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Explicit Bounds for the Sum of Reciprocals of Pseudoprimes and Carmichael Numbers

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Explicit Bounds for the Sum of Reciprocals of Pseudoprimes and Carmichael Numbers 1 2 Journal of Integer Sequences, Vol. 20 (2017), 3 Article 17.6.4 47 6 23 11 Explicit Bounds for the Sum of Reciprocals of Pseudoprimes and Carmichael Numbers Jonathan Bayless and Paul Kinlaw Husson University 1 College Circle Bangor, ME 04401 USA [email protected] [email protected] Abstract From a 1956 paper of Erd˝os, we know that the base-two pseudoprimes and the Carmichael numbers both have a convergent sum of reciprocals. We prove that the values of these sums are less than 33 and 28, respectively. 1 Introduction By Fermat’s little theorem, if p is a prime number then for all a Z we have ap a (mod p). However, a number p satisfying this property need not be a∈ prime. For all≡a Z, a base-a Fermat pseudoprime (or briefly, an a-pseudoprime) is a composite number ∈n such that gcd(a,n)=1 and an a (mod n). A Carmichael number is an odd composite number n for which an a (mod≡n) for all integers a, and so is an a-pseudoprime for all a with gcd(a,n)=1. ≡ Let P = 341, 561, 645, 1105,... be the set of 2-pseudoprimes, also called Poulet or 2 { } Sarrus numbers, and let P2(x)= n P2 : n x be the corresponding counting function. Let C = 561, 1105, 1729,... be|{ the∈ set of≤ Carmichael}| numbers and write its counting function as{ C(x) = n C :}n x . Erd˝os [11] proved that for sufficiently large x, we have |{ ∈ ≤ }| P2(x) < x/ exp(c1 log x log log x) p 1 and c log x log log log x C(x) < x/ exp 2 log log x for constants c1,c2 > 0. This implies that both sets have asymptotic density zero, as well as the stronger statement that both sets have a bounded sum of reciprocals. Pomerance [18] improved these bounds, showing that log x log log log x P (x) < x/ exp 2 2loglog x for all sufficiently large x, and that the constant c2 in the above bound on C(x) may be taken as 1. In the other direction, it is well known that there are infinitely many pseudoprimes with respect to a given base. In 1994 Alford, Granville and Pomerance [1] proved that there are infinitely many Carmichael numbers. In particular, they proved that C(x) > x2/7 for sufficiently large x. Their work has since been improved by Harman [13] to show that for large x, the inequality C(x) >xα holds for some constant α> 1/3. In this paper we determine explicit upper and lower bounds for the sum of reciprocals of 2-pseudoprimes, as well as for the sum of reciprocals of Carmichael numbers. This extends previous work [4, 5, 3, 14] on reciprocal sums. See [5] for a discussion of reciprocal sums and their importance in number theory. 2 Preliminary lemmas We state several preliminary lemmas. Throughout the paper m,n and k denote positive integers, p denotes a prime number, π(x) is the prime counting function, log x denotes the natural logarithm and P (n) denotes the largest prime factor of n. Put x0 = exp(100) and y0 = exp(10). We begin with Korselt’s criterion [7, Thm. 3.4.6] which provides a useful characterization of Carmichael numbers. Lemma 1 (Korselt’s Criterion). A positive integer n is a Carmichael number if and only if it is composite, squarefree, and for each prime p n we have p 1 n 1. | − | − It follows from Korselt’s criterion that Carmichael numbers are odd and have at least three different prime factors. The following lemma involves certain divisibility properties possessed by all 2-pseudoprimes. In particular, it shows that if a 2-pseudoprime n is divisible by p2 for a prime p, then p must be a Wieferich prime, that is, p 1 2 2 − 1 (mod p ). ≡ It follows that p is greater than or equal to 1093, the smallest Wieferich prime, see [7, p. 