当 x 足够大的时候,在 x 和 x+x^{12} 之间一定至少存在一个素数

There always exists at least one prime between
x and x+x1/2when x is sufficiently larger
Wan-Dong Xu
School of Science,Tianjin University,Tianjin,300072,China
e-mail:xwandong@
Abstract
In this paper one has shown that there always exists at least one pseudo
prime between x and x+x1/2when x is sufficiently larger for a pseudo
sequence of odd numbers,so it also is true that there always exists at
least one real prime for the real sequence of odd numbers.
Keywords:Prime,distribution of primes
1.Introduction
Generally,one considers that the Desbov conjecture,there always exists at least one prime between x2and(x+1)2for every integer x≥1,could be turn to another description,there always exists at least one prime between
x and x+x1/2for every sufficiently large integer x,with an uncomplicated transformation of a variable.That is a wrong think deeply.It couldn’t be done for the transformation of a variable for among intervals of the sequence
of natural number,owing to that the interval after transformation is the unsame one as before transformation.And the former conjecture and other two correlative ones has been shown in the earlier papers[1-3],the latter is
an independent conjecture so that it also is an important unsolved problem
in number theory for understanding the distribution of primes in sequence of natural number to show whether exist primes in such the sort of intervals.
In this paper one will try to show it with the method of contrasting a sequence of pseudo odd number,which is analogous as that using in the earlier papers.And one will use the same mathematical symbols as those papers.
2.There always exists at least one prime number between x and x+x1/2in{P x}for every integer x≥4
In order to research the number of prime numbers in the intervals between
x and x+x1/2for the sequence{O x}and that of pseudo prime numbers in the same intervals for the sequence{P x},one redivides both sequences{O x} and{P x}into infinitely many of intervals respectively(Fig.1).
Conveniently,one let,in Fig.1,order number n represent x and X[n]denote the interval[x+1, x+x1/2 ],for{O x}and{P x},and let|X[n]|denote the
1
Figure1.A sequence of interval X(n)extending and moving. number offigures in it,or say,the length of this interval.Now one let P X(n) denote the number of pseudo prime numbers in X[n]of{P x},and let O X(n) denote the number of real prime numbers in X[n]of{O x}.
Hence there is
|X[n]|= x1/2 ,
and there has
|X[n+1]|= (x+1)1/2 ,
and owing to that x1/2are strictly monotone increasing functions,and let X is the increasing length of the intervals when n→n+1,one has
X=|X[n+1]|−|X[n]|≥0,
and
|X[n+1]|≥|X[n]|
so that those intervals are monotone increasing sequence as n increases and after moving some intervals the odd number in X[n]would be adding one.
Let|X[n]|P
odd denote the number of the pseudo odd numbers in X[n]of{P x},
and there is|X[n+1]|P
odd −|X[n]|P odd=0or1,i.e.,it is unadding,or adding
one pseudo odd number,in X[n+1],but never subtracting,compared with that in X[n].
Theorem1.There always exists at least one pseudo prime between x and x+x1/2in{P x}for every integer x≥25
Proof.One can show this theorem to be true by the second mathematical induction.
2
At first,one can empirically know,from Fig.2of ref.1,that there is P X (24)=
0for the interval X [24]=[25, 28.9 ]and the Theorem 1is not certainly for all intervals below that one in {P x };For the interval X [25]=[26,30],one knows that there exists one prime
numbers 29by seeing that figure,that is, P X (25)=1,and so on;
For the interval X [97]=[98, 107.9 ],one knows that there appear all the pseudo prime numbers and the pseudo odd composite numbers in {P x }for x ≥98and there exists one pseudo prime numbers at the site 105for X [97]
by seeing that figure,that is, P X (97)=1,and so on.
Because the length of interval is x 1/2 ≥9when n >97,and there exist at least four odd numbers in every interval,and j =3,there are three continued pseudo odd composite numbers for every small region of 3-continued pseudo odd composite-prime group,so there must exist at least one pseudo prime number in every interval when 97≤n ≤169.
And the length is x 1/2 ≥13when n ≥169,and there exist at least six odd numbers in every interval,and j =4,there are four continued pseudo odd composite numbers for every small region of 4-continued pseudo odd composite-prime group,so there must exist at least one pseudo prime number in every interval when 169≤n ≤329,and so on.
