Visualising the distribution of primes

Transcription

1 University of Luxembourg BASI Mathematics Experimental Mathematics Lab Project Visualising the distribution of primes Authors: Jim Barthel Pietro Sgobba Fa Zhu Supervisors: Prof. Gabor Wiese Dr. Panagiotis Tsaknias Summer semester 2015

2 Contents 1 Introduction 2 2 The asymptotic distribution of prime numbers The prime-counting function The prime number theorem Approximation in terms of logarithmic integral Probabilistic interpretation of the Prime number theorem Estimate for the nth prime number Frequencies Ulam spiral History Our Ulam spiral Comparison with random distribution Lines of primes in the Ulam spiral Klauber triangle History Our Klauber triangle Comparison with random distribution Lines of primes in the Klauber triangle

3 1 Introduction The aim of this project is to study the distribution of prime numbers. We are going to observe how prime numbers are distributed in the set of natural numbers. At a rst look of the rst prime numbers, e.g. prime numbers less then 100, one may think that their distribution is random, although they tend to appear less frequently approaching 100. However, if we observe prime numbers on a much greater scale, we can notice that the apparently random distribution of primes follows some particular behaviours. In this project we are going to analyse two phenomena. In the rst part, we will describe the asymptotic behaviour of the prime-counting function π. In the second part, we will introduce the Ulam spiral and the Klauber triangle, which are simple models to display natural numbers and to notice some patterns of primes. Let us start from the basic denition of a prime number. Denition. A natural number p N is a prime number if p 2 and it has no positive divisors other than 1 and p. A rst question about prime numbers, which is also at the basis of our study, is whether there are innitely many primes. The answer was already given by Euclid more than 2200 years ago in his famous Elements (Book 9, Proposition 20). Before proving Euclid's theorem, let us state and prove the following lemma. Lemma. Let n 2 be a natural number. Then there exist prime numbers p 1, p 2,..., p k (which may be not all dierent) such that n = p 1 p 2 p k. Proof. Let us assume that there exists a positive integer 2 which cannot be written as a product of primes. Let n be the smallest such integer. Since n is obviously not prime, n has a positive divisor d which is neither 1 nor n such that n = md, for some m N. Note that both d and m are 2 and < n. Because n is the smallest positive integer which cannot be written as a product of primes and d and m are less than n, we have that m = p 1 p 2 p k and d = q 1 q 2 q l for some prime numbers p 1,..., p k, q 1,..., q l. Therefore, n = md = p 1 p 2 p k q 1 q 2 q l, i.e. n can be written as a product of some prime numbers. wrong. Hence, our assumption is We are now able to prove Euclid's theorem. 2

4 Theorem (Euclid's theorem). There exist innitely many primes. Proof. Let us assume that there are nitely many primes, say p 1, p 2, p 3,..., p n. Consider the positive integer m := p 1 p 2 p n + 1. (1) Using the previous Lemma, every positive integer 2 can be written as a product of prime numbers. Then, there is a prime which is a divisor of m, say p i, for some 1 i n. Now, this contradicts equation (1), which shows that dividing m by p i we get 1 as reminder. Hence, our assumption is wrong. 3

5 2 The asymptotic distribution of prime numbers 2.1 The prime-counting function The goal of this section is to study the asymptotic distribution of prime numbers. Let us start with the following denition. Denition. The prime-counting function is the function π : R + N dened by π(x) := #{p N p is a prime number and p x}. Prime numbers less than 100 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. Therefore, π(10) = #{2, 3, 5, 7} = 4, and so on, as we see in the following table: x π(x) Table 1: Some values of π Figure 1 shows the graph of π for 0 x 100. Of course, we are interested in the asymptotic behaviour of this function. Table 2 gives values of π for powers of 10 up to 10 9 and Figure 2 shows the graph of π for x in this interval, i.e. 0 x x π(x) Table 2: Values of π for powers of 10 4

