<?xml version="1.0" encoding="UTF-8"?><!DOCTYPE article  PUBLIC "-//NLM//DTD Journal Publishing DTD v3.0 20080202//EN" "http://dtd.nlm.nih.gov/publishing/3.0/journalpublishing3.dtd"><article xmlns:mml="http://www.w3.org/1998/Math/MathML" xmlns:xlink="http://www.w3.org/1999/xlink" dtd-version="3.0" xml:lang="en" article-type="research article"><front><journal-meta><journal-id journal-id-type="publisher-id">APM</journal-id><journal-title-group><journal-title>Advances in Pure Mathematics</journal-title></journal-title-group><issn pub-type="epub">2160-0368</issn><publisher><publisher-name>Scientific Research Publishing</publisher-name></publisher></journal-meta><article-meta><article-id pub-id-type="doi">10.4236/apm.2022.123018</article-id><article-id pub-id-type="publisher-id">APM-116294</article-id><article-categories><subj-group subj-group-type="heading"><subject>Articles</subject></subj-group><subj-group subj-group-type="Discipline-v2"><subject>Physics&amp;Mathematics</subject></subj-group></article-categories><title-group><article-title>
 
 
  The Computing Formula of Number of Primes No More than Any Given Positive Integer
 
</article-title></title-group><contrib-group><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Maoze</surname><given-names>Wang</given-names></name><xref ref-type="aff" rid="aff1"><sup>1</sup></xref><xref ref-type="corresp" rid="cor1"><sup>*</sup></xref></contrib><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Zhenxiang</surname><given-names>He</given-names></name><xref ref-type="aff" rid="aff2"><sup>2</sup></xref></contrib><contrib contrib-type="author" xlink:type="simple"><name name-style="western"><surname>Meiyi</surname><given-names>Wang</given-names></name><xref ref-type="aff" rid="aff3"><sup>3</sup></xref></contrib></contrib-group><aff id="aff3"><addr-line>Beijing Forestry University, Beijing, China</addr-line></aff><aff id="aff2"><addr-line>School of Cyberspace Security, Gansu Politics and Law College, Lanzhou, China</addr-line></aff><aff id="aff1"><addr-line>Lanzhou Institute of Technology, Lanzhou, China</addr-line></aff><pub-date pub-type="epub"><day>07</day><month>03</month><year>2022</year></pub-date><volume>12</volume><issue>03</issue><fpage>229</fpage><lpage>247</lpage><history><date date-type="received"><day>19,</day>	<month>December</month>	<year>2021</year></date><date date-type="rev-recd"><day>28,</day>	<month>March</month>	<year>2022</year>	</date><date date-type="accepted"><day>31,</day>	<month>March</month>	<year>2022</year></date></history><permissions><copyright-statement>&#169; Copyright  2014 by authors and Scientific Research Publishing Inc. </copyright-statement><copyright-year>2014</copyright-year><license><license-p>This work is licensed under the Creative Commons Attribution International License (CC BY). http://creativecommons.org/licenses/by/4.0/</license-p></license></permissions><abstract><p>
 
 
  In this paper, we give out the formula of number of primes no more than any given 
  <em>n</em> (
  <em>n </em>
  ∈ 
  <em>Z</em>
  <sup>+</sup>, 
  <em>n </em>&gt; 2). At the same time, we also show the principle, derivation process of the formula and application examples, it is usually marked with 
  <em>π</em>(
  <em>n</em>), which is:
  <inline-formula><inline-graphic xlink:href="dit_1c9aa3d5-4229-4a28-b9d1-5f2373c1b9e5.png" xlink:type="simple"/></inline-formula> that is:
  <inline-formula><inline-graphic xlink:href="dit_05cb9c72-b682-4093-937c-20476a9c8e40.png" xlink:type="simple"/></inline-formula>
   
  <inline-formula><inline-graphic xlink:href="dit_c5248c32-55c0-4e31-8089-71374630421c.png" xlink:type="simple"/></inline-formula> where “[ ]” denotes taking integer. 
  <em>r</em> = 1,2,3,4,5,6; 
  <em>s</em>
  <sub><em>x</em></sub> = 
  <em>s</em>
  <sub>1</sub>,
  <em>s</em>
  <sub>2</sub>,...,
  <em>s<sub>j</sub></em>,
  <em>s<sub>h</sub></em>; 
  <em style="white-space:normal;">s</em>
  <sub style="white-space:normal;">1</sub>
  ,
  <em style="white-space:normal;">s</em>
  <sub style="white-space:normal;">2</sub>
  ,...,
  <em style="white-space:normal;">s<sub>j</sub></em>
  ,,<em>s<sub>h </sub></em>
  <sub>= 0,1,2,3,....</sub>As 
  <em>i </em>≥ 2, 
  2 ≤ <em>s<sub>x </sub></em>≤ <em>i</em>-1 (
  <em>x</em>=1,2,...,
  <em>j</em>,
  <em>h</em>).
