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As it is well known that prime number is $2,3,5\cdots \cdots$, thus all these prime number are denoted by$p_{1},p_{2},\cdots \cdots ,p_{n}\cdots \cdots$. The prime maximal gap $\max_{p_{n+1}\leqslant N}(p_{n+1}-p_{n})$ means the maximum value of $ (p_{2}-p_{1},p_{3}-p_{2},\cdots \cdots ,p_{n+1}-p_{n})$. In 1937, Cramér gave a conjecture about the prime maximal gaps that $$\lim_{n\rightarrow \infty }sup\frac{p_{n+1}-p_{n}}{(logp_{n})^{2}}=1$$which is still an unproven conjecture.

I found a conjecture about the prime maximal gaps that $$\max_{p_{n+1}\leqslant N }(p_{n+1}-p_{n})\approx logN(logN-2loglogN)+2$$ when $N\geqslant 7$. My conjecture gives an approximate value of the prime maximal gap ,which is close to the actual value.

question: Has anyone a clue how to prove or disprove the above conjecture?


\begin{matrix} A& B & C & D & E & F & G\\\ 1&2&1&——& ——& ——& ——\\\ 2 & 3 & 2 &—— & —— & —— & ——\\\ 3 &7 &4 & 3 & 0.75 & 4 & 1.00\\\ 4 & 23 & 6 & 5 & 0.83 & 10 & 1.67\\\ 5& 89& 8& 9& 1.13& 20& 2.50\\\ 6& 113& 14& 10& 0.71& 22& 1.57\\\ 7& 523& 18& 18& 1.00& 39& 2.17\\\ 8& 887& 20& 22& 1.10& 46& 2.30\\\ 9& 1129& 22& 24& 1.09& 49& 2.23\\\ 10& 1327& 34& 25& 0.74& 52& 1.53\\\ 11& 9551& 36& 45& 1.25& 84& 2.33\\\ 12& 15683& 44& 51& 1.16& 93& 2.11\\\ 13& 19609& 52& 54& 1.04& 98& 1.88\\\ 14& 31397& 72& 61& 0.85& 107& 1.49\\\ 15& 155921& 86& 86& 1.00& 143& 1.66\\\ 16& 360653& 96& 100& 1.04& 164& 1.71\\\ 17& 370261& 112& 101& 0.90& 164& 1.46\\\ 18& 492113& 114& 106& 0.93& 172& 1.51\\\ 19& 1349533& 118& 127& 1.08& 199& 1.69\\\ 20& 1357201& 132& 127& 0.96& 199& 1.51\\\ 21& 2010733& 148& 135& 0.91& 211& 1.43\\\ 22& 4652353& 154& 154& 1.00& 236& 1.53\\\ 23& 17051707& 180& 186& 1.03& 277& 1.54\\\ 24& 20831323& 210& 191& 0.91& 284& 1.35\\\ 25& 47326693& 220& 213& 0.97& 312& 1.42\\\ 26& 122164747& 222& 240& 1.08& 347& 1.56\\\ 27& 189695659& 234& 253& 1.08& 363& 1.55\\\ 28& 191912783& 248& 253& 1.02& 364& 1.47\\\ 29& 387096133& 250& 275& 1.10& 391& 1.56\\\ 30& 436273009& 282& 279& 0.99& 396& 1.40\\\ 31& 1294268491 &288& 314& 1.09& 440& 1.53\\\ 32& 1453168141& 292& 318& 1.09& 445& 1.52\\\ 33& 2300942549& 320& 334& 1.04& 465& 1.45\\\ 34& 3842610773 &336& 352& 1.05& 487& 1.45\\\ 35& 4302407359& 354& 357& 1.01& 492& 1.39\\\ 36& 10726904659& 382& 390& 1.02& 533& 1.40\\\ 37& 20678048297& 384& 416& 1.08& 564& 1.47\\\ 38& 22367084959& 394& 419& 1.06& 568& 1.44\\\ 39& 25056082087& 456& 423& 0.93& 573& 1.26\\\ 40& 42652618343& 464& 445& 0.96& 599& 1.29\\\ 41& 127976334671& 468& 490& 1.05& 654& 1.40\\\ 42& 182226896239& 474& 505& 1.07& 672& 1.42\\\ 43& 241160624143& 486& 518& 1.07& 687& 1.41\\\ 44& 297501075799& 490& 527& 1.08& 698& 1.42\\\ 45& 303371455241& 500& 528& 1.06& 699& 1.40\\\ 46& 304599508537& 514& 528& 1.03& 699& 1.36\\\ 47& 416608695821& 516& 542& 1.05& 716& 1.39\\\ 48& 461690510011& 532& 547& 1.03& 721& 1.36\\\ 49& 614487453523& 534& 560& 1.05& 737& 1.38\\\ 50& 738832927927& 540& 568& 1.05& 747& 1.38\\\ 51& 1346294310749& 582& 596& 1.02& 780& 1.34\\\ 52& 1408695493609& 588& 598& 1.02& 783& 1.33\\\ 53& 1968188556461& 602& 614& 1.02& 801& 1.33\\\ 54& 2614941710599& 652& 628& 0.96& 818& 1.25\\\ 55& 7177162611713& 674& 678& 1.01& 876& 1.30\\\ 56& 13829048559701& 716& 711& 0.99& 916& 1.28\\\ 57& 19581334192423& 766& 729& 0.95& 937& 1.22\\\ 58& 42842283925351& 778& 771& 0.99& 985& 1.27\\\ 59& 90874329411493& 804& 812& 1.01& 1033& 1.28\\\ 60& 171231342420521& 806& 847& 1.05& 1074& 1.33\\\ 61& 218209405436543& 906& 861& 0.95& 1090& 1.20\\\ 62& 1189459969825483& 916& 961& 1.05& 1205& 1.32\\\ 63& 1686994940955803& 924& 982& 1.06& 1229& 1.33\\\ 64& 1693182318746371& 1132& 982& 0.87& 1230& 1.09\\\ 65& 43841547845541059& 1184& 1191& 1.01& 1468& 1.24\\\ 66& 55350776431903243& 1198& 1207& 1.01& 1486& 1.24\\\ 67& 80873624627234849& 1220& 1233& 1.01& 1516& 1.24\\\ 68& 203986478517455989& 1224& 1297& 1.06& 1589& 1.30\\\ 69& 218034721194214273& 1248& 1301& 1.04& 1594& 1.28\\\ 70& 305405826521087869& 1272& 1325& 1.04& 1621& 1.27\\\ 71& 352521223451364323& 1328& 1336& 1.01& 1632& 1.23\\\ 72& 401429925999153707& 1356& 1345& 0.99& 1643& 1.21\\\ 73& 418032645936712127& 1370& 1348& 0.98& 1646& 1.20\\\ 74& 804212830686677669& 1442& 1395& 0.97& 1700& 1.18\\\ 75& 1425172824437699411& 1476& 1437& 0.97& 1747& 1.18 \end{matrix} A:Serial numbe, B:Natural number, C:$\max_{p_{n+1}\leqslant N}(p_{n+1}-p_{n})$, D:$logN(logN-2loglogN)+2$, E:$\frac{logN(logN-2loglogN)+2}{\max_{p_{n+1}\leqslant N}(p_{n+1}-p_{n})}$, F:$ (logN)^{2}$, G:$\frac{(logN)^{2}}{\max_{p_{n+1}\leqslant N}(p_{n+1}-p_{n})}$