By the principle of the second mathematical induction one now assumes that there exists at least one pseudo prime number in a interval X [k ]of {P x },that is,there is P X (k )≥1,when k ≥98;
Then,it can be got that there must exist at least one pseudo prime number
in an interval X [k +1]of {P x },that is,there is P X (k +1)≥1,for all k ,here
k =98,99,···.
The reason for this is owing to that |X [k +1]|≥|X [k ]|and the pseudo primes are uniformly distribution in the interval of j -continued pseudo odd composite-prime groups,so that if there exist at least one pseudo prime in X [k ]there must exist at least one pseudo prime in X [k +1]while both are in an identical j -continued group.
On the other hand,if a interval X [k ]is,by coincidence,in the interval at the end region of the j -continued group and the X [k +1]is in the begin region of the (j +1)-continued group,due to there increases one pseudo odd composite in every small interval of later group,and this couldn’t be offset
when |X [k +1]|P odd is not changed compared with |X [k ]|P odd .
At this case,one should consider these intervals in other way.Because the increase of j is very slowly,it is yet adding 1after j 2repeatings of (j +1)
figures,but the |X [k +i ]|P odd is adding 1(Note:It will be appeared when √k +i − √k =2)after overgrowing some of intervals,X [k ] X [k +i ],
here i =1,2,···,and the symbol ‘ ’means X [k ]turn to X [k +i ]through X [k +1],X [k +2],···,so the increase of |X [k ]|P odd is largely fast than that of the former.And one can conclude that there must exist at least one pseudo prime in X [k +1]if there exists one pseudo prime in X [k ].3
From the Table1,one can see that P X(n)is very slowly and oscillatingly increased as n increases,and in that table p P r denote the sites at where appear some pseudo prime numbers.
One has now completed to show that P X(n)≥1for all n,here n=25,26,···,and that the Theorem1to be true.
Table1.The numbers of P X(n)in{P x}while n<5400.
n[x+1, x+x1/2 ]p P r P X(n)
98[99, 107.9 ]1051
103[104, 113.1 ]105,1132
105[106, 115.2 ]1131
111[112, 121.5 ]113,1212
113[114, 123.6 ]1211
119[120, 129.9 ]121,1292
134[135, 145.6 ]137,1452
137[138, 148.7 ]1451
258[259, 274.1 ]259,2692
259[260, 275.1 ]2691
1060[1061, 1092.5 ]1063,1077,10913
1257[1258, 1293.5 ]1261,12772
2057[2058, 2102.3 ]2061,20792
4057[4058, 4120.7 ]4069,4089,41093
5257[5258, 5329.5 ]5261,5283,5305,53274
3.There always exists at least one real prime between x and
x+x1/2in{O x}for every integer x≥116
Theorem2.There always exists at least one real prime between x and x+x1/2
in{O x}for every integer x≥116
Proof.Analogous as the proof of the Theorem3of ref.1,one also will show this theorem to be true in three cases.
Case one:There exists an continued odd composite group of thefirstly appeared local maximum in X[n].
Because there exist many of continued odd composite groups of thefirstly appeared local maximum in{O x}as in Fig.1of ref.1,so it is possible that these groups would be in many of X[n].This case is crucial for proving the Theorem2.Now,one can empirically list some data of L(s)in L[s]and of
| X[n] |O in X[n]while n(s)2<1.5×106in Table2(Note:The values of x could be obtained by solving a irrational equation x+x1/2=n(s)2for every pair of x and n(s)2with the method of iteration,and yet every n could be get.)
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Table 2.The value of L (s )and |[X [n ]|O while n (s )2≤1.5×106.