 
</p></abstract><kwd-group><kwd>Positive Integer Numbers Spectrum</kwd><kwd> Row</kwd><kwd> Column</kwd><kwd> Composition</kwd><kwd> Prime</kwd></kwd-group></article-meta></front><body><sec id="s1"><title>1. Introduction</title><p>In the long history of mathematics, it has always been believed that the distribution of prime numbers is irregular, which is sometimes more or less in positive integers. So is there really no law in the distribution of prime numbers [<xref ref-type="bibr" rid="scirp.116294-ref1">1</xref>] ?</p><p>For any given positive integer n, how many primes are there less than n? Until the 18th century, it was still not known [<xref ref-type="bibr" rid="scirp.116294-ref2">2</xref>]. This is one of the important and interesting essential problems, mathematicians have been puzzled since the Euclid era in 300 BC. It has been explored continuously, some mathematicians have only found some approximate formulas.</p><p>In the late eighteenth century, some mathematicians examined tables of prime numbers created by using hand calculations. With these values, they looked for functions that estimated π(n). In 1798, French mathematician Adrien-Marie Legendre used tables of primes up to 400,031, computed by Jurij Vega, to note that π(n) could be approximated by the function:</p><p>π ( n ) ≈ n ln n − 1.08366 [<xref ref-type="bibr" rid="scirp.116294-ref3">3</xref>].</p><p>The great German mathematician Karl Friedrich Gauss conjectured that π(n) increases at the same rate as the functions:</p><p>π ( n ) = n ln n and Li ( n ) ≈ ∫ 2 n d t ln t .</p><p>(where ∫ 2 x d t ln t represents the area under the curve y = 1/lnn and above the t-axis from t = 2 to t = x. Li is an abbreviation of logarithmic integral) [<xref ref-type="bibr" rid="scirp.116294-ref3">3</xref>].</p><p>Despite their efforts to explore, mathematicians finally failed to find an accurate formula for the calculation of prime numbers. However, Legendre, a famous mathematician at the time, wrongly asserted rational expression of π(n) did not exist!</p><p>The discovery of the positive integer spectrum finally solves this problem. It is a powerful tool to solute prime problems, which reflects the inner law of the distribution of prime numbers, the π(n) formula was derived from that.</p><p>For given 100, 000, 000, 000, 000, 000, 000, someone has calculated out the number of primes less than that is 2,220,819,602,560,918,840 by using the computer. If you program the same problem according to our given π(n) formula, it would save much time.</p><p>It is a great discovery to use continuous quantity to express discrete quantity, this determines the extraordinary theoretical significance of the formula. By using the π(n) formula, we can verify some proven theorems and conjectures, such as prime theorem, Bertrand conjecture. We might prove some unproved conjectures, such as Brocard conjecture, Crame conjecture, Jeboff conjecture (prime interval A conjecture) and Oberman conjecture (prime interval B conjecture), which can also clarify many other problems about prime distribution that was not understood before. For example, whether there is at least one</p><p>prime number between the successive triangle numbers n ( n + 1 ) 2 . It will become</p><p>a basic tool for anyone to study and solve the problems of the prime number and prime distribution. It may also be an important tool to solve Goldbach conjecture and twin prime conjecture.</p><p>At present, many scholars are studying the distribution law and number calculation of prime numbers [<xref ref-type="bibr" rid="scirp.116294-ref4">4</xref>] - [<xref ref-type="bibr" rid="scirp.116294-ref9">9</xref>]. Everyone uses different principles and tools, and the applicable situations of the formula may be different [<xref ref-type="bibr" rid="scirp.116294-ref10">10</xref>] [<xref ref-type="bibr" rid="scirp.116294-ref11">11</xref>]. In comparison, the π(n) formula we given is the most convenient computing formula that only depends on n without any other conditions.</p><p>In this paper, we take the positive integer spectrum as the basic tool and deduce the formula based on the properties.</p><p>The positive integer number spectrum is the special arrangement with infinite rows and six columns by all positive integers put in order of natural numbers.</p><p>The expression of the row, column and elements of the positive integer number spectrum.</p><p>The positive integer number expresses that any number X can be only expressed into the type of 6 ( n − 1 ) + r , ( n ∈ Z + , n ≠ 1 ; r = 1 , 2 , 3 , 4 , 5 , 6 ), n is row ordinal number, r is a column ordinal number.</p></sec><sec id="s2"><title>2. Main Conclusions</title><p>The whole process contains 3 parts. Part 1 is the positive integer number spectrum and its properties. Part 2 is the formula π(n). Part 3 is the examples of the π(n) formula application.</p><p>Part 1</p><p>In this part we introduce the positive integer number spectrum and its some properties firstly in the following.</p><sec id="s2_1"><title>2.1. The Positive Integer Number Spectrum and Its Some Properties</title><sec id="s2_1_1"><title>2.1.1. The Positive Integer Number Spectrum</title><p>1) Definition: The positive number spectrum is the special arrangement with infinite rows and six columns in all positive integers put in order of integer numbers.</p><p>2) The expression of columns, line and elements of the positive integer number spectrum</p><p>The positive integer number spectrum indicates that any number X can be only expressed into the type of 6 ( n − 1 ) + r , ( n ∈ Z + , n ≠ 1 ; r = 1 , 2 , 3 , 4 , 5 , 6 ), n is row ordinal number, r is column ordinal number.</p><p>A number of the r-th ( r = 1 , 2 , 3 , 4 , 5 , 6 ) is denoted with q r , the set of all numbers of the r-th column is denoted with Q r .