250 G. ALKAUSKAS and A. DUBICKAS Baker and Harman [2] proved that the sequence [ο p ] (where p runs over the primes) contains infinitely many prime nu

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1 Acta Math. Hungar. 105 (3) (2004), PRIME AND COMPOSITE NUMBERS AS INTEGER PARTS OF POWERS G. ALKAUSKAS and A. DUBICKAS Λ (Vilnius) Abstract. We studyprime and composite numbers in the sequence of integer parts of powers of a fixed real number. We first prove a result which implies that there is a transcendental number ο>1 for which the numbers [ο n! ], n =2; 3;:::; are all prime. Then, following an idea of Huxleywho did it for cubics, we construct Pisot numbers of arbitrarydegree such that all integer parts of their powers are composite. Finally, we give an example of an explicit transcendental number (obtained as the limit of a certain recurrent sequence) for which the sequence [ n ], n =1; 2;:::;has infinitelymanyelements in an arbitraryinteger arithmetical progression. 1. Introduction One of the oldest questions in number theory is to determine which integer sequences contain infinitely many prime numbers. In this paper, we are interested in sequences which are integer parts of powers of a fixed real number [ο n ], n 2 N. Usually, the sparser the sequence is the more difficult the problem becomes, but sometimes the situation is precisely opposite. The following theorem is a generalization of an old result of Mills [9] who showed that there is a w>1such that the numbers [w 3n ], n 2 N, are all prime. (See also [10] for another prime representing function.) Theorem 1. Let A 1 =1<A 2 <A 3 < be a sequence of positive integers satisfying A n > 2:1053A for n =2; 3;::: : Then there exists w>1 such that [w An ], n 2 N, are all prime. If, in addition, lim sup n!1 A n =A = 1 then w can be chosen to be transcendental. Corollary. There is a transcendental number w>1 such that the numbers [w n! ], n =2; 3;:::; are all prime. Λ This research was partially supported by thelithuanian State Science and Studies Foundation. Key words and phrases: transcendental numbers, Pisot numbers, prime and composite numbers Mathematics Subject Classification: 11A41, 11J81, 11R /4/$ cfl 2004 Akadémiai Kiadó, Budapest

2 250 G. ALKAUSKAS and A. DUBICKAS Baker and Harman [2] proved that the sequence [ο p ] (where p runs over the primes) contains infinitely many prime numbers for almost all ο>1. On the other hand, there are uncountably many ο such that [ο n ] are composite for all n 2 N [2]. Their method does not allow to give an explicit value for such ο. Nevertheless, Huxley [2, p. 80] pointed out that, for every n 2 N, [ 1=(2 1=3 1) 2 n] is either 6k + 2 or 6k + 3 with k 2 N, and so it is composite. In this direction, we construct algebraic numbers ff of arbitrary degree d such that [ff n ], n 2 N, are all even and so composite. (Evidently, one can take ff = 4 for d = 1.) Recall that ff>1 is a Pisot numberifitisan algebraic integer whose conjugates are all smaller than 1 in absolute value. Theorem 2. Let d = 2 be an integer and let ff be the largest positive root of z d 3Bz d 1 +2Bz d 2 +2=0, where B>3(2d) d 2 is an odd integer. Then ff is a Pisot number and the numbers [ff n ], n 2 N, are all even and = 14; so they all are composite. A result of Koksma [8] implies that the sequence [ο n ], n 2 N, contains infinitely many composite numbers for almost all ο>1. However, just a few explicit such ο are known. Forman and Shapiro [6] proved that the sequences (3=2) nλ and (4=3) nλ (n 2 N) both contain infinitely many composite numbers (see also Problem E19 in [7]). The second named author [5] proved that this is also the case with the sequence [ff n ], n 2 N, where ff is a Pisot or a Salem number. (Earlier, Cass [3] proved this result for quadratic Pisot numbers.) All these explicit examples are with algebraic numbers: no explicit transcendental numbers ο for which [ο n ] is composite for infinitely many n were known. Our next result gives an example of such anumber (obtained as the limit of a certain recurrent sequence). Using the lemma (see Section 2) below one can construct many similar examples. Theorem 3. (i) Let x 1 =2 and x n = x n +2 for n =2; 3;::: : Then lim n!1 x 1=n! n = ο = 2: ::: exists, it is transcendental, and the numbers [ο n! ] are even for every n 2 N. (ii) Let x 1 =2and x n = x n + n 1 for n =2; 3;::: : Then lim n!1 x 1=n! n = ν =2: ::: exists, it is transcendental, and the sequence [ν n ], n 2 N, contains infinitely many elements divisible by an arbitrary prime number p. Part (ii) of the theorem proves much more than required for the problem concerning composite numbers. However the next theorem is even more general. We construct a universal" explicit such that the set [ n ], n 2 N, has an infinite intersection with an arbitrary integer arithmetical progression. Theorem 4. Let a n be a sequence of positive integers such that a 1 =1 and a n = n a 1:::a for n =2; 3;::: : If x 1 =2 and x n = n!x an + n for