6 Figure 1: Graph of π for 0 x 100 Figure 2: Graph of π for 0 x

7 2.2 The prime number theorem By Euclid's theorem we know that there exist innitely many primes. Of course, this means that π is an increasing function and that π(x) as x. Intuitively we see that prime numbers appear to be more sporadic as they become larger, but we would like to describe more precisely the asymptotic behaviour of π. This is formalized by the following famous theorem. Theorem (Prime number theorem). Let π be the prime counting function and log be the natural logarithm. Then a good asymptotic approximation of π is given by the function This means that lim x x Equivalently, this can be written in the form π(x) x log x. π(x) x/ log x = 1. x log x as x. (2) Near the end of the 18th century, the mathematicians Carl F. Gauss and Adrien- Marie Legendre, through empirical considerations, conjectured equivalent statements to (2). During the following century, Bernhard Riemann gave an important contribution to the theory. Then the theorem was nally proven in 1896, independently by two mathematicians, Jacques Hadamard and Charles J. de la Vallée-Poussin. In the 20th century, simpler proofs were found by Atle Selberg and Paul Erd s (1949) and later by Donald J. Newman (1980). We will not prove the theorem here, but we will analyse the approximation of π. We have plotted both functions π and x on the same graph for x up to log x 109 (Figure 3). Moreover, we have plotted the quotient π(x)/( x ), again for x up to log x 109 and we have enlarged the graph for the interval 1 y 1.2 on the y-axis (Figure 4). As we can see, the convergence to 1 is very slow. 6

8 Figure 3: Functions π(x) (-) and x log x (-) for x up to 109 Figure 4: Quotient π(x)/( x log x ) for x up to 109, 1 y 1.2 7

9 In order to describe the convergence of the quotient π(x)/( x ), the following table log x gives values illustrating this behaviour. x π(x) x log x π(x)/( x log x ) Table 3: Values concerning (2) for powers of Approximation in terms of logarithmic integral There exists a better asymptotic approximation of the prime-counting function than For this, let us consider the Logarithmic integral function x. log x li(x) = x 0 dt, for x > 0, x 1. log t Now, the Eulerian logarithmic integral function is dened as Li(x) = x 2 dt log t = li(x) li(2), for x > 2. The Prime number theorem also states that the prime-counting function is asymptotically equivalent to Li. Then (2) is equivalent to π(x) Li(x) as x. (3) We have plotted the functions π, Li and x in the same graph (Figure 5): π and Li log x almost coincide. Moreover, it is sucient to plot the quotient π(x)/ Li(x) for x up to 10 5 (Figure 6). Indeed, we can see that the convergence to 1 is very quick. Figure 7 shows the comparison between the two quotients π(x)/ Li(x) and π(x)/ x for x up to log x

10 Figure 5: Functions π(x) (-), Li(x) (-) and x log x (-) for x up to 109 Figure 6: Quotient π(x)/ Li(x) for x up to 10 5, 0.8 y 1 9

11 Figure 7: Quotients π(x)/ x log x (-) and π(x)/ Li(x) (-) for x up to 109, 0.8 y 1.2 Table 4 shows values concerning statement (3), while Table 5 compares them to the respective results for the approximation by x (Table 3). The two gures together with log x both tables clearly point out that the Li function is a much stronger approximation for π than x. log x x π(x) Li(x) π(x)/ Li(x) Table 4: Values concerning (3) for powers of 10 10

12 x π(x) Li(x) x log x π(x)/ Li(x) π(x)/( x log x ) Table 5: Comparison of values concerning (2) and (3) for powers of Probabilistic interpretation of the Prime number theorem The Prime number theorem can be also understood using a simple probabilistic approach. Statement (2) is equivalent to π(x) x 1 log x as x. (4) The quotient π(x)/x represents the relative amount, i.e. the percentage, of how many prime numbers one can nd in the interval [1, x]. Then, (4) says that this value is asymptotically equal to 1/ log x. We can interpret this statement as follows: for x very large, the probability that k [1, x], where k N, is a prime can be approximated by 1/ log x. Analogously, statement (3) is equivalent to π(x) x Li(x) x as x. (5) Then, for x very large, a better approximation for the probability that k [1, x], k N, is a prime is given by Li(x)/x. Table 6 compares values concerning statements (4) and (5). Note that a strong consequence of these observations is that in particular, the probability for a large integer n to be prime can be approximated by 1/ log n and even better by Li(n)/n. 11

