CERIAS Tech Report The period of the Bell numbers modulo a prime by Peter Montgomery, Sangil Nahm, Samuel Wagstaff Jr Center for Education

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1 CERIAS Tech Reort The eriod of the Bell numbers modulo a rime by Peter Montgomery, Sangil Nahm, Samuel Wagstaff Jr Center for Education and Research Information Assurance and Security Purdue University, West Lafayette, IN

2 MATHEMATICS OF COMPUTATION Volume 00, Number 0, Pages S (XX) THE PERIOD OF THE BELL NUMBERS MODULO A PRIME PETER L. MONTGOMERY, SANGIL NAHM, AND SAMUEL S. WAGSTAFF, JR. Abstract. We discuss the numbers in the title, and in articular whether the minimum eriod of the Bell numbers modulo a rime can be a roer divisor of N = ( 1)/( 1). It is known that the eriod always divides N. The eriod is shown to equal N for most rimes below 180. The investigation leads to interesting new results about the ossible rime factors of N. For examle, we show that if is an odd ositive integer and m is a ositive integer and q = 4m is rime, then q divides m2 1. Then we exlain how this theorem influences the robability that q divides N. 1. Introduction The Bell exonential numbers B(n) are ositive integers that arise in combinatorics. They can be defined by the generating function e ex 1 = n=0 B(n) xn n!. See [5] for more background. Williams [11] roved that for each rime, the Bell numbers modulo are eriodic and that the eriod divides N = ( 1)/( 1). In fact the minimum eriod equals N for every rime for which this eriod is known. Theorem 1.1. The minimum eriod of the sequence {B(n) mod } is N when is a rime < 126 and also when = 137, 149, 157, 163, 167 or 173. Theorem 1.1 imroves the first art of Theorem 3 of [10]. The statements about the rimes = 103, 107, 109, 137, 149 and 157 are new here and result from calculations we did using the same method as in [10]. These calculations are ossible now because new rime factors have been discovered for these N. Table 1 lists all new rime factors discovered for N since [10], even when the factorization remains incomlete. Table 1 uses the same notation and format as Table 1 of [10]. The L and M in the table reresent ieces of algebraic factorizations exlained in [10] Mathematics Subject Classification. 11B73, 11A05, 11A07, 11A51. Key words and hrases. Bell numbers, eriod modulo. This work was suorted in art by the CERIAS Center at Purdue University. 1 c 1997 American Mathematical Society

3 2 PETER L. MONTGOMERY, SANGIL NAHM, AND SAMUEL S. WAGSTAFF, JR. Table 1. Some new rime factors New rime factors of N P P L P58 137L P85 149L P M P C L \ P C369 Theorem 1.1 is roved by showing that the eriod does not divide N /q for any rime divisor q of N. (We made this check for each new rime q in Table 1, including those written as Pxxx, and not just those for the for which N is comletely factored.) In [10], this condition was checked also for all airs (, q) of rimes for which < 1100, q < 2 31 and q divides N. It was conjectured there that the minimum eriod is always N. As early as 1979 [6] others wondered whether the minimum eriod is always N. See [3] for a summary of work on this conjecture u to We resent a heuristic argument below suorting the conjecture. Touchard s [8] congruence B(n+) B(n)+B(n+1) mod, valid for any rime and for all n 0, shows that any consecutive values of B(n) mod determine the sequence modulo after that oint. If N divides N, then one can test whether the eriod of the Bell numbers modulo divides N by checking whether B(N + i) B(i) mod for 0 i 1. The eriod divides N if and only if all of these congruences hold. A olynomial time algorithm for comuting B(n) mod has been known at least since 1962 [5]. Pseudocode for the algorithm aears in [10]. In the last section of this note we give a heuristic argument for the robability that the conjecture holds for a rime and estimate the exected number of rimes > 126 for which the conjecture fails. The most difficult iece of this heuristic argument is determining the robability that a given rime q divides N. We investigate this robability in the next section. The assumtions made in the heuristic argument are clearly labeled with the words assume or assuming. 2. How often does 2k + 1 divide N as varies? It is well known that every rime factor of N has the form 2k + 1 when is an odd rime. According to age 381 of Dickson [4], Euler roved this fact in On the following age Dickson writes that Legendre roved it again in A recent roof of a slightly more general result aears on age 642 of Sabia and Tesauri [7]. Here is a short roof. Suose q is rime and q N. The radix- exansion 1 1 N = 1 + i 1 + = 1 + ( 1) 1 mod ( 2 ) i=1 i=1