</p></sec><sec id="s2_1_2"><title>2.1.2. Multiplication Law and Some Properties of the Positive Integer Number Spectrum</title><p>1) Multiplication law of the positive integer number spectrum</p><p>We can easily prove the positive integer number spectrum satisfying the following laws according to it’s formula of general term 6 ( n − 1 ) + r .</p><p>Multiplication law:</p><p>q 1 ⋅ q 1 ∈ Q 1 , q 1 ⋅ q 2 ∈ Q 2 , q 1 ⋅ q 3 ∈ Q 3 , q 1 ⋅ q 4 ∈ Q 4 , q 1 ⋅ q 5 ∈ Q 5 , q 1 ⋅ q 6 ∈ Q 6</p><p>q 2 ⋅ q 2 ∈ Q 4 , q 2 ⋅ q 3 ∈ Q 6 , q 2 ⋅ q 4 ∈ Q 2 , q 2 ⋅ q 5 ∈ Q 4 , q 2 ⋅ q 6 ∈ Q 6</p><p>q 3 ⋅ q 3 ∈ Q 3 , q 3 ⋅ q 4 ∈ Q 6 , q 3 ⋅ q 5 ∈ Q 3 , q 3 ⋅ q 6 ∈ Q 6</p><p>q 4 ⋅ q 4 ∈ Q 4 , q 4 ⋅ q 5 ∈ Q 2 , q 4 ⋅ q 6 ∈ Q 6</p><p>q 5 ⋅ q 5 ∈ Q 1 , q 5 ⋅ q 6 ∈ Q 6</p><p>q 6 ⋅ q 6 ∈ Q 6</p><p>2) Some properties of the positive integer number spectrum</p><p>Property 1: Any composition of the first column can be written into</p><p>q 1 ⋅ q 1 or q 5 ⋅ q 5 .</p><p>Proof: Denoting the composition of the first column with H<sub>1</sub>, then:</p><p>H 1 = q 1 ⋅ q 1 or H 1 = q 5 ⋅ q 5 .</p><p>Because the composition 6 ( m − 1 ) + 1 can and only can be discomposed into [ 6 ( n − 1 ) + 1 ] [ 6 ( s − 1 ) + 1 ] ( n ≤ s &lt; m ) or [ 6 ( n − 1 ) + 5 ] [ 6 ( s − 1 ) + 5 ] ( n ≤ s &lt; m ); thereby it can not be discomposed into [ 6 ( n − 1 ) + a ] [ 6 ( s − 1 ) + b ] ( a , b = 2 , 3 , 4 , 6 ).</p><p>Therefore, H 1 = q 1 ⋅ q 1 or H 1 = q 5 ⋅ q 5 .</p><p>Property 2: Any composition of the fifth column can be written into p<sub>1</sub>p<sub>5</sub>.</p><p>Proof: Denoting the composition of the first column is H<sub>5</sub>, then H 5 = q 1 ⋅ q 5 .</p><p>Because the composition 6 ( m − 1 ) + 5 can and only can be discomposed into [ 6 ( n − 1 ) + 1 ] [ 6 ( s − 1 ) + 5 ] ( n ≤ s &lt; m ); thereby it can not be discomposed into [ 6 ( n − 1 ) + a ] [ 6 ( s − 1 ) + b ] ( a , b = 2 , 3 , 4 , 6 ).</p><p>Therefore, H 5 = q 1 ⋅ q 5 .</p><p>Part 2</p><p>In this part we introduce the π(n) formula.</p></sec></sec><sec id="s2_2"><title>2.2. The π(n) Formula and Its Computing Theorem and Proof</title><sec id="s2_2_1"><title>2.2.1. The Principle of Deducing π(n) and the Computing Theorem of the π(n) Formula</title><p>1) The principle of deducing π(n)</p><p>π(n) is equal to the sum of the number of primes of the first column and the number of primes of the fifth column and 2.</p><p>The number of primes of the first column is denoted with m<sub>1</sub>, the number of primes of the fifth column is denoted with m<sub>5</sub>. m<sub>1</sub> is equal to the number of the first column which is less than n abstracting the number of composite number and abstracting 1. m<sub>5</sub> is equal to the number of the fifth column which is less than n abstract the number of composite number.</p><p>2) The computing theorem of the π(n) formula</p><p>Number of primes no more than any given positive integer n is the sum of three parts, which are prime numbers of first column and fifth column, and the number 2 of number prime 2 and 3.</p></sec><sec id="s2_2_2"><title>2.2.2. The π(n) Formula and Its Conditions</title><p>1) π ( n ) = 2 [ n − 1 6 ] + [ r + 1 6 ] + 2 ( 2 &lt; n &lt; 25 );</p><p>2) π ( n ) = 2 [ n − 1 6 ] + [ r + 1 6 ] + 2                     + ∑ i = 1 [ log 5 n ] − 1 ( − 1 ) i ⋅ ∑ ∑ s x = i ( [ n 5 s 1 ⋅ 7 s 2 ⋯ ( 6 k − 1 ) s j ⋅ ( 6 k + 1 ) s h − R 6 ]                     + [ n 5 s 1 ⋅ 7 s 2 ⋯ ( 6 k − 1 ) s j ⋅ ( 6 k + 1 ) s h − R ′ 6 ] )     ( n ≥ 25 ) . .</p><p>where “[ ]” denotes taking integer.</p><p>r = 1 , 2 , 3 , 4 , 5 , 6 ; s x = s 1 , s 2 , ⋯ , s j , s h ; s 1 , s 2 , ⋯ , s j , s h = 0 , 1 , 2 , 3 , ⋯</p><p>As i ≥ 2 , 2 ≤ s x ≤ i − 1 ( x = 1 , 2 , ⋯ , j , h ).</p><p>2) The conditions of the π(n) formula</p><p>As i is determined, k takes n and n 5 s 1 ⋅ 7 s 2 ⋯ ( 6 k − 1 ) s j ⋅ ( 6 k + 1 ) s h − R ≥ 6 .</p><p>The regulations of R , R ′ in the following:</p><p>The base number of the power of the denominator of the fraction</p><p>n 5 s 1 ⋅ 7 s 2 ⋯ ( 6 k − 1 ) s j ⋅ ( 6 k + 1 ) s h are possible to take a part or all of the following</p><p>numbers 5 , 7 , ⋯ , ( 6 k − 1 ) , ( 6 k + 1 ) , setting the minimum of the part or the whole obtained numbers is p.</p><p>As p is the number of the fifth column, R = P − 6 , R ′ = P − 4 ;</p><p>As p is the number of the first column, R = P − 6 , R ′ = P − 2 .</p></sec><sec id="s2_2_3"><title>2.2.3. The Reasoning Process of the π(n) Formula</title><p>1) As 2 &lt; n &lt; 25, the expression of π(n) formula is:</p><p>π ( n ) = 2 [ n − 1 6 ] + [ r + 1 6 ] + 2.</p><p>2) As n ≥ 25, the expression of π(n) formula makes up of three parts:</p><p>π ( n ) = 2 [ n − 1 6 ] + [ r + 1 6 ] + 2       + ∑ i = 1 [ log 5 n ] − 1 ( − 1 ) i ⋅ ∑ ∑ s x = i ( [ n 5 s 1 ⋅ 7 s 2 ⋯ ( 6 k − 1 ) s j ⋅ ( 6 k + 1 ) s h − R 6 ]       + [ n 5 s 1 ⋅ 7 s 2 ⋯ ( 6 k − 1 ) s j ⋅ ( 6 k + 1 ) s h − R ′ 6 ] )</p><p>The first one is 2 [ ( n − 1 ) / 6 ] + [ ( r + 1 ) / 6 ] , we set its value to Z, the number of first column is n<sub>1</sub>, the number of fifth column is n<sub>5</sub>, then Z = n 1 + n 5 − 1 .</p><p>The second one is the later expression of sum, setting the value of that is H, H is equal to the number of the compositions of first column and the fifth column.</p><p>Set n = 6 ( k − 1 ) + r , where k is the row, r is column.</p><p>Z = 2 [ ( n − 1 ) / 6 ] + [ ( r + 1 ) / 6 ] = 2 [ ( 6 ( k − 1 ) + r − 1 ) / 6 ] + [ ( r + 1 ) / 6 ] = 2 [ ( k − 1 ) + ( r − 1 ) / 6 ] + [ ( r + 1 ) / 6 ] = 2 ( k − 1 ) + 2 [ ( r − 1 ) / 6 ] + [ ( r + 1 ) / 6 ]</p><p>1) As r = 1 , 2 , 3 , 4</p><p>[ r − 1 6 ] = 0 , [ r + 1 6 ] = 0 , then Z = 2 ( k − 1 ) .</p><p>Because the row is k, the numbers of the first column number is k in all, except1, the plus is k − 1, therefore the numbers of the first column number and the fifth column number which is less or equal to n except 1 is equal to ( k − 1 ) + ( k − 1 ) = 2 ( k − 1 ) , so the value of Z is right.</p><p>2) As r = 5 , 6</p><p>[ ( r − 1 ) / 6 ] = 0 , [ ( r + 1 ) / 6 ] = 1 , then Z = 2 ( k − 1 ) + 1 = 2 k − 1 .