31]. Lemma 2. Let n be a 2-pseudoprime. If p2 n then p is a Wieferich prime. Furthermore, if 7 n then n 1 (mod 3). | | ≡ 2 2 2 2 Proof. Suppose that p n and let k be the order of 2 in (Z/p Z)×. Thus k ϕ(p )= p(p 1). n 1 | n 1 2 | − Now 2 − 1 (mod n), so 2 − 1 (mod p ), and thus k n 1, so k does not divide p. ≡ p 1 ≡ 2 | − Thus k p 1, so that 2 − 1 (mod p ). This completes the proof of the first assertion. | − ≡ n 1 n 1 For the second assertion, we have 2 − 1 (mod n), so if 7 n then 2 − 1 (mod 7). The order of 2 modulo 7 is 3, so 3 n 1. ≡ | ≡ | − We will also use explicit versions of several classical theorems. The following modification of [2, Thm. 3.2] gives fairly sharp explicit bounds on the partial sums of the harmonic series. Lemma 3. For all x 1, ≥ 1 1 (log x + γ) < n − x n x ≤ X where γ =0.5772156649 ... denotes Euler’s constant. We will also use the following modification of [3, Lem. 6.2] to bound the sum of reciprocals of numbers in a certain interval and residue class. 9 Lemma 4. Let a Z and d =2 p2, where p denotes the i-th prime number. We have ∈ i i i=2 Y 1 56.25587 < . n d n a(d) 19X≡ 10 <n x0 ≤ Proof. Let b = (1019 d + 1)/d and c = x /d. Without loss of generality, we may assume − 0 that a 0,...,d 1 . By Lemma 3 we have ∈ { − } 1 1 1 1 56.25587 100 log(1019 d +1)+ + < . dk + a ≤ d − − c b d b<k c X≤ We will use Dusart’s bounds [10, Thm. 5.6] on the sum of reciprocals of prime numbers up to x. Lemma 5. For all x 2278383, we have ≥ 1 0.2 (log log x + B) < , p − 3 p x log x ≤ X where B =0.2614972128476 ... denotes the Mertens constant. We will also use Dusart’s bounds [10, Cor. 5.2 & Thm. 5.9]. 3 Lemma 6. For all x> 1 we have x 1 2.53816 π(x) 1+ + , ≤ log x log x log2 x and for all x 599 we have ≥ x 1 π(x) 1+ . ≥ log x log x Lemma 7. For all x 2278382, we have ≥ 1 exp( γ) 0.2 1 − 1 . − p ≥ log x − log3 x p x Y≤ We will also use the following result [9, Lem. 9.6]. Lemma 8. Let f be a multiplicative function such that f(n) 0 for all n, and such that there exist constants A and B such that for all x> 1, we have ≥ f (pα) f(p)log p Ax and log pα B. (1) ≤ pα ≤ p x p α 2 X≤ X X≥ Then, for x> 1, we have x f(n) f(n) (A + B + 1) . ≤ log x n n x n x X≤ X≤ The following lemma makes the implied constants in [9, Lem. 9.7] explicit and modifies [3, Lem. 2.4]. Lemma 9. Let f be a multiplicative function such that 0 f(pα) exp 2α for all primes ≤ ≤ 3 p and integers α 1, and such that f (pα) = p2α/(3 log y) for all p y. Then for all x x 0 and y y , we have≥ ≤ ≥ ≥ 0 1 f(pα) f(n) 9.68765388x 1 . ≤ − p pα n x p x α 0 ! X≤ Y≤ X≥ Proof. We wish to apply Lemma 8, and so we first establish values of A and B to use in inequality (1) above. We have f(p)log p exp(2/3)θ(x) 1.00000075 exp(2/3)x 1.94773551x, (2) ≤ ≤ ≤ p x X≤ using the bound θ(x) < 1.00000075x for all x> 0 [17, Cor. 2]. We also have y y and ≥ 0 f (pα)= p2α/(3 log y) 4 whenever p y. For brevity, let g(t)=(f(t) log t)/t. We have ≤ · g (pα)= g (pα)+ g (pα) . p α 2 p<y0 α 2 p>y0 α 2 X X≥ X X≥ X X≥ Starting to bound the sum with p = 2, observe that α/15 α α(log 2) 2 1/15 1 α g (2 ) · = log2 α 2 − ≤ 2α α 2 α 2 α 2 X≥ X≥ X≥ < 1.23661725. Here we used the fact that 2 α 1 2r r αr − = − (1 r)2 α 2 X≥ − for r < 1. By similar reasoning, | | g (pα) < 1.25448156. 3 p<y0 α 2 ≤X X≥ Let h(t)=(2 e2/3/t)(log t)/(t e2/3)2. By partial summation we have − − (2 e2/3/p)log p g (pα) e4/3 − ≤ (p e2/3)2 p>y0 α 2 p>y0 X X≥ X − 4/3 ∞ = e h(y )π(y ) π(t)h′(t) dt − 0 0 − Zy0 e2/3 ∞ π(t) 2(2 )log t dt <e4/3 h(y )π(y )+ · − t − 0 0 (t e2/3)3 Zy0 − ! e2/3 ∞ 2t 1.1253816(2 ) dt <e4/3 h(y )π(y )+ · − t − 0 0 (t e2/3)3 Zy0 − ! 10 2/3 4/3 1.1253816 4e 3e = e 0.00010166 + · −2 − (e2/3 e10) ! − < 0.00038973. Here we used Lemma 6. Furthermore, for all x x we have ≥ 0 1 1 1.781072775 1 .
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