4 THE PERIOD OF THE BELL NUMBERS MODULO A PRIME 3 shows gcd(n, 2 ) = 1, whence gcd(q, 2 ) = 1. In articular q is odd, q, and q ( 1). We have 1 mod q because q N. Let d be the smallest ositive integer for which d 1 mod q. We cannot have d = 1 because q does not divide 1. But d, so d =. By Fermat s little theorem, q 1 1 mod q, so (q 1). The quotient (q 1)/ must be even because both and q are odd. Thus, q = 2k + 1. For each 1 k 50 and for all odd rimes < , we comuted the fraction of the rimes q = 2k + 1 that divide N. For examle, when k = 5 there are 1352 rimes < for which q = 2k + 1 is also rime, and 129 of these q divide N, so the fraction is 129/1352 = This fraction is called Prob in Table 2 because it aroximates the robability that q divides N, given that and q = 2k + 1 are rime, for fixed k. Table 2. Probability that (2k + 1) N Odd k Even k k 1/(2k) Prob k 1/k Prob The first observation is that usually Prob is aroximately 1/k when k is even and 1/(2k) when k is odd. The greatest anomalies to this observation in the table are that Prob is about 2/k when k = 2, 18, 32 and 50, and that Prob is about 4/k when k = 8. Note that these excetional values of k have the form 2m 2 for 1 m 5. (These numbers arise also in chemistry as the row lengths in the eriodic table of elements.)

5 4 PETER L. MONTGOMERY, SANGIL NAHM, AND SAMUEL S. WAGSTAFF, JR. We will now exlain these observations. Suose k is a ositive integer and that both and q = 2k + 1 are odd rimes. Let g be a rimitive root modulo q. If 1 mod 4 or k is even (so q 1 mod 4), then by the Law of Quadratic Recirocity ( ) = q ( ) q = ( ) 2k + 1 = ( ) 1 = +1, so is a quadratic residue modulo q. In this case g 2s mod q for some s. Now, by Euler s criterion for ower residues, (2k + 1) ( 1) if and only if is a (2k)-ic residue of 2k + 1, that is, if and only if (2k) (2s). It is natural to assume that k s with robability 1/k because k is fixed and s is a random integer. If 3 mod 4 and k is odd (so q 3 mod 4), then ( q ) = ( q ) ( ) 2k + 1 = = ( ) 1 = 1, so is a quadratic nonresidue modulo q. Now g 2s+1 mod q for some s. Reasoning as before, (2k + 1) ( 1) if and only if (2k) (2s + 1), which is imossible. Therefore q does not divide N. (This statement is equivalent to Lemma 1.1(c) of [3].) Thus, if we fix k and let run over all rimes, then the robability that q = 2k+1 divides N is 1/k when k is even and 1/(2k) when k is odd because, when k is odd only those 1 mod 4 (that is, half of the rimes ) offer a chance for q to divide N. In fact, when k = 1 and 1 mod 4, q always divides N. This theorem must have been known long ago, but we could not find it in the literature. Theorem 2.1. If is odd and q = is rime, then q divides N if and only if 1 mod 4. Proof. We have just seen that q does not divide N when 3 mod 4. If 1 mod 4, then is a quadratic residue modulo q, as was mentioned above, so = (q 1)/2 +1 mod q by Euler s criterion. Finally, q is too large to divide 1, so q divides N. We now exlain the anomalies, beginning with k = 2. Theorem 2.2. If q = is rime, then q divides N. This result was an old roblem osed and solved more than 100 years ago. In [2] it was roosed as Problem by C. E. Bickmore and solved by him, by Nath Coondoo, and by others. Here is a modern roof. Proof. Since q 1 mod 4, there exists an integer I with I 2 1 mod q. Then (1 + I) 4 (2I) mod q. Hence ( ) 1 (1 + I) 4 (1 + I) 1 q 1 mod q by Fermat s theorem. Thus, q divides 1. But q = is too large to divide 1, so q divides N. Lemma 2.3. Suose q is rime and q 1 mod 4. If the integer l divides (q 1)/4, then l is a quadratic residue modulo q.