</p><p>Because the row is k, the numbers of the first column number is k in all, except 1, the plus is k − 1, the numbers of the fifth column number is k − 1 in all, therefore the numbers of the first column number and the fifth column number which is less or equal to n except 1 is equal to ( k − 1 ) + k = 2 k − 1 , so the value of Z is right.</p><p>Let us deduce the value of H in the following.</p><p>The expansion of H is:</p><p>H = − [ n 5 + 1 6 ] − [ n 5 − 1 6 ] − [ n 7 − 1 6 ] − [ n 7 − 5 6 ] − ⋯ + [ n 5 &#215; 7 + 1 6 ] + [ n 5 &#215; 7 − 1 6 ] + [ n 5 &#215; 11 − 1 6 ] + [ n 5 &#215; 11 − 1 6 ] + ⋯ + [ n 7 &#215; 11 − 1 6 ] + [ n 7 &#215; 11 − 5 6 ] + [ n 7 &#215; 13 − 1 6 ] + [ n 7 &#215; 11 − 5 6 ] + ⋯ + ⋯ − [ n 5 &#215; 5 &#215; 7 + 1 6 ] − [ n 5 &#215; 5 &#215; 7 − 1 6 ] − [ n 5 &#215; 5 &#215; 11 + 1 6 ] − [ n 5 &#215; 5 &#215; 11 − 1 6 ] − ⋯ − ⋯</p><p>+ [ n 5 &#215; 5 &#215; 5 &#215; 7 + 1 6 ] + [ n 5 &#215; 5 &#215; 5 &#215; 7 − 1 6 ] + [ n 5 &#215; 5 &#215; 5 &#215; 11 + 1 6 ] + [ n 5 &#215; 5 &#215; 5 &#215; 11 − 1 6 ] + ⋯ + ⋯ − ⋯ ⋯ + ( − 1 ) ( [ log 5 n ] − 1 ) [ n 5 s 1 ⋅ 7 + 1 6 ] + ( − 1 ) ( [ log 5 n ] − 1 ) [ n 5 s 1 ⋅ 7 − 1 6 ] + ⋯ − ⋯ + ( − 1 ) ( [ log 5 n ] − 1 ) [ n ⋯ ( 6 k − 1 ) s j ⋅ ( 6 k + 1 ) s h + R 6 ] + ( − 1 ) ( [ log 5 n ] − 1 ) [ n ⋯ ( 6 k − 1 ) s j ⋅ ( 6 k + 1 ) s h + R ′ 6 ] + ⋯</p><p>1) As i = 1, (set p ∈ P 5 ),</p><p>[ n p − ( p − 6 ) 6 ] + [ n p − ( p − 4 ) 6 ] expresses the number of compositions of the first column and fifth column that is less than or equal to n and contain p and more than or equal to p<sup>2</sup>.</p><p>Let [ n p ] = t , the t numbers that are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15,</p><p>16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, …, include these numbers 5, 11, 17, 23, …, the method of calculating the numbers is, every six numbers is cut into a segment from former to later with startingq, the last number or numbers from 1</p><p>to q as a last segment, then there are [ [ n p ] − ( p − 1 ) + 5 6 ] = [ n p − ( p − 6 ) 6 ] segment in all.</p><p>Therefore, the first number of every segment is the number of the fifth column, so the numbers of the composition of the fifth column which contain q 1 , q 2 , q 3 , ⋯ , q n and is less than or equal to n and is more than or equal to p<sup>2</sup> is:</p><p>[ n p − ( p − 6 ) 6 ] .</p><p>We can calculate the number of compositions of the first column and the fifth column, which is less or equal to n and more than or equal to p<sup>2</sup> and contain the factor p is:</p><p>[ [ n p ] − ( p + 1 ) + 5 6 ] = [ n p − ( p − 4 ) 6 ] in all.</p><p>Because of composition:</p><p>p ⋅ t ′ ( t ′ ∈ { 1 , 2 , ⋯ , t } )     factor     t ′ ≥ p ,     so   the   composition     p ⋅ t ′ ≥ p 2 .</p><p>Because for the factor t, We can prove the following conclusion with same method that [ n p − ( p − 6 ) 6 ] + [ n p − ( p − 2 ) 6 ] expresses the number of compositions</p><p>of the first column and fifth column that is less than or equal to n and contain p and more than or equal to p<sup>2</sup>, where p is the number of the first column, and p ≠ 1 .</p><p>2) As i ≥ 2 ,</p><p>{ p 1 , p 2 , ⋯ , p j , p h } ⊆ { 5 , 7 , 11 , 13 , ⋯ , ( 6 k − 1 ) , ( 6 k + 1 ) }</p><p>and:</p><p>p 1 &lt; p 2 &lt; ⋯ &lt; p j &lt; p h .</p><p>Set p<sub>1</sub> is the number of the fifth column number, then:</p><p>[ n p 1 s 1 ⋅ p 2 s 2 ⋯ p j s j ⋅ p h s h − ( p 1 − 6 ) 6 ] + [ n p 1 s 1 ⋅ p 2 s 2 ⋯ p j s j ⋅ p h s h − ( p 1 − 4 ) 6 ]</p><p>expresses the number of compositions that is less than or equal to n and contain p 1 , p 2 , ⋯ , p j , p h and more than or equal to p 1 s 1 + 1 ⋅ p 2 s 2 ⋯ p j s j ⋅ p h s h , where p<sub>1</sub> is the number of the first column.</p><p>Set [ n p 1 s 1 ⋅ p 2 s 2 ⋯ p j s j ⋅ p h s h ] = t , thet numbers that are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10,</p><p>11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, …, include these numbers 5, 11, 17, 23, …, the method of calculating the numbers is: every six numbers is cut into a segment from former to later with starting q, the last number or numbers from 1 to q as a last segment, then there are:</p><p>[ t − ( p − 1 ) + 5 6 ] = [ n p 1 s 1 ⋅ p 2 s 2 ⋯ p j s j ⋅ p h s h − ( p 1 − 6 ) 6 ] segments in all.</p><p>Therefore, the first number of every segment is the number of the fifth column, so the numbers of the composition of the fifth column which contain q 1 , q 2 , q 3 , ⋯ , q n , and is less than or equal to n and is more than or equal to p 1 s 1 + 1 ⋅ p 2 s 2 ⋯ p j s j ⋅ p h s h is:</p><p>[ n p 1 s 1 ⋅ p 2 s 2 ⋯ p j s j ⋅ p h s h − ( p 1 − 6 ) 6 ] in all.</p><p>We can prove the following conclusion with the same method as that:</p><p>[ n p 1 s 1 ⋅ p 2 s 2 ⋯ p j s j ⋅ p h s h − ( p 1 − 6 ) 6 ] + [ n p 1 s 1 ⋅ p 2 s 2 ⋯ p j s j ⋅ p h s h − ( p 1 − 2 ) 6 ]</p><p>Express the number of compositions that is less than or equal to n and contain p 1 , p 2 , ⋯ , p j , p h and more than or equal to p 1 s 1 + 1 ⋅ p 2 s 2 ⋯ p j s j ⋅ p h s h ,where p<sub>1</sub> is the number of the first column.</p><p>In the following we prove: the composition of the first column and the fifth column p 1 s 1 + 1 ⋅ p 2 s 2 ⋯ p j s j ⋅ p h s h which is less or equal to n is unique, that is to calculate one time, where</p><p>{ p 1 , p 2 , ⋯ , p j , p h } ⊆ { 5 , 7 , 11 , 13 , ⋯ , ( 6 k − 1 ) , ( 6 k + 1 ) }</p><p>and</p><p>p 1 &lt; p 2 &lt; ⋯ &lt; p j &lt; p h , s 1 , s 2 , ⋯ , s j , s h ∈ N .</p><p>Because of 1 ≤ i ≤ [ log 5 n ] − 1 , and as ∑ s x = i ( x = 1 , 2 , ⋯ , j , h ), any composition of the first column and fifth column p 1 s 1 ⋅ p 2 s 2 ⋯ p j s j ⋅ p h s h which satisfy the condition ∑ s x = i and which is less than or equal to n calculated, s 1 , s 2 , ⋯ , s j , s h ∈ N .</p><p>And because p 1 &lt; p 2 &lt; ⋯ &lt; p j &lt; p h . as it satisfy ∑ s x = i and the values of s 1 , s 2 , ⋯ , s j , s h are determined, the composition p 1 s 1 ⋅ p 2 s 2 ⋯ p j s j ⋅ p h s h is unique.