Bounds for the Eventual Positivity of Difference Functions of Partitions into Prime Powers

Transcription

1 Journal of Integer Seuences, Vol. (7), rticle 7..3 ounds for the Eventual Positivity of Difference Functions of Partitions into Prime Powers Roger Woodford Department of Mathematics University of ritish Columbia Vancouver, C V6T Z Canada rogerw@math.ubc.ca bstract In this paper we specialize work done by ateman and Erdős concerning difference functions of partition functions. In particular we are concerned with partitions into fixed powers of the primes. We show that any difference function of these partition functions is eventually increasing, and derive explicit bounds for when it will attain strictly positive values. From these bounds an asymptotic result is derived. Introduction Given an underlying set N, we denote the number of partitions of n with parts taken from by p (n). The k-th difference function p (k) (n) is defined inductively as follows: for k =, p (k) (n) = p (n). If k >, then p (k) (n) = p(k ) (n) p (k ) (n ). Let f (k) (x) be the generating function for p(k) (n). We have [5] the following power series identity:

2 f (k) (x) = n= p (k) (n)xn () = ( x) k p (n)x n () n= = ( x) k a xa. (3) This may be used to define p (k) (n) for all k Z, including k <. ateman and Erdős [5] characterize the sets for which p (k) (n) is ultimately nonnegative. Note that if k <, then the power series representation of ( x) k has nonnegative coefficients so that p (k) (n). For k, they prove the following: if satisfies the property that whenever k elements are removed from it, the remaining elements have greatest common divisor, then lim n p(k) (n) =. simple conseuence of this is the fact that p (n) is eventually monotonic if k >. No explicit bounds for when p (k) (n) becomes positive are included with the result of ateman and Erdős [5]. y following their approach but specializing to the case when = {p l : p is prime}, (4) for some fixed l N, we shall find bounds for n depending on k and l which guarantee that p (k) (n) >. For the remainder, shall be as in (4), with l N fixed. In a subseuent paper, ateman and Erdős [6] prove that in the special case when l =, p () (n) for all n. That is, the seuence 67 of partitions of n into primes is increasing for n. Our result pertains to more general underlying sets; partitions into suares of primes (9677), cubes of primes, etc. In a series of papers (cf. []- [3]), L.. Richmond studies the asymptotic behaviour for partition functions and their differences for sets satisfying certain stronger conditions. The results none-the-less apply to the cases of interest to us, that is, where is defined as above. (n). Unfortunately, his formula is not (n) must be positive, since, as is customary, he does not include explicit constants in the error term. Furthermore, his asymptotic formula includes functions such as α = α(n) defined by Richmond proves [] an asymptotic formula for p (k) useful towards finding bounds for when p (k) n = a a e αa k e α. s we are seeking explicit constants, a direct approach will be cleaner than than attempting to adapt the aforementioned formula.

3 nother result worthy of comment from Richmond [] pertains to a conjecture of ateman and Erdős [5]. His result applies to as defined in (4), and states that p (k+) (n) p (k) (n) = O(n / ), as n. We omit the subscript and write p (k) (n) in case the underlying set is, and omit the superscript if k =. The letter will always be used to denote a finite subset of. We shall also write ζ n for the primitive n-th root of unity e πi/n. The letters ζ and η shall always denote roots of unity. We shall denote the m-th prime by p m. Our main results are Theorems. and.. The former is established in Sections and 3. Theorem.. Let k be a nonnegative integer, let = π π ( ) 3(k+) t = 6 (k + ) 8lk+l+3. and let Then there are positive absolute constants a,...,a 7 such that if ( ) t a a llog t 3 N = N(k,l) = a t + a 5 t ( 3 a 6 (a 7 t) 6l) t, a (k+3)l 4 then n N implies that p (k) (n) >. Remark. The values of the constants are approximately a.4866, a , a , a , a , a , a For the remainder of the paper, shall be as defined in Theorem.. Note that Furthermore, Define F(k,l) = min {N N : n N implies that p (k) (n) > }. We show in Section 4 that Theorem. yields the following asymptotic result: Theorem.. Fix l N. Then as k, log F(k,l) = o ( (k + ) 8lk). Following ateman and Erdős [5], we first tackle the finite case. 3