13 x π(x) π(x) x 1 log x Li(x) x Table 6: Values concerning (4) and (5) for powers of Estimate for the nth prime number An important consequence of the Prime number theorem is the asymptotic approximation of the nth prime number. Corollary. Let us denote by p n the nth prime number. Then the following statement holds: p n n log n, as n. Proof. If p n is the nth prime number and π is the prime-counting function, as above, we have that π(p n ) = n. Now, the Prime number theorem implies that n p n, as n. (6) log p n Let us prove the following claim By the above we have log n log log p n log n, as n. ) = log p n log log p n, as n. (7) log p n ( pn Hence, if we call m := p n and use that m if n, we have that lim n log n (7) log m log log m = lim log p n m log m 1 = lim 1 m log m 1 m 1 m log log m = lim 1 m log m = lim m 1 1 log m = 1 where we have used L'Hôpital's rule in the second last step. So the claim is proven. Finally, using 6 and the claim we obtain p n n log p n n log n, as n. 12

14 Figure 8 compares the two functions n p n and n log n. Figure 8: p n (-) and n log n (-) for n up to Frequencies We have read the article An observation on the distribution of primes (1967) by the mathematicians Myron Stein and Stanislaw Ulam. In this article, they study the following problem. For each positive integer k consider the interval I k := [k, k E(log k)], where E(log k) stands for the nearest integer to log k. One can count how many primes are in that interval, e.g. for some k's there will be no prime in it, for some other k's there will be exactly one prime in it, for other ones two primes, etc. Let n be a positive integer. Consider all the intervals I k, for 1 k n, and count how many primes are in each interval. We now dene the functions Γ m (n) := #{k n I k contains m primes}, n N, m = 0, 1, 2, 3,.... So Γ 0 (n) gives the number of intervals I k, k n, containing no prime, Γ 1 (n) gives the number of intervals I k, k n, containing exactly one prime, and so on. Now, we consider the frequencies of these classes by taking the ratio to n of every Γ m and dene γ m (n) := Γ m(n) n. 13

15 So γ 0, γ 1, γ 2,... will be our frequencies. Here are the values that we computed for these frequencies for m = 0, 1,..., 7. Note that in the computation we considered E as the oor function, i.e. the largest previous integer to the real number at which it is evaluated, in our case log k. n Γ 0 (n) Γ 1 (n) Γ 2 (n) Γ 3 (n) Γ 4 (n) Γ 5 (n) Γ 6 (n) Γ 7 (n) Table 7: Values of Γ m n γ 0 (n) γ 1 (n) γ 2 (n) γ 3 (n) γ 4 (n) γ 5 (n) γ 6 (n) γ 7 (n) e e-3 0.2e e e e e e e e e e e e e e e e e e e e e e e e-5 2.5e e e e-5 7.0e e e e-5 8.0e e e e-5 8.2e e e e-5 8.8e-8 Table 8: Values of γ m Now every function taking values in [0, 1] can converge to some a [0, 1] or diverge by uctuating in the interval. We cannot state that the γ m 's have nite limits for n, but by observing the data we can conjecture, as Ulam and Stein do in their article, that they converge. If this is true, we would be able to predict to which probability a given interval I k, k N, contains a certain number m of prime numbers. This would be another signicant characterization of the distribution of primes. 14

16 3 Ulam spiral 3.1 History In 1963, the mathematician Stanislaw Ulam discovered a method to visualize prime numbers. Ulam put the integers on a spiral where 1 stands in the middle and the other integers circle in some way around it. With this visualization, called today Ulam spiral, he noticed that prime numbers seemed to appear mostly on some diagonal lines. He recognized that these lines were given by some quadratic polynomials which indeed generate large numbers of primes. To produce a big picture of the spiral, Ulam with collaborators Myron Stein and Mark Wells used MANIAC II at Los Alamos Scientic Laboratory and created a spiral up to numbers. Martin Gardner rst wrote about the Ulam spiral in his Mathematical Games column in Scientic American, and the spiral featured on the front cover of the magazine. 3.2 Our Ulam spiral We have reproduced the Ulam spiral in Matlab in the following way. At rst we placed 1 in the centre of the spiral. Then we placed 2 on the right side of 1 and 3 above 2. Hence we directed our spiral counter-clockwise, such that it ends at the right lower corner. Figure 9 shows the procedure we have used Figure 9: Ulam spiral up to 25 Therefore we have plotted spirals with an odd number of elements on each side and hence a total odd number of elements on the spiral. Indeed, if n is the number of integers on each side, then the spiral contains N = n 2 integers. We have plotted several spirals with dierent size. One can reproduce every gure by using the Matlab code "Ulam_spiral". Here, you can see some of our results, in which we indicate the values of n and N as dened above. 15