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closed as not a real question by Steven Landsburg, Andy Putman, Felipe Voloch, Yemon Choi, Zev Chonoles Jun 20 '13 at 20:56

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

1  
I can't understand your table. Is B=N? What type of answer do you seek? If it seems to hold for all really large N, then yes, otherwise, clearly no. Mathoverflow is not a place to "publish" conjectures, but to seek advice on problems. A better question would be "has anyone seen a similar estimate before?" or, "This inequality seems to hold, is there any reason why this should be true?" –  Per Alexandersson Jun 20 '13 at 19:27
1  
How can we possibly tell if your conjecture is a good one or a bad one without any description of the heuristics that led you to it? And where is the mathematical question here? –  Steven Landsburg Jun 20 '13 at 19:28
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This doesn't seem like a real question, just a random speculation backed up by essentially no evidence. I've voted to close. –  Andy Putman Jun 20 '13 at 19:36
1  
What does ≈ mean? –  The User Jun 20 '13 at 19:42
2  
What does "almost equal to" mean? –  Steven Landsburg Jun 21 '13 at 15:58

1 Answer 1

This conjecture seems, at least from what you presented, not like a convincing conjecture. While it seems in line with the conjecture often attributed to Cramér that $$\limsup_n \frac{p_{n+1} - p_n}{(\log p_n)^2} = 1 $$ and one might thus consider it as some sort of refiniement thereof, there are reasons to believe that in fact $$\limsup_n \frac{p_{n+1} - p_n}{(\log p_n)^2} > 1$$ in contrast to your conjecture. Specifically, Granville pointed out that a certain natural refinement of Cramér's reasoning suggest a lim sup of (at least) $2 e^{-\gamma}$, where $\gamma$ the Euler-Mascheroni constant, which yields about $1.12$. (However, note that, contrary to some accounts, Granville did not present a conjecture.)

Since you do not present any supporting evidence for the plausibility of your conjecture beyond numerics, which are well-known no to go far enough for convincing predictions in this context, and it is in contradiction to arguments based on plausible heuristics, I see not reason to consider it a convincing conjecture.

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What Granville didn't do was conjecture that the lim sup was $2e^{-\gamma}$. He did conjecture that the lim sup was $\ge2e^{-\gamma}$, though. –  Charles Jun 20 '13 at 19:44
    
"Then we are led to conjecture that there are infinitely many primes $p_n$ with $p_{n+1}-p_n>2e^{-\gamma}\log^2p_n,$ contradicting Cramér's conjecture!" (Unexpected irregularities in the distribution of prime numbers, p. 7 in the PDF) –  Charles Jun 20 '13 at 19:53
    
@Charles: I am not sure. The reason why I am extra careful not to attribute a conjecture to Granville is that recently he pointed out on MO specifically that he did not conjecture a value (contrary to what I claimed) mathoverflow.net/questions/114399/… Now, you say still something else, however from what he said following the link my impression was that there is no conjecture at all, but it isn't very clear either. Yet then the phrase just before the one you quote is 'only' that something is plausible... –  quid Jun 20 '13 at 20:25
    
...so one might say that only under this plausible assumption one is lead to conjecture this. Yet the assumption is only plausible and not something conjectured to be true. Anyway, my point to avoid attributing a conjecture was that I thought (perhaps based on a misunderstanding to the extent) that he specifically does not want this to be done. But you see, if the situation is as you say he might have said something like 'I only conjectured inequality not equality' and not what he said. –  quid Jun 20 '13 at 20:29
    
@quid,@Charles: $\max_{p_{n+1}\leqslant N }(p_{n+1}-p_{n})\approx logN(logN-2loglogN)+2$ do not mean that $$\lim_{n\rightarrow \infty }\frac{\max_{p_{n+1}\leqslant N}(p_{n+1}-p_{n})}{ logN(logN-loglogN)+2}=1$$. Please think it carefully. –  Wenlong DU Jun 21 '13 at 12:28

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