4 Finite subsets of Lemma.. Let be a finite subset of of size r, and suppose k < r. The function p (k) (n) can be decomposed as follows: p (k) (n) = g(k) (n) + ψ(k) (n), where g (k) (n) is a polynomial in n of degree r k with leading coefficient ((r k )! ), and ψ (k) (n) is periodic in n with period. Proof. We use partial fractions to decompose the generating function f (k) (x) as follows: f (k) (x) = ( x) r k j= ζ j x = α x + α ( x) α r k ( x) r k + j= (5) β(ζ j ) ζ j x, (6) where the α i, and β(ζ j ) are complex numbers that can be determined. Note that ( α r k = ). The power series expansion for ( x) h is given by Hence, if and then the lemma is proved. ( x) = h n= h= ( n + h h ) x n. r k g (k) (n) = ( ) n + h α h, h ψ (k) (n) = j= β(ζ j )ζ jn, For the remainder of this paper, g (k) (n) and ψ(k) (n) shall be as in Lemma., for a given finite set which shall be clear from the context. Remark. Let be as in Lemma.. We wish to know the precise value of β(ζ j ). To simplify notation a little, we will freuently write β ζ instead, when ζ is clear from the context. 4

5 In particular, suppose η = ζ j, where, and < j <. Then ( ηx)f (k) (x) = ( x)k + ηx + + (ηx) = β η + ( ηx) ( p p x p α x + + α r k ( x) r k + ζ η β ζ ζx ), hence, β η = ( η)k p p η p. We shall freuently make use of the ineuality θ θ cos θ + θ4 4, which holds for all values of θ. Note that so for 3 θ 3, In particular, for θ π, θ e iθ = ( cosθ), θ θ eiθ θ. θ eiθ θ. (7) Lemma.. Suppose that ζ is a -th root of unity,. Then ζ. Proof. Clearly ζ. On the other hand, by euation (7), ζ e πi/ π. 4π 5

6 Lemma.3. Suppose that k < r, and, satisfies = r. Suppose further that η = ζ, j for some, j {,..., }. Then for k, k r, if k ; b β η r r k, if k <. b r k Proof. Making use of Remark and Lemma. we have that for k : β η = k k ζ j k ( p p ζ p p ) r = k r b r. similar arguments works for the case when k <. ζ jp Theorem.. Suppose that k < r, and, satisfies = r. Then k ψ (k) (n) b r r, if k ; r k, if k <. b r k Proof. First assume that k. y Lemmas.3 and., ψ (k) (n) = β(ζ j )ζ jn similar argument works for k <. j= = k b r k b r β(ζ) j j= 6 j= k r b r ( ) r r.

7 To obtain bounds for the coefficients α h, we will use Laurent series. Lemma.4. Suppose that k < r, and, satisfies = r. Denote the largest element of by Q, and suppose further that < r < ζ Q. Let d (r ) = ( ζ j r ). j= Then α h r r k h d (r ). Proof. Let γ be the circle z = r. From euations (5) and (6), and the Laurent expansion theorem, we have that α h = f (k) πi (z)(z )h dz π γ γ z r k h+ r r k h d (r ). j= ζ j z dz For the following proposition, we define several new constants: ) c = ( π, 3 4 k c = k=3 ( π k=3 3 4 k ) k, c 3 = 4π 6 c, c 4 = log π log, ( π ) / c 5 = c, ( ) π / /4 c 6 = ( ) π, 48 sin 4π/5 c 7 = 4π/5, b = c 5 c 6 = , b = c 3 c 6 = , b 3 = c 4 = , b 4 = c 7 π/ =