17 Figure 10: Ulam spiral with n = 201 and N = Figure 11: Ulam spiral with n = 251 and N =

18 Figure 12: Ulam spiral with n = 401 and N =

19 3.3 Comparison with random distribution In order to verify the pattern of the prime distribution in the Ulam spiral, we have created a spiral with random values that are distributed with the same probability as the prime numbers, which we will compare with our initial Ulam spiral. We specify brie y how we have established this random spiral. At rst we deleted every even number to get at least some pattern in the spiral. Then we chose some numbers to be highlighted such that the highlighted numbers have the same probability distribution as the prime numbers. For this, we used for each number n the probability p = log2 n to be highlighted. Indeed the probability for an integer 1 k n, and in particular for n, to be prime can be approximated by log1 n. Moreover, this value has to be doubled since we deleted all even numbers. Figure 13 shows our result for the random spiral with 401 elements on the side. Figure 13: Random spiral with n = 401 and N =

20 In order to compare both spirals, Figure 14 shows the Ulam spiral on the left and the random spiral on the right, both with the side length of 251 elements. Figure 14: Comparison between the Ulam spiral (left) and the random spiral (right) with n = 251 and N = We notice that the distribution in the random spiral is very chaotic, while in the Ulam spiral prime numbers can be found mostly along diagonal lines, i.e. they seem to follow a non-random distribution. In fact, one cannot predict a number to be prime or not by a formula (until now) but the primes on those lines can be given by some quadratic functions. One can reproduce a random spiral by using rst the Matlab code "Probability_vector_ of_primes" and then "Random_Ulam_Spiral". 3.4 Lines of primes in the Ulam spiral As we have mentioned above, the lines of primes we have observed on the Ulam spiral can be given by some functions, in particular some quadratic polynomials. We will now discuss how we have found the functions for some of the most dense lines and we will describe them. We show here again Figure 9 as it will be useful to the reader in the following steps

21 We will rst notice some basic facts. Let us consider the Ulam spiral as a sequence of squares one inside the other, which are numbered in the following way: n = 0 corresponds to the small square containing 1, n = 1 corresponds to the square containing 1,..., 9, n = 2 corresponds to the square containing 1,..., 25, and so on. Then the side of the nth square contains 2n + 1 elements. Indeed this is true for n = 0, 1, 2 as one can see on the gure above, and by induction, if it is true for n, the side of the (n + 1)st square contains the same amount of elements as the side of the nth square and 2 more, i.e. (2n + 1) + 2 = 2n + 3 = 2(n + 1) + 1. Thus, this is true also for n + 1. A direct consequence of this statement is that the nth square contains (2n+1) 2 elements. This means that the largest element in the nth square, i.e. the element in the right lower corner, is the square of an odd number, in particular of the (n+1)st odd number. Moreover, the diagonal line (in fact, the ray or half-line) starting at 1 and containing 9, 25,... is given by the function f(n) = (2n + 1) 2 = 4n 2 + 4n + 1, n N. Now we can also nd the function for the diagonal line containing 1, 3, 13,..., in the following way. Let us consider the largest integer of the (n 1)st square, i.e. (2n 1) 2, for n = 1, 2,.... If we want to obtain the smallest following integer which lies on that line, we need to add the number of elements corresponding to the side of the nth square minus 1, i.e. (2n 1) 2 + (2n + 1) 1 = 4n 2 4n n = 4n 2 2n + 1. Thus, the integers on the line are given by the function g(n) = 4n 2 2n + 1, n N. This line can be translated to the left by adding a positive constant to the function g. In particular, one of the most prime-dense lines is given by adding 40 to g, a constant which we computed manually. So we nally get one of the quadratic polynomials we were looking for: p 1 (n) = 4n 2 2n + 41, n N. (8) Note that the translated line to the left will only start from the nth square such that the integer p 1 (n) = g(n) + 40 lies on the same horizontal side as g(n), i.e. from the nth square such that 40 2n n 20. In other words, all the integers p 1 (n) for n = 0,..., 19 do not lie on that line. Figure 15 shows where the primes generated by p 1 are located in the spiral. 20

