Upper Bounds for Partitions into k-th Powers Elementary Methods

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1 Int. J. Contemp. Math. Sciences, Vol. 4, 2009, no. 9, Upper Bounds for Partitions into -th Powers Elementary Methods Rafael Jaimczu División Matemática, Universidad Nacional de Luján Buenos Aires, Argentina Abstract Let N (x = [x] i= p (i, where p (i is the number of partitions of i into -th powers. Therefore p (n =p(n is the number of partitions of n in positive integers, p 2 (n is the number of partitions of n in squares, and so. E. M. Wright wrote three papers on the asymptotics of particular generating functions... The point I wish to mae in the section is that Wright s third paper on partitions into powers IS UNIQUE in the history of this subject. Its starting point and fundamental philosophy are different from anything that has come before or since (G. E. Andrews []. E. M. Wright [3] obtained beau asymptotic expansions for p (n using analitical methods, these formulas are not simple. They have a main term ( a series and an error term O(exp(c + n, where the error term is of exponentially lower order of magnitude than the main term. G. E. Andrews [] explores the asymptotic expansions of E. M. Wright and mentions that Wright suggest that each of the main terms in his expansions has order exp ( c + 3 n + c 4 log n In this article we prove using very elementary methods the simple inequality N (x < exp(c log x + x + c 2 log x where c and c 2 are positive constants. We also prove the simple inequalities p (n < exp(c log n + n + c 2 log n <e n + ɛ Mathematics Subject Classification: P8 Keywords: Partitions, upper bounds, -th powers <e n

2 434 R. Jaimczu Introduction. Preliminary theorems Let r n be a sequence of positive numbers such that: r <r 2 <r 3 <... ( r n (2 Let us consider the infinite linear inequality ( x fixed r i x i x (x 0 (3 i= A solution to this inequality is a vector (x,x 2,..., where the x i (i =, 2, 3,... are non negative integers, which satisfies the inequality. Note that (see (2 a vector solution has only a finite number of positive x i and for each value of x there are a finite number of solutions. Let N(x be the number of solutions to the inequality (3. The function N(x (x 0 is increasing. Note that N(0 =, since in this case we have the unique solution (0, 0, 0,... Consequently N(x. Let us consider the finite linear inequality (x fixed n r i x i x (x 0 (4 i= A solution to this inequality is a vector (x,x 2,...,, where the x i (i =, 2,...,n are non negative integers, which satisfies the inequality. Let S n (x be the number of solutions to the inequality (4. The following lemma is well nown [2] ( the proof is elementary, mathematical induction and combinatory. Lemma. The following asymptotic formula holds, Theorem.2 We have where f(x. S n (x n! r r 2...r n (5 N(x =x f(x = e f(x log x (x> (6 Proof. If (x,x 2,...,+ is a solution to the inequality n+ i= r i x i x

3 Partitions into -th powers 435 then (x,x 2,...,+, 0, 0,... is a solution to the inequality (3. Hence S n+ (x N(x and S n+ (x N(x (7 Lemma. imply S n+ (x (8 (7 and (8 give N(x (9 Clearly, if x> we can write N(x =x f(x. From (9 we obtain N(x lim x = lim x x f(x n = (0 Limit (0 imply the set A of values of x such that f(x n 0 is bounded, since if x A we have x f(x n. Consequently there exists x 0 such that if x x 0 we have f(x n>0, that is f(x. The theorem is proved. Theorem.3 If x [r n,r n+, N(x =S n (x and N(x n. Proof. If x [r n,r n+ there exists a one-to-one correspondence between the solutions to (4 and the solutions to (3. If (x,x 2,..., is a solution to (4 then (x,x 2,...,, 0, 0,... is a solution to (3. On the other hand if (x,x 2,x 3,... is a solution to (3 then + = 0,+2 =0,... since in contrary case i= r i x i r n+. Finally, note that (see ( (0,...,,...,0 where is the i-th coordinate (i =,...,n is a solution to (4. The theorem is proved. Theorem.4 If x [r n,r n+ then ( x N(x r ( x rn =+ ( n x r i + ( n2 x 2 r i r j ( nn r...r n Proof. Clearly, we have ([ ] x S n (x r ([ ] ( ( x x x rn r rn On the other hand, N(x =S n (x (theorem.3. The theorem is proved.