8 Proposition.. Let satisfy = r, and suppose that all the elements of are odd. Let Q = max{}, and T = log Q. If then Furthermore, if d = d (r ), then r = min { ζ j ζ j :,j =,...,T }, < b 4 Q r < ζ Q π Q. d b ( ) r b Q b 3 e log Q log. Proof. Observe that T satisfies Now, let so that T < Q < T. δ = ( ζ j r ), j= d = δ. Note that since each element of is odd, we have that Conseuently, Now δ T k= j< k k T ( π k= 4 k ( π = ( π 4 48 ( π 4 ( π δ = ( ζ j r ). j= ζ k 3 4 k )) 4 )) ( π 4 48 )) k k T ( π k=3 T ( π k=3 4 k ( π 4 k 3 4 k )) ( π k k )) k k 8

9 T ( π k=3 So we have that 4 k ( π 3 4 k d )) k + (( c3 = b π(t ) 4 (T ) 4π 4 Q log π log Qe log Q log (π ( 4 T ) T+ T ( π ( 4 Q ) 4 4 ) c c 4π 4 c Q log π log e log Q log = c 3 Q c 4 e log Q log c 5 c 6 ) ) (c 5 c 6 ) Q c 4 e log Q log ( b Q b 3 e log Q log ) r ) c c ( ) π / c π / Our next task is to bound r from below. For, j {,...,T }, let Then by the mean value theorem, ζ j f(x) = ( cos x), θ = θ(j) = π, and j ε = ε(,j) = π j π j. ζ j = f(θ + ε) f(θ) ε sin c =, ( cos c) for some c (θ,θ + ε). It is easily seen that π / j / can be no greater than 4π/5. On the interval [, 4π/5], we have sin x c 7 x. Therefore f(θ + ε) f(θ) εc 7c c = εc 7. (8) Choose a N such that (a ) j + a j. Then clearly a T j. Furthermore, 9

10 Hence, by (8) we conclude that ε πa a j π j π = j /a π j π π j j T j π T π T π Q. r b 4 Q. ounding r from above is a far simpler matter. y definition, r < ζ Q π Q. (9) 3 Infinite subsets of N and Proposition 3.. For k, and D D N, we have p (k) D (n) p (k) D (n). Proof. This follows immediately from euation and the fact that for k, the power series expansion for ( x) k has nonnegative coefficients. For the sake of clarity and the comprehensiveness of this exposition we include the following theorem of ateman and Erdős [5] suitably adapted to our needs. Theorem 3.. Let D N be an infinite set. For any t N, we have that p D (n) D (n) t + p ( ) + (t ) t + n t 3 p ( ) D (n). Proof. Denote by P (n), the number of partitions of n into parts from D such that there are exactly distinct parts. P (n) has generating function If R (n) is defined by P (n)x n = n= R (n)x n = n= {a,...a } D {a,...a } [n] x a x a x a x a. x a x a x a x a,

11 where [n] = {,...,n}, then P (n) R (n). There are ( n ) subsets of [n] of size. lso, the coefficient of x n in (x a + x a + ) (x a + x a + ) is less than or eual to the coefficient of x n in so (x + x + ) = P (n) m= ( )( ) n n ( ) m x m, n. ny partition n = n a + n a, where a,...,a, gives rise to a partition of n a i for i =,...,, namely n a = (n )a + n a + + n a, n a = n a + (n )a + + n a,. n a = n a + n a + + (n )a. Note that no two distinct partitions of n can give rise to the same partition of any m < n in this way, and so n n P (n) p D (m). Now if t N, then and the theorem is proved. = = m= n p ( ) D (n) = p D (m) m= n p D (n) + P (n) = (t + )p D (n) + n ( t)p (n) = t (t + )p D (n) (t ) P (n) = (t + )p D (n) (t ) n t 3, For the remainder of this section, we follow the approach of ateman and Erdős [5] and simultaneously make the results explicit by applying them to the special case under consideration. To be consistent, we shall assume k, and use the following notation:

12 Notation. Let be the least k+ elements of, and let C = \. = {p l k+3,pl k+4,...,pl k+t } be the least t elements of C, where t is determined as in the statement of Theorem. from the values of k and l in uestion. Furthermore, let ( ) k+ g = (k + ) l(k+)+, and () ( ) k+ h = 3(k + ) l(k+) g = 3 (k + ) 4lk+6l+. Finally, let us remark that numerous constants shall be defined in the subseuent argument. Their definitions shall remain consistent throughout. Note that the right hand side of () is increasing in k and l, so g 6/ = Lemma 3.. For all n, we have p (k) (n) g. () Proof. Since = k +, g (k) (n) is linear in n, and g(k+) (n) is a constant function. y Lemma., g (k) (n) = h= ( ) n + h α h h where α = (p p k+ ) l. Substituting x = into (5) and (6) gives Hence, so α + α + β(ζ) j =. j= g (k) (n) = (p p k+ ) l n + β(ζ), j j=

13 p (k) Now, it is easy to see that (n) = g(k) (n) + ψ(k) (n) = (p p k+ ) l n + ( ζ jn )β(ζ) j j= (p p k+ ) l n + ζ jn β(ζ) j j= (p p k+ ) l n + β(ζ) j j= ( ) k+ (p p k+ ) l n + ( ) k+ k+. k+ k+ Q k+ (k + ) (k + ) l(k+)+, where Q = max (). From this, the Lemma follows. Lemma 3.. For all n, we have Proof. Note that Making use of Theorem., we see that g p (k+) + p (k+) (n) < g. () g (k+) (n) = (p p k+ ) l. (n) g (p p k+ ) l ψ (k+) (n) g (p p k+ ) l k i= b k+ k+ ( k+ ((k ) + ) l k+ ( ) p l k+ ) i (p p k+ ). Observe that for a fixed i, i k +, the expression ( (k + ) l ) k+ 3 ( ) p l k+ i,

14 is positive and increasing in l, for l, so, Hence, ( (k + ) l ) k+ ( ) p l k+ ( ) i (k + ) k+ ( pi ( ) (k + ) k+ (k + ) p k+ 4 + (k + ) p k+ 4. ) k+ ( ) k+ pk+ g p (k+) (n) (k + ) 4 (p p k+ ) = >, that is, + p (k+) (n) < g. Corollary 3.. p (k) (n) + p (k+) (n) Proof. It follows from the proof of Lemma 3. that p (k) (n) (p p k+ ) l n + g > (p p k+ ) l n g. if n h. (3) The Corollary follows from this, together with Lemma 3. and the fact that p p k+ (k + ) (k+). Lemma 3.3. There is an h N such that such that n h implies (t ) t + n t 3 (t ) p ( ) C (n) t + n t 3 (n) t +. (4) p ( ) 4

15 Proof. The first ineuality is a conseuence of Proposition 3.. Note that Hence ( t 6 ) 3(k+) (k + ) 8lk+l+3 ( ) 3k+3 6 8k b 3 ( ) k. b 3 ( ) ( ) 3 6 log t log + k log. since log t t/e, if we let M = e log ( /b 3 ), then b 3 k t M. Choose r and d with respect to the set as in Proposition., and let Q = p l k+t = max ( ). It is clear that t 6(/ ) 3 3 = 9, which we denote by t. y euation (9), we have that b 3 We also have that and r π Q π (t) l π t. r b 4 Q b 4 (k + t) 4l, ( ) t d b b Q b 3 e log Q log (b (k + t) b 3l (k + t) b (k+3)l (t ) (b b (k+3)l (k + t) b 3l = (k + t) l log (k+t) log ) t l log (k+t) log ) t Let us now bound p ( ) (n) from below. ssume that n t, and let b 5 = /M + = , and b 7 = t /(t π). Making use of Theorem., we have 5