22 Figure 15: Prime numbers generated by p1 in the Ulam spiral with 200 squares (401 elements on each side) 21

23 The following table gives the integers generated by p 1 for n = 0,..., 100, highlighting in red non-prime numbers. As we can see most of the integers in the table are prime, while 21 among the 101 generated integers are not prime. n p 1 (n) n p 1 (n) n p 1 (n) n p 1 (n) n p 1 (n) Table 9: Integers generated by p 1 for n = 0,..., 100. Non-prime numbers are highlighted in red. 22

24 Another quadratic polynomial generating many prime numbers is given by adding 100 to g : p2 (n) = 4n2 2n + 101, n N. In this case, only the integers p2 (n) with n 50 lie on the line. Figure 16: Prime numbers generated by p2 in the Ulam spiral with 200 squares (401 elements on each side) 23

25 Another prime-dense line corresponds to the parallel line to p1 in the lower part of the spiral. In order to nd the corresponding function, we proceed in a similar way to p1. Consider the largest integer of the nth square, i.e. (2n + 1)2. Then the integers on the line starting at 1 and containing 7, 21,... are given by subtracting 2n from (2n + 1)2. The equation we get is h(n) = 4n2 + 2n + 1, n N. Now, we translate this line to the right by adding a positive constant to h. In our case, the line we want to obtain is given by adding 40: p3 (n) = 4n2 + 2n + 41, n N. As we have discussed above, the integers lying on the line are generated for n 20. Figure 17: Prime numbers generated by p3 in the Ulam spiral with 200 squares (401 elements on each side) 24

26 Finally, a last example of quadratic polynomial that we want to discuss concerns a line in the lower part of the spiral which is parallel to the one described by f. Indeed we can translate this line to the right by adding 58 to f and we get the polynomial p4 (n) = 4n2 + 4n + 59, n N. In this case, p4 describes the line we have considered if n 28. Figure 18: Prime numbers generated by p4 in the Ulam spiral with 200 squares (401 elements on each side) All pictures can be reproduced with the Matlab codes Ulam_Spiral_Lines_fα with α=1,2,3,4. 25

27 4 Klauber triangle 4.1 History Before Ulam, the American herpetologist Laurence M. Klauber proposed already in 1932 a triangle which visualizes the prime numbers. In the original version of this triangle, he begins with 1 at the top and in every row he extends the number of elements by two, such that the nth row contains the numbers (n 1) to n 2. As in the Ulam spiral, the Klauber triangle shows that when prime numbers are highlighted a specic pattern can be observed in their distribution. Indeed, they seem to appear mostly on some lines. 4.2 Our Klauber triangle We have produced a slightly modied Klauber triangle in Matlab. We placed the integers 1, 2 and 3 on the rst row. Then every consecutive row is extended by two elements. Hence the nth row contains 2n + 1 elements, in particular the integers between n 2 and (n + 1) 2 1. See the picture below: Figure 19: Klauber triangle up to 24 We have plotted several triangles with dierent size. One can reproduce every gure by using the Matlab code "Triangle". Here you can see some of our results: 26