6 THE PERIOD OF THE BELL NUMBERS MODULO A PRIME 5 Proof. The hyothesis imlies gcd(q, l) = 1. In articular l 0. Factor (2.1) l = ±l 1... l ν where each l j is rime. The hyotheses that q is rime and q 1 mod 4 imly that ±1 are quadratic residues modulo q. We claim each l j is a quadratic residue modulo q, so their roduct (2.1) (or its negative) is also a quadratic residue. If l j = 2, then l is even and q 1 mod 8. Since q is rime, 2 is a quadratic residue modulo q. If instead l j is odd, then we can use quadratic recirocity: ( ) ( ) ( ) lj q 1 = = = +1, q which comletes the roof. l j Theorem 2.4. Let be an odd ositive integer and m be a ositive integer. If q = 4m is rime, then q divides m2 1. Proof. As in the roof of Theorem 2.2, q 1 mod 4, so there is an integer I with I 2 1 mod q and (1 + I) 4 4 mod q. By Lemma 2.3, m is a quadratic residue modulo q, so 4m 2 (1 + I) 4 m 2 mod q is a fourth ower modulo q, say r 4 4m 2 mod q. Then ( ) (q 1)/4 q 1 m2 = 4m 2 (( 4m 2 ) 1 ) (q 1)/4 = r 1 q 1 mod q, which roves the theorem. Of course, Theorem 2.2 is the case m = 1 of Theorem 2.4. We now aly Theorem 2.4. As before, let g be a rimitive root modulo q and let a = g (q 1)/m2 mod q. Then a j, 0 j < m 2, are all the solutions to x m2 1 mod q. Let b = mod q. By the theorem, b m2 1 mod q, so b a j mod q for some 0 j < m 2. It is natural to assume that the case j = 0, that is, q N, haens with robability 1/m 2. In the case m = 2, that is, k = 8, we can do even better. Theorem 2.5. If q = is rime, then q divides 2 1. Proof. As in the roof of Theorem 2.2, there is an integer I with I 2 1 mod q and (1 + I) 4 4 mod q. Therefore, (1 + I) / mod q and so ( ) (1 + I) 16 (1 + I) 1 q 1 mod q, which roves the theorem. Thus, a rime q = 2k + 1 divides ( 1)( + 1) when k = 8. Assuming that q has equal chance to divide either factor, the robability that q divides 1 is 1/2. So far, we have exlained all the behavior seen in Table 2. Further exeriments with q = 2m lead us to the following result, which generalizes Theorems 2.4 and 2.5. l j