</p><p>We prove the maximum of i is [ log 5 n ] − 1 .</p><p>Because n ≥ 5 s 1 ⋅ 7 s 2 ⋯ ( 6 k − 1 ) s j ⋅ ( 6 k + 1 ) s h &gt; 5 s 1 ⋅ 5 s 2 ⋯ 5 s j ⋅ 5 s h = 5 ∑ s x ,</p><p>Therefore the maximum of i is [ log 5 n ] − 1 .</p><p>We prove the code of every term is ( − 1 ) i in the following:</p><p>Because the composition with three factors can be regarded as the product of two factors, at the same cause, the composition with i factors can be regarded as the product of i − 1 factors. Therefore, we should take the composite of three factors into account when calculating a composite number containing two factors, and we should plus the number of composition with three factors. At the same cause, we should take the number of compositions with i factors into account when calculating the number of compositions with i − 1 factors, and we should plus the number of composition with i factors.</p><p>Therefore, asi is an odd, the code is negative, as i is an even, the code is positive, we can write them into the union form ( − 1 ) i , so the code of every terms is ( − 1 ) i .</p></sec><sec id="s2_2_4"><title>2.2.4. The Last Conclusion</title><p>π ( n ) = { 2 [ n − 1 6 ] + [ r + 1 6 ] + 2         ( 2 ≤ n &lt; 25 ) ; 2 [ n − 1 6 ] + [ r + 1 6 ] + 2 + ∑ i = 1 [ log 5 n ] − 1 ( − 1 ) i ⋅ ∑ ∑ s x = i ( [ n 5 s 1 ⋅ 7 s 2 ⋯ ( 6 k − 1 ) s j ⋅ ( 6 k + 1 ) s h − R 6 ] + [ n 5 s 1 ⋅ 7 s 2 ⋯ ( 6 k − 1 ) s j ⋅ ( 6 k + 1 ) s h − R ′ 6 ] )           ( n ≥ 25 ) .</p><p>that is:</p><p>1) π ( n ) = 2 [ n − 1 6 ] + [ r + 1 6 ] + 2 ( 2 ≤ n &lt; 25 );</p><p>2) π ( n ) = 2 [ n − 1 6 ] + [ r + 1 6 ] + 2                     + ∑ i = 1 [ log 5 n ] − 1 ( − 1 ) i ⋅ ∑ ∑ s x = i ( [ n 5 s 1 ⋅ 7 s 2 ⋯ ( 6 k − 1 ) s j ⋅ ( 6 k + 1 ) s h − R 6 ]                     + [ n 5 s 1 ⋅ 7 s 2 ⋯ ( 6 k − 1 ) s j ⋅ ( 6 k + 1 ) s h − R ′ 6 ] )     ( n ≥ 25 ) .</p><p>where “[ ]” denotes taking integer.</p><p>r = 1 , 2 , 3 , 4 , 5 , 6 ;</p><p>s x = s 1 , s 2 , ⋯ , s j , s h ;</p><p>s 1 , s 2 , ⋯ , s j , s h = 0 , 1 , 2 , 3 , ⋯</p><p>As i ≥ 2 , 2 ≤ s x ≤ i − 1 ( x = 1 , 2 , ⋯ , j , h ).</p><p>As i is determined, k take n and:</p><p>n 5 s 1 ⋅ 7 s 2 ⋯ ( 6 k − 1 ) s j ⋅ ( 6 k + 1 ) s h − R ≥ 6 .</p><p>The regulations of R , R ′ in the following:</p><p>The base number of the power of the denominator of the fraction n 5 s 1 ⋅ 7 s 2 ⋯ ( 6 k − 1 ) s j ⋅ ( 6 k + 1 ) s h are possible take a part or all of the following numbers 5 , 7 , ⋯ , ( 6 k − 1 ) , ( 6 k + 1 ) , setting the minimum of the part or whole obtained numbers is p.</p><p>As p is the number of the fifth column, R = P − 6 , R ′ = P − 4 ;</p><p>As p is the number of the first column, R = P − 6 , R ′ = P − 2 .</p><p>Part 3</p><p>In this part we introduce some examples of the π(n) formula application.</p></sec></sec><sec id="s2_3"><title>2.3. Examples of Applying the π(n) Formula Calculation</title><sec id="s2_3_1"><title>2.3.1. Example No. 1</title><p>π ( 100 ) = 2 [ ( 100 − 1 ) / 6 ] + 2 + [ ( 4 + 1 ) / 6 ]   − [ ( 100 / 5 + 1 ) / 6 ] − [ ( 100 / 5 − 1 ) / 6 ]   − [ ( 100 / 7 − 1 ) / 6 ] − [ ( 100 / 7 − 5 ) / 6 ] = 34 − 9 = 25</p></sec><sec id="s2_3_2"><title>2.3.2. Example No. 2</title><p>π ( 200 ) = 2 [ ( 200 − 1 ) / 6 ] + 2 + [ ( 2 + 1 ) / 6 ]   − [ ( 200 / 5 + 1 ) / 6 ] − [ ( 200 / 5 − 1 ) / 6 ]   − [ ( 200 / 7 − 1 ) / 6 ] − [ ( 200 / 7 − 5 ) / 6 ]   − [ ( 200 / 11 − 5 ) / 6 ] − [ ( 200 / 11 − 7 ) / 6 ]</p><p>  − [ ( 200 / 13 − 7 ) / 6 ] − [ ( 200 / 13 − 11 ) / 6 ]   + [ ( 200 / 5 &#215; 7 + 1 ) / 6 ] + [ ( 200 / 5 &#215; 7 − 1 ) / 6 ] = 68 − 23 + 1 = 46</p></sec><sec id="s2_3_3"><title>2.3.3. Example No. 3</title><p>π ( 400 ) = 2 [ ( 400 − 1 ) / 6 ] + 2 + [ ( 4 + 1 ) / 6 ]   − [ ( 400 / 5 + 1 ) / 6 ] − [ ( 400 / 5 − 1 ) / 6 ]   − [ ( 400 / 7 − 1 ) / 6 ] − [ ( 400 / 7 − 5 ) / 6 ]   − [ ( 400 / 11 − 5 ) / 6 ] − [ ( 400 / 11 − 7 ) / 6 ]   − [ ( 400 / 13 − 7 ) / 6 ] − [ ( 400 / 13 − 11 ) / 6 ]</p><p>  − [ ( 400 / 17 − 11 ) / 6 ] − [ ( 400 / 17 − 13 ) / 6 ]   − [ ( 400 / 19 − 13 ) / 6 ] − [ ( 400 / 19 − 17 ) / 6 ]   + [ ( 400 / 5 &#215; 7 + 1 ) / 6 ] + [ ( 400 / 5 &#215; 7 − 1 ) / 6 ]   + [ ( 400 / 5 &#215; 11 + 1 ) / 6 ] + [ ( 400 / 5 &#215; 11 − 1 ) / 6 ] = 134 − 62 + 6 = 78</p></sec><sec id="s2_3_4"><title>2.3.4. Example No.4</title><p>π ( 1000 ) = 2 [ ( 1000 − 1 ) / 6 ] + 2 + [ ( 4 + 1 ) / 6 ]   − [ ( 1000 / 5 + 1 ) / 6 ] − [ ( 1000 / 5 − 1 ) / 6 ]   − [ ( 1000 / 7 − 1 ) / 6 ] − [ ( 1000 / 7 − 5 ) / 6 ]   − [ ( 1000 / 11 − 5 ) / 6 ] − [ ( 1000 / 11 − 7 ) / 6 ]   − [ ( 1000 / 13 − 7 ) / 6 ] − [ ( 1000 / 13 − 11 ) / 6 ]</p><p>− [ ( 1000 / 17 − 11 ) / 6 ] − [ ( 1000 / 17 − 13 ) / 6 ] − [ ( 1000 / 19 − 13 ) / 6 ] − [ ( 1000 / 19 − 17 ) / 6 ]</p><p>− [ ( 1000 / 23 − 17 ) / 6 ] − [ ( 1000 / 23 − 19 ) / 6 ] − [ ( 1000 / 25 − 19 ) / 6 ] − [ ( 1000 / 25 − 23 ) / 6 ] − [ ( 1000 / 29 − 23 ) / 6 ] − [ ( 1000 / 29 − 25 ) / 6 ]</p><p>− [ ( 1000 / 31 − 25 ) / 6 ] − [ ( 1000 / 31 − 29 ) / 6 ] + [ ( 1000 / 5 &#215; 7 + 1 ) / 6 ] + [ ( 1000 / 5 &#215; 7 − 1 ) / 6 ] + [ ( 1000 / 5 &#215; 11 + 1 ) / 6 ] + [ ( 1000 / 5 &#215; 11 − 1 ) / 6 ] + [ ( 1000 / 5 &#215; 13 + 1 ) / 6 ] + [ ( 1000 / 5 &#215; 13 − 1 ) / 6 ] + [ ( 1000 / 5 &#215; 17 + 1 ) / 6 ] + [ ( 1000 / 5 &#215; 17 − 1 ) / 6 ]</p><p>+ [ ( 1000 / 5 &#215; 19 + 1 ) / 6 ] + [ ( 1000 / 5 &#215; 19 − 1 ) / 6 ] + [ ( 1000 / 5 &#215; 23 + 1 ) / 6 ] + [ ( 1000 / 5 &#215; 23 − 1 ) / 6 ] + [ ( 1000 / 5 &#215; 25 + 1 ) / 6 ] + [ ( 1000 / 5 &#215; 25 − 1 ) / 6 ] + [ ( 1000 / 5 &#215; 29 + 1 ) / 6 ] + [ ( 1000 / 5 &#215; 29 − 1 ) / 6 ] + [ ( 1000 / 5 &#215; 31 + 1 ) / 6 ] + [ ( 1000 / 5 &#215; 31 − 1 ) / 6 ]</p><p>+ [ ( 1000 / 5 &#215; 35 + 1 ) / 6 ] + [ ( 1000 / 5 &#215; 35 − 1 ) / 6 ] + [ ( 1000 / 5 &#215; 37 + 1 ) / 6 ] + [ ( 1000 / 5 &#215; 37 − 1 ) / 6 ] + [ ( 1000 / 7 &#215; 11 − 1 ) / 6 ] + [ ( 1000 / 7 &#215; 11 − 5 ) / 6 ] + [ ( 1000 / 7 &#215; 13 − 1 ) / 6 ] + [ ( 1000 / 7 &#215; 13 − 5 ) / 6 ] + [ ( 1000 / 7 &#215; 17 − 1 ) / 6 ] + [ ( 1000 / 7 &#215; 17 − 5 ) / 6 ]</p><p>  + [ ( 1000 / 7 &#215; 19 − 1 ) / 6 ] + [ ( 1000 / 7 &#215; 19 − 5 ) / 6 ]   − [ ( 1000 / 5 &#215; 5 &#215; 7 + 1 ) / 6 ] − [ ( 1000 / 5 &#215; 5 &#215; 7 − 1 ) / 6 ] = 334 − 200 + 35 − 1 = 168</p></sec><sec id="s2_3_5"><title>2.3.5. Example No. 5</title><p>π ( 5000 ) = 2 [ ( 5000 − 1 ) / 6 ] + 2 + [ ( 2 + 1 ) / 6 ]   − [ ( 5000 / 5 + 1 ) / 6 ] − [ ( 5000 / 5 − 1 ) / 6 ]   − [ ( 5000 / 7 − 1 ) / 6 ] − [ ( 5000 / 7 − 5 ) / 6 ]   − [ ( 5000 / 11 − 5 ) / 6 ] − [ ( 5000 / 11 − 7 ) / 6 ]   − [ ( 5000 / 13 − 7 ) / 6 ] − [ ( 5000 / 13 − 11 ) / 6 ]</p><p>− [ ( 5000 / 17 − 11 ) / 6 ] − [ ( 5000 / 17 − 13 ) / 6 ] − [ ( 5000 / 19 − 13 ) / 6 ] − [ ( 5000 / 19 − 17 ) / 6 ] − [ ( 5000 / 23 − 17 ) / 6 ] − [ ( 5000 / 23 − 19 ) / 6 ] − [ ( 5000 / 25 − 19 ) / 6 ] − [ ( 5000 / 25 − 23 ) / 6 ] − [ ( 5000 / 29 − 23 ) / 6 ] − [ ( 5000 / 29 − 25 ) / 6 ]</p><p>− [ ( 5000 / 31 − 25 ) / 6 ] − [ ( 5000 / 31 − 29 ) / 6 ] − [ ( 5000 / 35 − 29 ) / 6 ] − [ ( 5000 / 35 − 31 ) / 6 ]</p><p>− [ ( 5000 / 37 − 31 ) / 6 ] − [ ( 5000 / 37 − 35 ) / 6 ] − [ ( 5000 / 41 − 35 ) / 6 ] − [ ( 5000 / 41 − 37 ) / 6 ] − [ ( 5000 / 43 − 37 ) / 6 ] − [ ( 5000 / 43 − 41 ) / 6 ]</p><p>− [ ( 5000 / 47 − 41 ) / 6 ] − [ ( 5000 / 47 − 43 ) / 6 ] − [ ( 5000 / 49 − 43 ) / 6 ] − [ ( 5000 / 49 − 47 ) / 6 ] − [ ( 5000 / 53 − 47 ) / 6 ] − [ ( 5000 / 53 − 49 ) / 6 ] − [ ( 5000 / 55 − 49 ) / 6 ] − [ ( 5000 / 55 − 53 ) / 6 ] − [ ( 5000 / 59 − 53 ) / 6 ] − [ ( 5000 / 59 − 55 ) / 6 ]</p><p>− [ ( 5000 / 61 − 55 ) / 6 ] − [ ( 5000 / 61 − 59 ) / 6 ] − [ ( 5000 / 65 − 59 ) / 6 ] − [ ( 5000 / 65 − 61 ) / 6 ] − [ ( 5000 / 67 − 61 ) / 6 ] − [ ( 5000 / 67 − 65 ) / 6 ] + [ ( 5000 / 5 &#215; 7 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 7 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 11 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 11 − 1 ) / 6 ]</p><p>+ [ ( 5000 / 5 &#215; 13 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 13 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 17 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 17 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 19 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 19 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 23 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 23 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 25 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 25 − 1 ) / 6 ]</p><p>+ [ ( 5000 / 5 &#215; 29 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 29 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 31 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 31 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 35 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 35 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 37 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 37 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 41 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 41 − 1 ) / 6 ]</p><p>+ [ ( 5000 / 5 &#215; 43 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 43 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 47 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 47 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 49 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 49 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 53 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 53 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 55 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 55 − 1 ) / 6 ]</p><p>+ [ ( 5000 / 5 &#215; 59 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 59 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 61 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 61 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 65 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 65 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 67 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 67 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 71 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 71 − 1 ) / 6 ]</p><p>+ [ ( 5000 / 5 &#215; 73 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 73 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 77 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 77 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 77 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 77 − 1 ) / 6 ]</p><p>+ [ ( 5000 / 5 &#215; 79 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 79 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 83 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 83 − 1 ) / 6 ]</p><p>+ [ ( 5000 / 5 &#215; 85 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 85 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 91 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 91 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 95 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 95 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 97 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 97 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 101 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 101 − 1 ) / 6 ]</p><p>+ [ ( 5000 / 5 &#215; 103 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 103 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 107 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 107 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 109 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 109 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 103 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 103 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 107 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 107 − 1 ) / 6 ]</p><p>+ [ ( 5000 / 5 &#215; 109 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 109 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 113 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 