On the Distribution of Perfect Powers

Transcription

1 Journal of Integer Sequences, Vol. 4 (20), rticle.8.5 On the Distribution of Perfect Powers Rafael Jakimczuk División Matemática Universidad Nacional de Luján Buenos ires rgentina jakimczu@mail.unlu.edu.ar In memory of my sister Fedra Marina Jakimczuk ( ) bstract Let N() be the number of erfect owers that do not eceed. In this article we obtain asymtotic formulae for the counting function N(). Introduction natural number of the form m n where m is a ositive integer and n 2 is called a erfect ower. The first few terms of the integer sequence of erfect owers are, 4, 8, 9, 6, 25, 27, 32, 36, 49, 64, 8, 00, 2, 25, 28,..., and they are sequence in Sloane s Encyloedia. Let N() be the number of erfect owers that do not eceed. M.. Nyblom [3] roved the following asymtotic formula, N(). M.. Nyblom [4] also obtained a formula for the eact value of N() using the inclusioneclusion rincile (also called the rincile of cross-classification). In this article we obtain more recise asymtotic formulae for the counting function N(). For eamle, we rove N() o ( 7 ). Consequently + 3 < N() <

2 2 Preliminary Results Let be a set. The number of elements in we denote in the form. We need the following results. Lemma. (Inclusion-eclusion rincile) Let us consider a given finite collection of sets, 2,..., n. The number of elements in n i i is n i i n ( ) k+ i ik, k i < <i k n where the eression i < < i k n indicates that the sum is taken over all the k-element subsets {i,...,i k } of the set {, 2,...,n}. Proof. See for eamle either [, age 233] or [2, age 84]. Let n () (n 2) be the set {k n : k N,k n }, that is, the set of erfect owers whose eonent is n that do not eceed. Lemma 2. We have where. denotes the integer-art function. n () n, Proof. We have k n k n. M.. Nyblom [4] roved the following Lemma and the following Theorem. Lemma 3. For any set consisting of m 2 ositive integers {n,...,n m } all greater than unity, we have the set equality m ni () [n,...,n m](), i where [n,...,n m ] denotes the least common multile of the m integers n,...,n m. Let n be the n-th rime. Consequently we have, 2, 2 3, 3 5, 4 7, 5, 6 3,... Theorem 4. If 4 and, 2,..., m denote the rime numbers that do not eceed, then for k m the number of erfect owers that do not eceed is k N() i (). Besides i () {} for i m +. i 2

3 We also need the following two well-known results. The binomial formula n ( ) n (a + b) n a n k b k, () k k0 and the following roerty of the absolute value where a,a 2,...,a r are real numbers. a + a a r a + a a r, (2) 3 Main Results Theorem 5. Let n be the n-th rime number with n 2, where n is an arbitrary but fied ositive integer. Then n N() ( ) k+ k i < <i k n i ik < n i ik + g() n, (3) where lim g() and the inner sum is taken over the k-element subsets {i,...,i k } of the set {, 2,...,n } such that the inequality i ik < n holds. Proof. Let k + +, where 4. If, 2,..., m denote the rime log numbers that do not eceed, then we have < < m < + k m+ < m+2 < (4) Note that if i m + we have < i k < 2. That is, < i < 2. Consequently if i m + (see Lemma 2) i (). That is, i () {}. Note also that k and m are increasing functions of. On the other hand n 2 is an arbitrary but fied ositive integer. Equation (4) gives m < k. (5) On the other hand Therefore (5) and (6) give and consequently log m < m. (6) m < k, (7) m < k. (8) There eist three ossible relations between m, k, n and n + and consequently between m, k, n and n+. 3

4 First relation. and hence Second relation. and hence Third relation. and hence m < k n < n + m < k n < n+. m n < n + k m n < n+ k. n < n + m < k n < n+ m < k. If we define S() ma{n +,k} then these three relations, Theorem 4 and Lemma 2 give us ( 4) i () N() < S() i () + i () i i in+ S() i () + i i in+ i () + (S() n) n+. (9) Note that there eists 0 such that if 0 the third relation holds. Note also that S() n is either equal to or k n < k either case S() n log i log log, and so in as 4. Consequently (9) and (0) give i () N() < i () + log n+. () i Lemma gives n i () ( ) k+ i k On the other hand, Lemma 3 gives i i < <i k n (0) i () ik (). (2) i () ik () [i,..., ik ] () i ik (). 4