7 6 PETER L. MONTGOMERY, SANGIL NAHM, AND SAMUEL S. WAGSTAFF, JR. Theorem 2.6. Suose, m, t are ositive integers, with t a ower of 2 and t > 1. Let k = (2m) t /2 and q = 2k+1 = (2m) t +1. If q is rime, then (a) is a (2t)-th ower modulo q, and (b) k/t 1 mod q. Proof. To rove art (a), note that since q 1 mod 2 t, the cyclic multilicative grou (Z/qZ) of order q 1 has an element ω of order 2 t. Then ω 2t 1 1 mod q so I = ω 2t 2 satisfies I 2 1 mod q. Now m t = (q 1)/ (2 t ), so m is a quadratic residue modulo q by Lemma 2.3. We will show that 1 (1 q)/ = (2m) t mod q is a (2t)-th ower modulo q. If t = 2, then (2m) t (2Im) 2 = (1 + I) 4 m 2 mod q is a fourth ower modulo q. If t > 2, then t 4 because t is a ower of 2. Then (q 1)/4 = 2m((2m) t 1 /4) is divisible by 2m. Hence 2m is a quadratic residue modulo q by Lemma 2.3. Therefore, (2m) t is a (2t)-th ower modulo q. Finally, 1 is a (2 t 1 )-th ower modulo q because 2 t 1 divides (q 1)/2. Hence 1 is a (2t)-th ower modulo q because 2t 2 t 1 when t 4. For art (b), aly art (a) and choose r with r 2t mod q. Observe that 2t divides 2 t which divides q 1 = 2k. Hence, This comletes the roof. 1 r q 1 (r 2t ) 2k/2t k/t mod q. When t = 2, the theorem is just Theorem 2.4. When t = 4, Theorem 2.6 says that if q = (2m) = 16m is rime, then q divides 2m4 1. Theorem 2.5 is the case m = 1 of this statement. When t = 8, Theorem 2.6 says that if q = (2m) = 256m is rime, then q divides 16m8 1. The first case, m = 1, of this statement is for k = 128, which is beyond the end of Table 2. We now aly Theorem 2.6. As above, let g be a rimitive root modulo q and let a = g (q 1)t/k mod q. Then a j, 0 j < k/t, are all the solutions to x k/t 1 mod q. Let b = mod q. By the theorem, b k/t 1 mod q, so b a j mod q for some 0 j < k/t. It is natural to assume that the case j = 0, that is, q N, haens with robability 1/(k/t) = t/k. When k is an odd ositive integer, define c(k) = 1/2. When k is an even ositive integer, define c(k) to be the largest ower of 2, call it t, for which there exists an integer m so that k = (2m) t /2. Note that c(k) = 1 if k is even and not of the form 2m 2. Also, c(k) 2 whenever k = 2n 2 because if k = (2m) t /2 with t 2, then k = 2n 2 with n = 2 (t 2)/2 m t/2. Note that c(k) = 1/2 if k is odd, 1 if k is even and not of the form 2m 2, O(log k) if k = 2m 2 for some ositive integer m. Hence the average value of c(k) is 3/4 because the numbers 2m 2 are rare. We have given heuristic arguments which conclude that, for fixed k, when and q = 2k + 1 are both rime, the robability that q divides N is c(k)/k. Emirical evidence in Table 2 suorts this conclusion. We have exlained all the behavior shown in Table 2. We tested many other values of k and found no further anomalies beyond those listed in this section.

8 THE PERIOD OF THE BELL NUMBERS MODULO A PRIME 7 3. Is the conjecture about the eriod of the Bell numbers true? We follow, in rincile, the heuristic argument on age 386 of [9]. According to the Bateman-Horn conjecture [1], for each ositive integer k the number of x for which both and 2k + 1 are rime is asymtotically x 2C 2 f(2k) (log x) log(2kx), where C 2 = q odd rime ( 1 (q 1) 2 ), f(n) = k A<2k+1 B q n q odd rime q 1 q 2. Thus, by the Prime Number Theorem, if is known to be rime and k is a ositive integer, then the robability that 2k + 1 is rime is 2C 2 f(2k)/ log(2k). Now we aly the results of the revious section. If is rime and k is a ositive integer, then the robability that 2k + 1 is rime and divides N is (2C 2 f(2k)/ log(2k)) (c(k)/k). For a fixed rime and real numbers A < B, let F (A, B) denote the exected number of rime factors of N between A and B. Then 2C 2 f(2k)c(k) F (A, B). k log(2k) The anomalous values of c(k) occur when k is twice a square, and these numbers are rare. The denominator k log(2k) changes slowly with k. If B A is large, so that there are many k in the sum, then we may ignore the anomalies and relace c(k) by its average value 3/4. This change makes little difference in the sum. Thus, F (A, B) k A<2k+1 B k A<2k+1 B 3C 2 f(2k) 2k log(2k). Just as in the heuristic argument on age 386 of [9] we may relace C 2 f(2k) by 1 and find 3 F (A, B) 2k log(2k) 3 ( ) log B 2 log. log A We can now estimate the exected value of the number d of distinct rime factors of N. (Question: Is N always square free?) The exected value of d is F (2, N ) 3 ( ) log 2 log N = 3 ( ) log(2) 2 log log N 3 log. log (2) 2 Now we are ready to comute the robability that the conjecture holds for a rime. If the conjecture fails for, then there is a rime factor q of N such that the eriod of the Bell numbers modulo divides N = N /q. The eriod will divide N if and only if B(N + i) B(i) mod for all i in 0 i 1. Assume that the numbers B(N + i) mod for 0 i 1 are indeendent random variables uniformly distributed in the interval [0, 1]. Then the robability that the eriod divides N is because, for each i, there is one chance in that B(N + i) will have the needed value B(i) mod. The robability that the eriod does not divide N is 1. Assume also that the robabilities that the eriod divides N = N /q for different rime divisors q of N are indeendent. Then the robability that the minimum