3 PRIME AND COMPOSITE NUMBERS 251 n =2; 3;:::; then lim n!1 x 1=a 1:::a n n = exists. The number is transcendental and the sequence [ n ], n 2 N, has infinitely many elements in an arbitrary integer arithmetical progression. In case if a n in the theorem is defined by a n = n a 1:::a for every n = 2 we have = 3: ::: : We conclude with the most general theorem which shows that arithmetical progressions can be replaced by arbitrary infinite sets of positive integers. Naturally, as in case of Theorem 1, this is an existence result. Theorem 5. Let L k = fs k1 <s k2 <s k3 < g, k 2 N, be a countable set of arbitrary infinite sequences of positive integers. Then there is a transcendental number fl such that [fl n ], n 2 N, has infinitely many elements in every L k. On the other hand, there exists a countable set of infinite sequences L k, k 2 N, such that for any algebraic fl>1the set [fl n ], n 2 N, has no elements in at least one L k. Indeed, let fi i, i =1; 2;:::; be all real algebraic numbers greater than 1. Φ For every i, let k i be a positive integer Ψ such that fi k i i (fi i 1) > 2. Set L i = [fi k i i ]+1;[fi k i+1 i ]+1;[fi k i+2 i ]+1;:::. If fl = fi k then no element of[fl n ], n 2 N, belongs to L k. In the next section we give the proof of our main lemma which is used for all other proofs (except for the proof of Theorem 2 which is based on different ideas). In Section 3 we prove the theorems. 2. Transcendency of the limit of a recurrent sequence Lemma. Let a 1 =1and let a 2 ;a 3 ;::: be a sequence of rational numbers greater than 1 such that A n = a 1 :::a n 2 N for every n 2 N. Suppose that x 1 ;x 2 ;x 3 ;::: are positive integers such that x an 5 x n 5 x an + a nx bn 1; where b n are some real numbers satisfying 0 5 b n 5 a n 1 for n =2; 3;::: : Then lim n!1 x1=an n = ο; exists and [ο An ]=x n for every n 2 N. Moreover, if x an <x n for infinitely many n and lim sup n!1 (a n b n ) = 1, then ο is a transcendental number.

4 252 G. ALKAUSKAS and A. DUBICKAS Proof. Clearly, the sequence x 1=An n is nondecreasing in n, whereas the sequence (1 + x n ) 1=An is decreasing, because (1 + x ) an >x an + a nx a = x an + a nx bn = 1+x n : Since the nth element of the first sequence is smaller than the nth element of the second, every element of the first sequence is strictly smaller than an arbitrary element of the second. In particular, every element of the first sequence is bounded above by x 1 +1, whereas every element of the second sequence is bounded from below by x 1. Both sequences thus tend to certain limits, say, ο and ο Λ. The inequalities x 1=An n 5 ο 5 ο Λ < (1 + x n ) 1=An imply that x n 5 ο An < 1+x n for every n 2 N, so[ο An ]=x n. We are left to show that with the additional conditions ο is transcendental. For the contradiction, assume that ο is an algebraic number of Q degree d. Recall that the Mahler measure M(ο) of ο is defined as u(ο) max i f1; jο(i) jg, where the product runs over every conjugate ο (i) of ο and where u(ο) 2 N is the leading coefficient of the minimal polynomial of ο. Consider the positive algebraic number fi n = x 1=A. Let D be the degree of fi n over Q and let fi (j) n, 1 5 j 5 D, be conjugate to fi n. They are all of equal modulus. Clearly, fi n <ο. Hence no conjugate of fi n is equal to ο. Consider the resultant ofο and fi n (which, by the above, is nonzero) R(ο;fi n )=u(ο) D u(fi n ) d Y i;j jο (i) fi (j) n j; where the product which contains Dd terms is taken over every pair of conjugates and where n is so large that fi n = x 1=A >ο ffi>1+ffi for some absolute positive constant ffi. The polynomial X A x vanishes at fi n. Hence it is divisible by the minimal polynomial of fi n and D 5 A. By the multiplicative property of the Mahler measure for polynomials, we have M(fi n ) 5 x. Since ο>fi n,by estimating every difference jο (i) fi (j) n j (except for jο fi n j)by 2 max f1; jο (i) jg max f1; jfi (j) n jg, we deduce that 1 5 fi fi R(ο;fin ) fi fi 5 jο fin j2 dd 1 M(ο) D M(fi n ) d < jο fi n j 2 d M(ο) A x d : Using the inequality (1+x n )x an 5 1+a n x bn an,we obtain that 0 < jο fi n j = ο fi n < (1 + x n ) 1=An x 1=A