28 Figure 20: Klauber triangle with 100 rows Figure 21: Klauber triangle with 200 rows 27

29 Figure 22: Klauber triangle with 300 rows 28

30 4.3 Comparison with random distribution Here again, we will check, as for the Ulam spiral, if there is a certain pattern in the Klauber triangle. We used the same method as in the previous named comparison, brie y said, a random distribution of highlighted numbers with the same probability as the primes. The triangle is then implemented in the same way as the Klauber triangle is. One can reproduce a random triangle by using rst the Matlab code "Probability_vector _of_primes" and then "Random_Triangle". This is our random triangle plotted up to 250 rows: Figure 23: Random triangle with 250 rows In order to compare both spirals, Figure 24 shows the Klauber triangle on the left and the random triangle on the right, both with 200 rows. Again we can see that prime numbers seem to follow a certain pattern in their distribution. If we observe our Klauber triangle, we clearly see some straight lines, but in the random triangle, we cannot recognize any precise distribution. Then one could again nd some quadratic formulas describing the lines in our triangle. 29

31 Figure 24: Comparison between the Klauber triangle (left) and the random triangle (right) with 200 rows 4.4 Lines of primes in the Klauber triangle We have looked for quadratic polynomials, with integer input, to describe some of the lines which can be seen in the triangle. First of all, let us calculate the last number in each row, which corresponds to the total number of elements in the triangle, up to that row. As the rst row begins with 3 elements and then every row has two elements more than the previous one, the ith row contains 2i + 1 integers and the total number of elements for the triangle of n 1 rows is given by n X (2i + 1). i=1 Now, this formula can be simpli ed: n X i=1 (2i + 1) = 2 n X i+n=2 i=1 2 n(n + 1) 2 +n = n + 2n = n2 + 2n = (n + 1)2 1. Hence, the last element of the nth row is given by (n + 1)2 1, as we have mentioned while introducing the Klauber triangle. Then, it follows that the rst element of the nth row, which is given by the last element of the previous row plus 1, is equal to n2. In order to get the integers on the middle line, we notice that the middle value of each row is exactly the average of the rst and the last value of that row: n2 + (n + 1)2 1 n2 + n2 + 2n = = n2 + n. 2 2 Hence the integers in the middle vertical line are given by the polynomial: ϕ(n) = n2 + n, 30 n 0.

32 Now, every vertical line of the triangle can be computed by shifting the middle line to the right or to the left by adding a constant c Z to ϕ. To get for example the line on the right side of the triangle which begins at the last element of the row 41, going straight down, we choose the formula q 1 (n) = n 2 + n + 41, n 0. The coloured integers in Figure 25 are only the primes given by this function. Figure 25: Prime numbers given by q 1 in the Klauber triangle with 200 rows One can see that this function does not only give the vertical line we wished to represent, but also the diagonal line beginning at the second last element of the 14th row (i.e. at 223) and ending at the rst element of the 41st row (i.e. at 1681). This diagonal line goes from the beginning to the end always 2 elements left in the next row, which can be empirically veried. Indeed if we want to calculate the polynomial generating this line, we consider the vertical line starting at the last element of the 13th row, i.e. n 2 + n + 13, and we subtract 2(n 14), i.e. every row we climb down from the 13th one, we go two integers left. Then n 2 + n (n 14) = n 2 + n n + 28 = n 2 n + 41, and after a change of variables from n to n + 1, we get (n + 1) 2 (n + 1) + 41 = n 2 + 2n + 1 n = n 2 + n + 41, 31

33 which is exactly q 1 for 13 n 40. After these computations, one can observe that for every vertical line given by n 2 +n+c, where c Z is an odd number, we get one or more diagonal lines for n smaller than c. Here one can see two other lines calculated in the same way as the rst example. One is given by q 2 (n) = n 2 + n 107 Figure 26: Prime numbers given by q 2 in the Klauber triangle with 200 rows The other one is given by See Figure 27. q 3 (n) = n 2 + n

34 Figure 27: Prime numbers given by q 3 in the Klauber triangle with 200 rows At last, we want to represent a line which is not vertical and which cannot be constructed using the same method of the previous lines. We choose a line which is parallel to the left side of the triangle. A line with this property is easy to calculate as the distance between each elements on this line and the rst element of the same row is constant. This kind of lines are given by n 2 + d where d is a positive integer. An example is q 4 (n) = n , which is less prime-dense than the lines we have already discussed, as q 4 (2k) is even for every k N. See Figure

35 Figure 28: Prime numbers given by q 4 in the Klauber triangle with 200 rows All pictures can be reproduced with the Matlab codes Triangle_Lines_f α with α= 1,2,3,4. 34