9 8 PETER L. MONTGOMERY, SANGIL NAHM, AND SAMUEL S. WAGSTAFF, JR. eriod is N is (1 ) d, where d is the number of distinct rime factors of N. Using our estimate for d, we find that this robability is (1 ) 3(log )/2. When is large, this number is aroximately 1 (3 log )/(2 ) by the binomial theorem. This shows that the heuristic robability that the minimum eriod of the Bell numbers modulo is N is exceedingly close to 1 when is large. Finally, we comute the exected number of rimes > x for which the conjecture fails. When x > 2, this number is >x 3 log 2 < 1 x >x x t 1 x dt = x2 x x 2. By Theorem 1.1, the conjecture holds for all rimes < 126. Taking x = 126, the exected number of rimes for which the conjecture fails is < /124 < Thus, the heuristic argument redicts that the conjecture is almost certainly true. References [1] P. T. Bateman and R. A. Horn. A heuristic asymtotic formula concerning the distribution of rime numbers. Math. Com., 16: , [2] C. E. Bickmore. Problem Math. Quest. Educ. Times, 65:78, [3] M. Car, L. H. Gallardo, O. Rahavandrainy, and L. N. Vaserstein. About the eriod of Bell numbers modulo a rime. Bull. Korean Math. Soc., 45(1): , [4] L. E. Dickson. History of the Theory of Numbers, volume 1: Divisibility and Primality. Chelsea Publishing Comany, New York, New York, [5] J. Levine and R. E. Dalton. Minimum eriods, modulo, of first-order Bell exonential numbers. Math. Com., 16: , [6] W. F. Lunnon, P. A. B. Pleasants, and N. M. Stehens. Arithmetic roerties of Bell numbers to a comosite modulus I. Acta Arith., 35:1 16, [7] J. Sabia and S. Tesauri. The least rime in certain arithmetic rogressions. Amer. Math. Monthly, 116: , [8] J. Touchard. Proriétés arithmétiques de certains nombres recurrents. Ann. Soc. Sci. Bruxelles, 53A:21 31, [9] S. S. Wagstaff, Jr. Divisors of Mersenne numbers. Math. Com., 40: , [10] S. S. Wagstaff, Jr. Aurifeuillian factorizations and the eriod of the Bell numbers modulo a rime. Math. Com., 65: , [11] G. T. Williams. Numbers generated by the function e ex 1. Amer. Math. Monthly, 52: , Microsoft Research, One Microsoft Way, Redmond, WA address: montgom@cwi.nl Deartment of Mathematics, Purdue University, 150 North University Street, West Lafayette, IN address: snahm@urdue.edu Center for Education and Research in Information Assurance and Security, and Deartments of Comuter Science and Mathematics, Purdue University, 305 North University Street, West Lafayette, IN address: ssw@cerias.urdue.edu