113 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 115 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 115 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 119 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 119 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 121 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 121 − 1 ) / 6 ]</p><p>+ [ ( 5000 / 5 &#215; 125 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 125 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 127 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 127 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 131 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 131 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 133 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 133 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 137 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 137 − 1 ) / 6 ]</p><p>+ [ ( 5000 / 5 &#215; 139 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 139 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 143 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 143 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 145 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 145 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 149 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 149 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 151 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 151 − 1 ) / 6 ]</p><p>+ [ ( 5000 / 5 &#215; 155 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 155 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 157 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 157 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 161 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 161 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 163 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 163 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 167 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 167 − 1 ) / 6 ]</p><p>+ [ ( 5000 / 5 &#215; 169 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 169 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 173 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 173 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 175 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 175 − 1 ) / 6 ]</p><p>+ [ ( 5000 / 5 &#215; 179 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 179 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 181 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 181 − 1 ) / 6 ]</p><p>+ [ ( 5000 / 5 &#215; 185 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 185 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 187 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 187 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 191 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 191 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 193 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 193 − 1 ) / 6 ] + [ ( 5000 / 5 &#215; 197 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 197 − 1 ) / 6 ]</p><p>+ [ ( 5000 / 5 &#215; 199 + 1 ) / 6 ] + [ ( 5000 / 5 &#215; 199 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 11 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 11 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 13 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 13 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 17 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 17 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 19 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 19 − 5 ) / 6 ]</p><p>+ [ ( 5000 / 7 &#215; 23 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 23 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 25 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 25 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 29 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 29 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 31 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 31 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 35 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 35 − 5 ) / 6 ]</p><p>+ [ ( 5000 / 7 &#215; 37 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 37 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 41 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 41 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 43 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 43 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 47 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 47 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 49 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 49 − 5 ) / 6 ]</p><p>+ [ ( 5000 / 7 &#215; 53 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 53 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 55 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 55 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 59 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 59 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 61 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 61 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 65 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 65 − 5 ) / 6 ]</p><p>+ [ ( 5000 / 7 &#215; 67 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 67 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 71 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 71 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 73 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 73 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 77 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 77 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 79 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 