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Let $S_n$ be a sequence of consecutive $2n-1$ primes starting from $2$.

For example $S_3=(2,3,5,7,11)$.

Denote $p_{2n-1}$ as the ${2n-1} ^{th}$ (last) number in this sequence, then $p_n$ denotes median value in this growing sequence.

In the example above $\ \ p_3=5, p_5=11 $.

For other $n =\{4,5,6,7,8, \dots\}$ we have appropriately $(p_n,p_{2n-1})$: $(7,17),(11,23),(13,31),(17,41),(19,47), \dots$

For bigger numbers, for example $n=200 \ \ (1217,2731)$, $n=400 \ \ (2741,6131)$.

It's clearly visible in these examples that for $n>2: \ \ \ \ \ $ $2p_n<p_{2n-1}$ and the pattern seems to be quite elementary.

  • Is it possible to prove for any $n$ this pattern in a relatively simple way?
Widawensen
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  • @Peter Only brute force can prove this quite simple proposition ? There is no any theory for series of primes? – Widawensen Sep 20 '17 at 12:57
  • What exactly do you mean? For instance, "Let $S_4$ be a series of 4 consecutive 7 primes starting from 2." – Randall Sep 20 '17 at 12:58
  • @Randall O.k $4$ is here not necessary (even misleading), I removed it.. – Widawensen Sep 20 '17 at 13:01
  • @peter Up to $n=400$ it seems to be correct. Is this number big enough to suspect that the claim is correct ? – Widawensen Sep 20 '17 at 13:05
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    Not at all, counterexamples can be much bigger. We at least need a heuristical argument. But be a bit patient, I will continue my search. – Peter Sep 20 '17 at 13:07
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    Numerical calculations gave $\large \frac{p_{2n-1}}{p_n}\approx 2.13$ for some "middle-large" values $n$, so I am not convinced yet. – Peter Sep 20 '17 at 13:14
  • @Peter We can also notice that generally difference $2p_n-p_{2n+1}$ is growing.. what confirms coefficient $2.13$ – Widawensen Sep 20 '17 at 13:24
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    But I get smaller and smaller values and according to Wolfram alpha we can get even below $2.05$ (with $n=10^{12}$) – Peter Sep 20 '17 at 13:26
  • @Peter $2.05$ ? for what number ? Maybe 2.00000...1 can't be crossed .. – Widawensen Sep 20 '17 at 13:27
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    For large $n$, we have $$p_n\approx n(\ln(n)+\ln(ln(n)))$$ – Peter Sep 20 '17 at 13:34
  • Using this approximation, it seems we can get arbitary close to $2$, but maybe we never reach $2$ itself or smaller numbers. The conjecture turns out to be very interesting and probably very difficult to prove or disprove. – Peter Sep 20 '17 at 13:42
  • @Peter Thank you Peter for you expertise, at least situation was checked up to $10^{12}$. Conjecture seems quite simple, only involves median of prime series, I'm a little astonished that even for such case we can't be sure that median is less than half of the last number in series... – Widawensen Sep 20 '17 at 13:48
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    @Peter I'm removing the second question because even the first one seems to be too hard.. – Widawensen Sep 20 '17 at 13:49
  • @Peter I see, thank you Peter for your effort.. – Widawensen Sep 20 '17 at 13:51
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    @Peter Hmm, at least leave upvoted comments.. – Widawensen Sep 20 '17 at 13:53

1 Answers1

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This isn't "relatively simple," but it is a proof:

By results of Pierre Dusart, we have that

$$n\left(\ln n+\ln\ln n-1+\frac{\ln\ln n - 2.1}{\ln n}\right) < p_n < n\left(\ln n+\ln\ln n-1+\frac{\ln\ln n - 2}{\ln n}\right)$$

for large enough $n$ (something in the $600000$s).

So

$$\frac{p_{2n-1}}{p_n} > \frac{2n-1}{n}\frac{\ln (2n-1)+\ln\ln (2n-1)+\frac{\ln\ln (2n-1) - 2.1}{\ln (2n-1)}}{\ln n+\ln\ln n-1+\frac{\ln\ln n - 2}{\ln n}}.$$

Asymptotically, this grows like $2\left(1+\frac{1}{\log_2(n)}\right) > 2$, so for large enough $n$ this is $>2$ (I'm not sure what the threshold is, but it can probably be calculated). From there, it just remains to check "small" cases. Numerically, this should hold for all $n$ where the prime bounds themselves hold.

Carl Schildkraut
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  • Isn't the limit exactly $2$ ? – Peter Sep 20 '17 at 14:03
  • @Peter Yes; the limit is exactly $2$, but it grows like $2(1+1/(\log_2 n))$, so it should grow quickly enough. – Carl Schildkraut Sep 20 '17 at 16:33
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    The right side is a proven lower bound for, lets say , $n\ge 10^6$,right ? This means, to prove the conjecture for $n\ge 10^6$, we "only" have to show that on the right side, there is a function that can never be less than $2$, in particular, it is sufficient to prove that the right side is decreasing. Together with the limit, the proof would be virtually finished. I am just doing brute force, for $3\le n\le 2\cdot 10^5$, the conjecture is true. – Peter Sep 20 '17 at 17:32
  • If the conjecture is actually true, the inequality is actually optimal, we cannot replace $2$ by any larger number. – Peter Sep 20 '17 at 17:38
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    I stopped the brute-force-routine and arrived at $n=4\cdot 10^5$. No counterexample upto this $n$ – Peter Sep 20 '17 at 19:05
  • @Peter At the same time $2p_n<p_{2n-1}$ and $p_{2n-1}<3p_n$ ..? – Widawensen Dec 15 '19 at 14:08