5 PRIME AND COMPOSITE NUMBERS 253 = fi n( (1 + xn )x an 1=An 1) 5 fi n( Set c =2 d M(ο). It follows that <fi n (a n =A n )x bn an 5 fi n x bn an : 1 <fi n x bn an+d c A = fi n fi b n a n+d n 1+an x bn an 1=An 1) A c 5 fi bn an+d+1 A n c : But fi n > 1+ffi and lim sup n!1 (a n b n )=1, so the right hand side will be smaller than 1 for certain large integer n, a contradiction. 3. Proofs Proof of Theorem 1. Let x 1 be a sufficiently large prime number. It was shown in [1] that there is always a prime in the interval (x; x + x 0:525 ) for every sufficiently large x. Set a n = A n =A and b n =0:525a n. Then a n 1 > 0:525a n = b n. It follows that there is a sequence of prime numbers x 1 <x 2 <x 3 < such that x an <x n <x an + a nx bn 1: By applying the lemma we get the first part. The second part also follows from the lemma, because lim sup n!1 (a n b n ) = lim sup n!1 0:475A n =A = 1. Λ Proof of Corollary. Set A n =(n + 1)!=2 and replace w by w 2 in Theorem 1. Λ Proof of Theorem 2. Set P (z) =z d 3Bz d 1 +2Bz d On the circle jzj =1=2d, j2bz d 2 j > jz d 3Bz d 1 +2j, sobyrouché's theorem P (z) has d 2 roots in the circle jzj 5 1=2d, say ff 3 ;:::;ff d. Note that P (2=3) > 0 and P d 21B 1 1 d 2 4d 4d 8 4d +2B 1 1 d 2 +2< 3 5B 4d d 2 4d which is negative (check this for d =2 and for d = 3). Hence the polynomial P (z) has a root in the interval (2=3; 1 1=4d), say, ff 2. Similarly, using

6 254 G. ALKAUSKAS and A. DUBICKAS P (3B 1) = 2 (B 1)(3B 1) d 2 < 0 and P (3B) > 0, we see that it has a root in (3B 1; 3B) which we denote by ff = ff 1. Since ff is a unique root of P (z) outside the unit circle, it is a Pisot number. (Moreover, it is a strong Pisot number in the sense of [4].) It follows that [ff] =3B 1iseven and = 14. Set S k = ff k + ff k 2 + ffk ffk d. We claim that [ffk ]=S k 1 for every k = 2. Indeed, jff k 3 + +ffk d j 5 (d 2)(2d) k and (2=3) k <ff k 2 < (1 1=4d)k. It is easy to check that (d 2)(2d) k < (2=3) k and (d 2)(2d) k +(1 1=4d) k < 1 for k = 2, so S k ff k is positive and < 1 which completes the proof of the claim, because S k 2 N. Finally, we will show that S k, k =2; 3;:::; are all odd. Write P (z) = z d ff 1 z d 1 + ff 2 z d 2 +( 1) d ff d. By Waring's formula, we can express S k as S k = X `k1;:::;k d ff k 1 1 :::ffk d d ; where k 1 +2k dk d = k, `k;0;:::;0 = 1 and `k1;:::;k d are integers. Every term in the above expression except for ff1 k contains at least one ff j with j = 2. But ff j,25 j 5 d, arealleven, so S k ff k 1 = S k (3B) k is even. Hence S k is odd for every k = 2. Λ Remark 1. By the same method we can show, for instance, that the numbers h 5+ p 17 2 n i, n 2 N, are all even. However it is not known whether there h are finitely or infinitely many primes in the Fibonacci sequence f n = 1p p n i , n 2 N. 2 Proof of Theorem 3. Part (i) follows immediately from the lemma setting a n = n, b n =1. Then A n = n! and [ο n! ]=x n are all even. For (ii), setting a n = n, b n =0,A n = n! and applying the lemma, we see that lim n!1 x 1=n! n = ν is transcendental. Since [ν n! ]=x n, it suffices to show that the sequence x 1 ;x 2 ;x 3 ;::: contains infinitely many elements divisible by an arbitrary prime number p. Set n = p(p 1)k, where k 2 N. We claim that either x or x n is divisible by p. Indeed, if x is not divisible by p, then by the Little Fermat Theorem x n = xp(p 1)k is equal to 1 modulo p. Hence x n =(x n 1) + p(p 1)k is divisible by p. Λ Proof of Theorem 4. We cannot apply the lemma directly because of the factor n!, but follow its proof. Set A n = a 1 :::a n. Firstly, the sequence x 1=An n is increasing, because x n >x an. Secondly, the sequence (1 + x n) 1=An is decreasing for n = 6. Indeed, we have that

