On a Smarandache problem concerning the prime gaps

Transcription

1 O a Smaradache problem cocerig the prime gaps Felice Russo Via A. Ifate Avezzao (Aq) Italy felice.russo@katamail.com Abstract I this paper, a problem posed i [] by Smaradache cocerig the prime gaps is aalyzed. Let's p be the -th prime umber ad d the followig ratio: d p p + = where If we idicate with becomes: g = p + p the gap betwee two cosecutive primes, the previous equatio d = g I [], Smaradache posed the followig questios:. Does the sequece d cotai ifiite primes?. Aalyze the distributio of d First of all let's observe that d is a ratioal umber oly for =, beig p =, p = 3. For >, istead, the ratio is always a atural umber sice the gap of prime umbers g is a eve umber []. Moreover let's observe that the gap g ca be as large as we wat. I fact let's be ay iteger greater tha oe ad let's cosider the followig sequece of cosecutive itegers:! +,! + 3,! + 4,...,! + Notice that divides the first, 3 divides the secod,..., divides the -st, showig all of these umbers are composite. So if p is the largest prime smaller tha!+ we have g >. This proves our assertio. Now let s check the first terms of sequece d : d p

2 Here p is the smallest prime relative to the gap d. As we ca see, for the first 0 terms of sequece d we have 4 primes regardless if those are repeated or ot. O the cotrary, if we cosider oly how may distict primes we have the this umber is. So, the Smaradache questio ca be split i two sub-questios:. How may times the sequece d takes a prime value?. How may distict primes the sequece d cotais? Provig both the questios is a very difficult task. Ayway, we ca try to uderstad the behaviour of sequece d by usig a computer search ad the get a heuristic argumet o the umber of primes withi it. Thaks to a Ubasic code, the coutig fuctios p ( N ) ad p ( have bee calculated for N up 9 to 0. p ( N ) deotes how may times d takes a prime value for N while p ( N ) deotes the umber of distict primes i d, always for N. I table, the results of the computer search ca be foud. I the third colum, the umber of distict primes are reported whereas i the secod oe the umber of all primes regardless of the repetitios are show. N # primes # distict primes Table. Number of primes i d for differet N values Let s aalyze the data of colum. It is very easy to verify that those data grow liearly with N, that is: c( N () A estimatio of c( ca be obtaied usig the followig asymptotic relatioship give i [3]: h N ( d c ) l N ( p d, p> p e p d l( N ) where hn ( d ) / N is the frequecy of d for N ad p ay prime umber. The costat c is the twi prime costat defied i the followig way: c =.3003 ( ) p > p...

3 By defiitio of p ( ) fuctio we have: N dmax c N d l ( ) = N p d, p> p e p d l( N ) = () where the above summatio is exteded o all prime values of d up to d max. But the largest gap d for a give N ca be approximated by [],[3]: max d max l ( ad the () ca be rewritte as: where the fuctio : l ( N ) d = d l( N ) N p ( e (3) l ( J ( d ) = p d, p> p p has small values of order ad the has bee replaced by its mea value Sice, as N goes to ifiity, the summatio: c [3]. l ( N ) d = N e d l( N ) is the umber of primes i the rage to l ( ), we ca write: l ( N ) d = e d l( N ) π ( l ( ) where π ( is the coutig fuctio of prime umbers []. Usig the Gauss approximatio [] for it, we have: 3

4 l ( N ) d = e d l( N ) l ( l( l ( N )) ad the: by usig () ad (3), that implies: c( N N l l ( c ( l l ( Accordig to those experimetal data the followig cojecture ca be posed: Cojecture A: The sequece d takes ifiite times a prime value. Let s ow aalyze the data reported i table, colum 3. By usig the least square method, we ca clearly see that the best fit is obtaied usig a logarithmic fuctio like: N N ) c( l( ) (4) where c( ca be estimated usig the followig approximatio: π (0.5 l ( ) beig p ( N ) the umber of primes i the rage to max Therefore: d. l ( c( l( l(0.5 l ( ) l( c( l(0.5 l ( ) I table, the compariso of (4) with calculated values p ( N ) show i table (colum 3) is reported. Notice the good agreemet betwee p ( N ) ad its estimatio as N icrease. Accordig to those data, also p ( N ) like p ( goes to the ifiity as N icrease, although p ( N ) more slowly the p ( ). The this secod cojecture ca be posed: N Cojecture B: The sequece d cotais a ifiite umber of distict primes 4

5 N P ( ) ratio Table. Compariso of p ( N ) with the approximated formula c( l( N ). I the third colum the ratio / c( N ) l( ) N Let's aalyze ow the distributio of d, as always requested by Smaradache. Thaks to a Ubasic code the frequecy of prime gaps up to N= have bee calculated. The plot of those frequecies versus d for > is reported i Fig. It shows a clear jigsaw patter superimposed oto a expoetial decay. The jigsaw patter is due to a double populatio that is clearly visible i the two plots of fig. The frequecy of d for beig a multiple of 3 ( or equivaletly for multiple of 6 for g ) is always larger tha adjacetes differeces. Prime gap distributio for N<= Frequecy (%) d=g/ Fig.: Plot of prime gap distributio i the explored rage. 5

6 4 Prime gap distributio for N<= Frequecy % d=g/ Not multiples of 3 Multiples of 3 Frequecy % Prime gap distributio for N<= d=g/ Not m ultiples of 3 M ultiples of 3 Fig. Prime gap distributio. The secod plot uses a logarithmic scale for the Y-axis. Accordig to the cojecture reported i [3] ad already metioed above, the umber of pairs p, p N with d + < p p + = is give by: h N ( d c ) l N ( p d, p> p e p d l( N ) Let's p ( p) = p f where p is ay prime umber greater tha. As it ca be see i fig 3. this fuctio approaches quickly, with the maximum value at p=3. 6

7 f(p) f(p) versus p for p> p Fig.3: Plot of fuctio f(p) versus p Beig f(p) maximum for p=3 meas that h N ( d ) has a relative maximum every time d has 3 as prime factor, that is whe d is a multiple of 3. This explais the double populatio see i the Fig ad the the jigsaw patter of the fig. I fig. 4, the distributio of d obtaied by computer search ad the oe estimated with the use of h d ) formula is reported. Notice the very good agreemet betwee them. N ( Prime gap distributio for N<= Frequecy (%) d=g/ Experimet h_n(d) Fig 4: Prime gap distributio compariso. The good agreemet betwee the experimetal ad the estimated data has to be oticed. 7

8 Refereces: [] F. Smaradache, Oly problems ot solutios, Xiqua, 99. [] E. Weisstei, CRC Cocise Ecyclopedia of Mathematics, CRC Press, 999. [3] M. Wolf, Some cojectures o the gaps betwee cosecutive primes, preprit at 8