4 436 R. Jaimczu ( ni Notation. = n(n...(n i+ denotes the number of summands in each sum i! and [α] denotes the largest integer that does not exceed α. Let C(x be the function defined in the following way, if x [r n,r n+ then ( ( x x C(x = =+ x + x ( r rn ( n r i ( n2 r i r j ( nn r...r n Theorem.4 and ( give N(x C(x. We shall call to C(x a trivial upper bound to N(x. We can write C(x =x g(x (2 Consequently ( see theorem.2 f(x g(x and g(x. Let R(x bea function such that if x x 0 we have N(x R(x. If we write R(x =x s(x (3 then f(x s(x. We shall call to R(x a nontrivial upper bound to N(x if and only if (s(x/g(x 0. Note that is definition imply (R(x/C(x 0 since g(x. 2 Main Results Let us consider the sequence r n = n where is a positive integer. In this case, we have N(x =N (x = [x] i= p (i, where p (i is the number of partitions of i into -th powers. Theorem 2. There exists x 0 such that if x x 0 the following inequality holds N (x < exp(c log x + x + c 2 log x (4 where R(x = exp(c log x + x + c 2 log x is a nontrivial upper bound and, ( c = + ( + + c 2 > + Proof. Let us consider the inequality We have that n = n (n + (5 n 0 x dx + H (n = + n + + H (n (6

5 Partitions into -th powers 437 where 0 <H (n <n. From (6 we obtain that inequality (5 holds if n = [ ( + + (n + + ] + (7 Since < 2 <...<n, any subset in the set {, 2,...,n } with h (h n elements satisfies inequality (5. Note that ( n ( n2 ( nn < <...< since (n /n 0. Hence if x [r n,r n+ =[n, (n + we have (see theorem.3 and theorem.4 N (x = S n (x < + ( n < + ( n x + x r i + ( n2 ( n2 x < n n = ( log n log x + ( + x 2 r i r j ( nn r i...r m ( nn ( nn <n That is ( see (7 ( ( N (x < exp log x + (( + + (n Now ( log x (n + + log xn + 0 Consequently ( ( N (x < exp log x + (( + + n o( (8 From (8 we obtain (see (3 N (x < exp(c log x + x + c 2 log x =R(x =x s(x (9 That is, inequality (4. Let us consider the sequence r n = n.ifx [n, (n + we have (see ( and (2 C(x =x g(x > 2...n = xl(x Therefore log + log log n g(x >l(x =n n log x log + log log n log n

6 438 R. Jaimczu Now log + log log n = n where 0 <H 2 (n < log n. Consequently log xdx+ H 2 (n =n log n n ++H 2 (n g(x >n n log n n ++H 2(n log n > n log n 2 (20 Inequalities (9 and (20 give s(x g(x < c (n + n 2 log n + + c2 That is (s(x/g(x 0. Therefore the upper bound R(x is a nontrivial upper bound. The theorem is proved. Corollary 2.2 There exists n 0 such that if n n 0 then p (n < exp(c log n + n + c 2 log n <e n + ɛ <e n (0 <ɛ< Corollary 2.3 The following limit holds N (x x Corollary 2.4 The following limit holds p (n n References [] G. E. Andrews, Partitions: at the interface of q-series and modular forms, The Ramanujan Journal, 7 (2003, [2] R. Jaimczu, A note on integers of the form p s p s p s where p,p 2,...,p are distinct primes fixed, International Journal of Contemporary Mathematical Sciences, 27 (2007, [3] E. M. Wright, Asymptotic partitions formulae III. Partitions into -th powers, Acta. Math., 63 (934, Received: September, 2008