2 p 3.039593967977556 47 62 15 p/log p 12.20732420209676 p/log^2 p 3.170612003725475 53 69 16 p/log p 13.34914438384008 p/log^2 p 3.362257656237909 59 76 17 p/log p 14.46951764677999 p/log^2 p 3.548592219160638 61 79 18 p/log p 14.83869415980414 p/log^2 p 3.609620399478779 67 86 19 p/log p 15.93457740311697 p/log^2 p 3.789712791282478 71 91 20 p/log p 16.65618860623435 p/log^2 p 3.907445336428887 73 94 21 p/log p 17.01449483472118 p/log^2 p 3.965658006585665 79 101 22 p/log p 18.08008761459324 p/log^2 p 4.137842634827439 83 106 23 p/log p 18.7832074896648 p/log^2 p 4.250709440961442 89 113 24 p/log p 19.82784807433869 p/log^2 p 4.417343362461308 97 122 25 p/log p 21.20352530592732 p/log^2 p 4.634943148444331 p p + n n  From Dusart's results, we need check only for$p < 4000.$In that range, we always get$p_{n+1} \leq p_n + n,$with equality only at$p_{n+1} = 3,5,11.$As you can see in the output,$n$is about$p / \log p$and much larger than$p / \log^2 p,$even for fairly small numbers. Indeed, from Rosser and Schoenfeld (1962) we have, for$n \geq 6,$$$p_n > n \log n,$$ but $$p_n < n \log n + n \log \log n$$ yes. First, this is reasonable, as$n \approx \frac{p}{\log p}.$That is, you are asking, more or less, whether we have a prime between$p$and$p + \frac{p}{\log p} $We have the result of Dusart, page 8, theorem 6.8, that there is a prime between$x$and $$x + \frac{x}{25 \log^2 x}$$ as long as$x \geq 396738.$Here we go, Pierre Dusart in his Ph. D. dissertation, for a lower bound gives a milder outcome that is good enough, there is prime between$x$and $$x + \frac{x}{2 \log^2 x}$$ as long as$x \geq 3275.$• But we have$p_n \approx n \log (n)$approximately, how to know that there is not some large$p$that deviates a lot from its approximated value? – Farewell Jun 5 '16 at 22:12 • @Farewell see links to references – Will Jagy Jun 5 '16 at 22:33 • Does it follow from them that we will also have a prime strictly between$p_n$and$p_n + \sqrt{n}$for all sufficiently large$n\$? I know that this is a brand new question, but maybe you know the answer or can point me in the right direction about this one? – Farewell Jun 5 '16 at 22:36