7 PRIME AND COMPOSITE NUMBERS x n =1+n!x an + n 5 2n!xan <nn x an : We need to show that this is less than (1 + x ) an which would follow from the inequality a n > 2x n log n. In order to prove this inequality note that x n 5 2n!x an < (2x ) an. Hence log x n <a n (log 2 + log x ), giving log x n <A n (1 + 1=A 1 +1=A =A ) log 2 <A n log 6: Consequently, x n < 6 An. We are left to show that a n = n A > 6 A 2n log n for n = 7 which follows from A > 13n log n (for n = 7). This proves that lim n!1 x 1=An n = exists. By the above >x A. In order to show that is transcendental it suffices to prove that, given any absolute positive constants c and d, ( x 1=A )c A x d < 1 for n sufficiently large. Note that x 1=A < 6 (2n!) 1=An 1 which is less than n 2 =A n < 1=a n for n sufficiently large. But c A x d < (c6d ) A <n A 5 an for large n, giving the required inequality. This shows that is transcendental and [ An ]=x n for every n = 6. Let uy + v, y =1; 2;:::; be an integer arithmetical progression with integers u>0 and v = 0. Setting n = uy + v with y sufficiently large, we see that x n = n!x an + n is of the form uy0 + v, where y 0 2 N, so it belongs to the same arithmetical progression. This completes the proof. Remark 2. The inequality a n = n a 1:::a for the sequence a n, n =2; 3;::: (in Theorem 4) can be replaced by lim n!1 (log a n )=a 1 :::a = 1. Proof of Theorem 5. Take x 1 2 L 1, x 2 2 L 1, x 3 2 L 2, x 4 2 L 1, x 5 2 L 2, x 6 2 L 3 ;:::; where x an <x n <x an+2. Here a 1 =1, a n, n =2; 3;:::; is a sequence of positive integers such that a n = n a 1:::a and x an+2 < (1 + x ) an 1. The possibility tochoose x n in any L k exists, because the intervals (x s ;xs+2 ), s = N;N +1;N +2;:::; overlap and their union contains all sufficiently large integers. Setting A n = a 1 :::a n, we see that the sequence x 1=An n, n =1; 2;:::; is increasing and the sequence (1 + x n ) 1=An, n =1; 2;:::; is decreasing. Set fl = lim n!1 x 1=An n. Clearly, x 1 <fl<x Since Λ fl<(1 + x n ) 1=An < 1+x an+2 1=An <x (an+3)=an = x 1=A 1+3=an

8 256 G. ALKAUSKAS and A. DUBICKAS: PRIME AND COMPOSITE NUMBERS and x 1=A <x 1 +1,wehave that fl x 1=A < (x 1 +1) (x 1 +1) 3=an 1 : For n sufficiently large, this is less than c 0 =a n, where c 0 is an absolute constant. As in the proof of Theorem 4, c A x d <ga with an absolute ca x d constant g. Hence 0 < fl x 1=A < 1 for n sufficiently large, so fl cannot be algebraic. Thus it is transcendental which completes the proof. Λ References [1] R. C. Baker, G. Harman and J. Pintz, The difference between consecutive primes. II, Proc. London Math. Soc., 83 (2001), [2] R. C. Baker and G. Harman, Primes of the form [c p ], Math. Zeitschr., 221 (1996), [3] D. Cass, Integer parts of powers of quadratic units, Proc. Amer. Math. Soc., 101 (1987), [4] A. Dubickas, A note on powers of Pisot numbers, Publ. Math. Debrecen, 56 (2000), [5] A. Dubickas, Integer parts of powers of Pisot and Salem numbers, Archiv der Math., 79 (2002), [6] W. Forman and H. N. Shapiro, An arithmetic propertyof certain rational powers, Comm. Pure Appl. Math., 20 (1967), [7] R. K. Guy, Unsolved Problems in Number Theory, Springer Verlag (New York, 1994). [8] J. F. Koksma, Ein mengen-theoretischer Satz über Gleichverteilung modulo eins, Compositio Math., 2 (1935), [9] H. W. Mills, A prime representing function, Bull. Amer. Math. Soc., 53 (1947), 604. [10] E. M. Wright, A prime representing function, Amer. Math. Monthly, 58 (1951), (Received May13, 2003) DEPARTMENT OF MATHEMATICS AND INFORMATICS VILNIUS UNIVERSITY NAUGARDUKO 24 LT VILNIUS LITHUANIA S: GIEDRIUS.ALKAUSKAS@MAF.VU.LT ARTURAS.DUBICKAS@MAF.VU.LT