79 − 5 ) / 6 ]</p><p>+ [ ( 5000 / 7 &#215; 83 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 83 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 85 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 85 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 89 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 89 − 5 ) / 6 ]</p><p>+ [ ( 5000 / 7 &#215; 91 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 91 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 95 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 95 − 5 ) / 6 ]</p><p>+ [ ( 5000 / 7 &#215; 97 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 97 − 5 ) / 6 ] + [ ( 5000 / 7 &#215; 101 − 1 ) / 6 ] + [ ( 5000 / 7 &#215; 101 − 5 ) / 6 ] + [ ( 5000 / 11 &#215; 13 − 5 ) / 6 ] + [ ( 5000 / 11 &#215; 13 − 7 ) / 6 ] + [ ( 5000 / 11 &#215; 17 − 5 ) / 6 ] + [ ( 5000 / 11 &#215; 17 − 7 ) / 6 ] + [ ( 5000 / 11 &#215; 19 − 5 ) / 6 ] + [ ( 5000 / 11 &#215; 19 − 7 ) / 6 ]</p><p>+ [ ( 5000 / 11 &#215; 23 − 5 ) / 6 ] + [ ( 5000 / 11 &#215; 23 − 7 ) / 6 ] + [ ( 5000 / 11 &#215; 25 − 5 ) / 6 ] + [ ( 5000 / 11 &#215; 25 − 7 ) / 6 ] + [ ( 5000 / 11 &#215; 29 − 5 ) / 6 ] + [ ( 5000 / 11 &#215; 29 − 7 ) / 6 ] + [ ( 5000 / 11 &#215; 31 − 5 ) / 6 ] + [ ( 5000 / 11 &#215; 31 − 7 ) / 6 ] + [ ( 5000 / 11 &#215; 35 − 5 ) / 6 ] + [ ( 5000 / 11 &#215; 35 − 7 ) / 6 ]</p><p>+ [ ( 5000 / 11 &#215; 37 − 5 ) / 6 ] + [ ( 5000 / 11 &#215; 37 − 7 ) / 6 ] + [ ( 5000 / 11 &#215; 41 − 5 ) / 6 ] + [ ( 5000 / 11 &#215; 41 − 7 ) / 6 ] + [ ( 5000 / 13 &#215; 17 − 7 ) / 6 ] + [ ( 5000 / 13 &#215; 17 − 11 ) / 6 ] + [ ( 5000 / 13 &#215; 19 − 7 ) / 6 ] + [ ( 5000 / 13 &#215; 19 − 11 ) / 6 ] + [ ( 5000 / 13 &#215; 23 − 7 ) / 6 ] + [ ( 5000 / 13 &#215; 23 − 11 ) / 6 ]</p><p>+ [ ( 5000 / 13 &#215; 25 − 7 ) / 6 ] + [ ( 5000 / 13 &#215; 25 − 11 ) / 6 ] + [ ( 5000 / 13 &#215; 29 − 7 ) / 6 ] + [ ( 5000 / 13 &#215; 29 − 11 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 7 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 7 − 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 11 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 11 − 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 13 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 13 − 1 ) / 6 ]</p><p>− [ ( 5000 / 5 &#215; 5 &#215; 17 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 17 − 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 19 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 19 − 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 23 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 23 − 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 25 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 25 − 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 29 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 29 − 1 ) / 6 ]</p><p>− [ ( 5000 / 5 &#215; 5 &#215; 31 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 31 − 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 35 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 35 − 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 37 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 37 − 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 41 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 41 − 1 ) / 6 ] − [ ( 5000 / 5 &#215; 7 &#215; 7 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 7 &#215; 7 − 1 ) / 6 ]</p><p>− [ ( 5000 / 5 &#215; 7 &#215; 11 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 7 &#215; 11 − 1 ) / 6 ] − [ ( 5000 / 5 &#215; 7 &#215; 13 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 7 &#215; 13 − 1 ) / 6 ] − [ ( 5000 / 5 &#215; 7 &#215; 17 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 7 &#215; 17 − 1 ) / 6 ]</p><p>− [ ( 5000 / 5 &#215; 7 &#215; 19 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 7 &#215; 19 − 1 ) / 6 ] − [ ( 5000 / 5 &#215; 7 &#215; 23 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 7 &#215; 23 − 1 ) / 6 ]</p><p>− [ ( 5000 / 5 &#215; 7 &#215; 25 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 7 &#215; 25 − 1 ) / 6 ] − [ ( 5000 / 5 &#215; 11 &#215; 11 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 11 &#215; 11 − 1 ) / 6 ] − [ ( 5000 / 5 &#215; 11 &#215; 13 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 11 &#215; 13 − 1 ) / 6 ] − [ ( 5000 / 5 &#215; 11 &#215; 17 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 11 &#215; 17 − 1 ) / 6 ] − [ ( 5000 / 5 &#215; 11 &#215; 19 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 11 &#215; 19 − 1 ) / 6 ]</p><p>− [ ( 5000 / 5 &#215; 13 &#215; 13 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 13 &#215; 13 − 1 ) / 6 ] − [ ( 5000 / 7 &#215; 7 &#215; 11 + 1 ) / 6 ] − [ ( 5000 / 7 &#215; 7 &#215; 11 − 1 ) / 6 ] − [ ( 5000 / 7 &#215; 7 &#215; 13 + 1 ) / 6 ] − [ ( 5000 / 7 &#215; 7 &#215; 13 − 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 5 &#215; 7 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 5 &#215; 7 − 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 5 &#215; 11 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 5 &#215; 11 − 1 ) / 6 ]</p><p>  − [ ( 5000 / 5 &#215; 5 &#215; 7 &#215; 7 + 1 ) / 6 ] − [ ( 5000 / 5 &#215; 5 &#215; 7 &#215; 7 − 1 ) / 6 ] = 1668 − 1441 + 495 − 54 + 1 = 699</p></sec></sec></sec><sec id="s3"><title>Acknowledgements</title><p>We would like to express our deep appreciation to Professor Yongbao Yang and Professor Husheng Qiao of Northwest Normal University, Senior English teacher Zhengrong Yin of Longxi County Education Committee, Vice-professor Chang Feng of Can-Su University of Traditional Chinese Medicine for their great help!</p></sec><sec id="s4"><title>Conflicts of Interest</title><p>The authors declare no conflicts of interest regarding the publication of this paper.</p></sec><sec id="s5"><title>Cite this paper</title><p>Wang, M.Z., He, Z.X. and Wang, M.Y. 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