s [n (s )1,n (s )2]L (s )n |X [n ]|O L (s )|X [n ]|O
1[25,27]32350.6
2[91,95]58690.5555
3[115,125]1111610 1.1
4[525,539]15517220.6818
5[889,905]17876290.5862
6[1131,1149]191116330.5757
7[1329,1359]311323360.8611
8[12855,12887]33127741130.2920
9[15685,15725]41156011240.3306
10[19611,19659]49195201390.3525
11[31399,31467]69312911760.3920
12[155923,156005]831556113940.2107
13[360655,360747]933601476000.155
14[370263,370371]1093697636080.1793
15[492115,492225]1114915247010.1583
16[1349535,1349649]115134848811610.0990
17[1357203,1357331]129135616711640.1108
All of ones know that according to the well-known Sieve of Eratosthenes
[4-5],and from the Table 2,and as the same reason was narrated in Case one of ref.1,it is obviously that the relatively ability of primes ≤√x +x 1/2on continuously exact division to odd numbers in X [n ]of {O x },C (s )|X [n ]|O odd ,i.e.,L (s )|X [n ]|O ,is rapidly and oscillatingly decreased as √x +x 1/2increases that there is C (s )|X [n ]|O odd <0.7when n is not sufficiently larger except the interval X [116],and there is C (s )|X [n ]|O odd 0.5when n is sufficiently larger.Then,one has X [n ]L [s ]=|X [n ]|−L (s )≥3for every pair of n and s ,here n ≥517and s ≥4,and there will be some of rest primes remaining below L [s ]but in X [n ]1.And one can now conclude that there always exists at least one prime in X [n ]when n ≥517,and s ≥4,i.e., O X (n )≥1,in which there is a continued odd composite group of the firstly appeared local maximum,such that it is certainly that the Theorem 2is true in this case.1Note:For the pair n =23and s =1there is X [23]L [1]=2<3,and for the pair n =116and s =3there is X [116]L [3]=−1<3,so that there isn’t any prime in those two intervals.5
Case two:There exist some of the continued odd composite groups of local maximum of lower order in X[n].
It is possible that there are some of the continued odd composite groups of local maximum of lower order in X[n+i],i=1,2,···.As the number of continued odd composites in those groups is slightly less than that in the continued odd composite groups of thefirstly appeared local maximum,and interval X[n]is monotone increasing sequence as n increases,the|X[n+i]|
will be adding1while n→n+i,thus,there also has|X[n+i]|O
odd ≥|X[n]|O odd,
and there always exists at least one prime in X[n]of{O x},in which there is the continued odd composite groups of thefirstly appeared local maximum,so that there must always exist at least one prime in X[n+i]of{O x},in which there are some of the continued odd composite groups of local maximum of lower order,such that it also is certainly that the Theorem2is true in this case.
Case three:There exists neither the continued odd composite groups of thefirstly appeared local maximum nor some of the continued odd composite groups of local maximum of lower order in X[n].
Except of two cases mentioned above,the distribution of primes in{O x}is approximate uniformly decreased as x increases(Fig.1in ref.1).It is possible that there appear infinitely many of intervals,X[n],in{O x}falling in this case.According to the Euclid prime theorem,there must exist infinitely many
of prime numbers in interval(x2,∞).And by the formula(10)of ref.1,one know that the number of primes in{O x}is largely larger than that of pseudo primes in{P x}when x is sufficiently larger,for example,when x+1≥10000,
or say,when n≥9999.
By the Theorem1,there always exists at least one pseudo prime in X[n]of {P x}for every n,here n≥25.As a result of appearing the Case one and the Case two in{O x},there would be a little primes in some intervals in which there appear a continued odd composite group of thefirstly appeared local maximum or some that of local maximum of lower order,then there will be some more primes in all intervals where would have neither the former nor the latter.Because there are P X(n)≥1for n≥25in{P x}andπO(x)
πP(x)(formula(10)in ref.1),of course,one can now conclude that there is
O X(n)≥ P X(n)and there must always exist at least one prime in X[n]of {O x},i.e., O X(n)≥1,when n≥10000.
For the intervals of n≤9999,one can empirically investigate everyone
of X[n].From the Table3it is certainly that the Theorem2is true when 117≤n≤9999.
From that table,one can also see that the number of prime number, O X(n),
in X[n]of{O x}is slowly and oscillatingly increased as n increases.
Up to now,one has completed the proof of the Theorem2.
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Table3.The distribution of primes in X[n]while n≤9999 n[x+1, x+x1/2 ]|X[n]|p r O X(n)
112[113, 122.5 ]101131
113[114, 123.6 ]10none0
114[115, 124.6 ]10none0
115[116, 125.7 ]10none0
116[117, 126.8 ]10none0
117[118, 127.8 ]101271
118[119, 128.8 ]101271
119[120, 129.9 ]101271
120[121, 130.9 ]101271
121[122, 132.1 ]11127,1312
125[126, 136.2 ]11127,1312
126[127, 137.2 ]11127,131,1373
127[128, 138.3 ]11131,1372
129[130, 140.4 ]11131,137,1393
130[131, 141.4 ]11131,137,1393
875[876, 904.6 ]29877,881,883,8874
1115[1116, 1148.4 ]331117,1123,1293
1322[1323, 1358.4 ]3613271
9995[9996, 10094.9 ]9910007,···,1009310
9999[10000, 10098.9 ]9910007,···,1009310 REFERENCE
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x1/2log x,Waited for published.
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7。