5 Therefore (Lemma 2) we obtain i () ik () i ik () Substituting (3) into (2) we find that n i () ( ) k+ i k n ( ) k+ k n ( ) k+ k i < <i k n i < <i k n i < <i k n i ik i ik ( i ik i ik. (3) ) i ik. (4) Now 0 i ik i ik <. (5) Consequently (see () and (2)) n ( ) ( ) k+ i ik i ik n k i < <i k n k i < <i k n n ( ) n n ( ) n ( + ) n 2 n. (6) k k That is, k k0 n ( ) k+ k i < <i k n ( Equations (4) and (7) give n i () ( ) k+ i k i ik i < <i k n ) i ik O(). (7) i ik + O() (8) If and C() B() n ( ) k+ k n ( ) k+ k i < <i k n i ik < n+ i < <i k n i ik > n+ i ik i ik (9) ) o ( n+ (20) 5

6 then n ( ) k+ k i < <i k n i ik B() + C(). (2) Equations (8) and (2) give i () B() + C() + O(). (22) Equations (22) and () give Therefore, i B() + C() + O() N() < B() + C() + O() + log n+. Equations (23) and (20) give C() + O() N() B() < C() + O() + log n+. (23) ǫ < N() B() < + ǫ (ǫ > 0). log n+ That is, ) N() B() + O (log n+. (24) Note that (see (9)) if k,...,n then, i < <i k n i ik < n+ Now, if k,...,n then and if k 2,...,n then i ik i < <i k n i ik < n i < <i k n n i ik < n+ i < <i k n i ik < n i ik i ik i ik + i < <i k n i ik < n i < <i k n n< i ik < n+ i < <i k n n i ik < n+ i ik. (25) i ik, (26) i ik (27) On the other hand, if k then i n n i < n+ i n (28) 6

7 Equations (9), (25), (26), (27) and (28) give That is, B() n ( ) k+ k + n + i < <i k n i ik < n n ( ) k+ k2 i ik i < <i k n n< i ik < n+ i ik B() n ( ) k+ k + n + o ( n i < <i k n i ik < n ) i ik (29) Finally, equations (29) and (24) give (3). 4 Eamles If n 2 then Theorem 5 becomes where lim g(). If n 3 then Theorem 5 becomes where lim g(). If n 4 then Theorem 5 becomes N() + g() 3, (30) N() g() 5, N() g() 7, where lim g(). Consequently + 3 < N() < If n 5 then Theorem 5 becomes N() g(), where lim g(). To finish, we shall establish two simle theorems. 7

8 Theorem 6. Let us consider the n oen intervals (0, 2 ), ( 2, 2 2 ),...,((n ) 2,n 2 ). Let S(n) be the number of these n oen intervals that contain some erfect ower. Then S(n) lim n n 0. Therefore, almost all the oen intervals are emty. Proof. We have (Nyblom s asymtotic formula) where lim f() 0. Consequently N() + f(), N(n 2 ) n + f(n 2 )n, where n are the n squares 2, 2 2,...,n 2. Therefore 0 S(n) N(n 2 ) n f(n 2 )n. That is 0 S(n) n f(n2 ). Using equation (30) we can obtain a more strong result. Theorem 7. Let us consider the n oen intervals (0, 2 ), ( 2, 2 2 ),...,((n ) 2,n 2 ). Let F(n) be the number of erfect owers in these n oen intervals. Then F(n) n 2 3. Proof. Equation (30) gives N(n 2 ) n + g(n 2 )n 2 3, where n are the n squares 2, 2 2,...,n 2. Therefore F(n) N(n 2 ) n g(n 2 )n 2 3 n cknowledgements The author would like to thank the anonymous referee for his/her valuable comments and suggestions for imroving the original version of this article. The author is also very grateful to Universidad Nacional de Luján. 8

9 References [] G. H. Hardy and E. M. Wright, n Introduction to the Theory of Numbers, Oford, 960. [2] W. J. LeVeque, Toics in Number Theory, ddison-wesley, 958. [3] M.. Nyblom, counting function for the sequence of erfect owers, ustral. Math. Soc. Gaz. 33 (2006), [4] M.. Nyblom, Counting the erfect owers, Math. Sectrum 4 (2008), Mathematics Subject Classification: Primary 99; Secondary B99. Keywords: Perfect owers, counting function, asymtotic formulae. (Concerned with sequence ) Received May 28 20; revised version received ugust Published in Journal of Integer Sequences, Setember Return to Journal of Integer Sequences home age. 9