16 p ( ) Observe that (n) + ψ ( ) (n) ( ) t n + t ( ) n + h α t α h t h (n) =g ( ) ψ ( ) (n) (p k+3 p k+t ) l n t (t )! b t k+t m=k+3 p l(t ) m (k + t) l(t ) n t (t )! b t (b 5t) l(t ) n t (t )! h= ( n + t 3 t 3 (t )(k + t) l(t ) (n + t 3)t 3 (t 3)! nt 3 t 3 (t 3)! (t )(k + t)l(t ) b t (b 5t) l(t ) n t (t )! t(b 5t) l(t ) b t. b 7n t 3 t 3 (t 3)! ) t α h h= t h= ( r )r t d r h r t d ((k + t) (4 b l log (k+t) 3)l+ log b 4 b b (k+3)l ) t (k + t) (4 b l log (k+t) 3)l+ log (b 5 t) (4 b l log Ct 3)l+ log = e l(log b 5+log t)((4 b 3 + log C log )+ log t log ) = e l( log log t+(4 b log b 5 log ) log t+(4 b log b 5 log ) log b 5) e b 6llog t, where b 6 is a constant determined as follows. Let x = log t. Then we may take b 6 = ( log + 4 b log b ) ( b log b ) 5 log b5 log x log x =

17 It suffices to select h such that for n h, p ( ) (n)n (t 3) (t ). (5) Observe that (5) is implied by (b 5 t) l(t ) n (t )! b 7 (t 3)! ( ) t e b 6llog t t(b 5t) l(t ) b 4 b b (k+3)l n t 3 b t (t ), which is euivalent to n (b 5 t) l(t ) ( ) t e b b 6llog t 7 (t ) + t(b 5t) l(t ) (t )! b 4 b b (k+3)l n t 3 b t + (t ) (t )!. (6) The ineuality holds for n 7. This implies that e n n! n n+, (t )! (t) t 3 (t) 3 e t (t ). Thus, since n t by assumption, (6) is implied by where n (b 5 t) l(t ) ( ), ( e b 6llog t = b 7 t b 4 b b (k+3)l = 6t4 (b 5 t) l(t ) (t )e t b t 3 = (t ) (t) t. (t )e t ) t The term 3 is negligible relative to, yet and are not easily compared since one or the other may dominate depending on the values chosen for k, and l. None the less, we may simplify matters a little by absorbing 3 into in the following way: 7

18 + 3 t 3 (b 5 t) l(t ) /(e t b t ) 6t t + = 6t t + 6t t + b t (t) t t(t )(b 5 t) (t ) ( ) ( 4t b t b 5t ( ) ( 4t b t b 5t ) t ) t So, let Then we may take = t3 (b 5 t) l(t ) e t b t. h = (b 5 t) l(t ) ( + ). Remark 3. Note that ( ) 3(k+) t = 6 (k + ) 8lk+l+3 = g h, and so t + (g + )(g )h. Lemma 3.4. There exists an N = N(k, l) >, such that if n N, then p (k) (n) p ( ) C (n) >. Proof. The second ineuality is obvious. For the first, by Proposition 3., Theorem 3., Lemma 3.3 and Remark 3, we have that for n h, p C (n) C (n) t + p ( ) + (t ) t + (n)n (t 3) (g + )(g )h. (7) p ( ) Now, using, (), (), (3), (7), and the identity we have that for n h + h, p (k) (n) = n m= p (k) (n m)p C(m), 8

19 p (k) (n) m n h (g ) ( + p (k+) (n m) )p C (m) n h<m n ( + p (k+) (n m) )p C (m) n ( + p (k+) (n m) )p C (m) (g + )(g ) m= n ( + p (k+) (n m) )p C (m) m= ( n ) (g p C (m) ) n h<m n p ( ) C ( (m) ) (g p C (m) ) n h<m n p ( ) C (m) n ( + p (k+) (n m) )p C (m) m= n p C (m) m= =p ( ) C (n). p ( ) C (n) n h<m n p C (m) n ( + p (k+) (n m) )p C (m) m= Proof of Theorem.. y the proofs of Lemmas 3.3 and 3.4, it suffices us to choose N h+h = h+(b 5 t) l(t ) ( + ). First observe that t. The uantity (b 5 t) l(t ) may be simplified further with an upper bound. Let and Then b 9 = log b 5 x + x + b 6 = , = b b 4. 9

20 Denote ( (b 5 t) l(t ) (b 5 t) l e b 6llog t = b 7 t b (k+3)l ) t ( e l(log b 5+log t+b 6 log t) = b 7 t ( e b 9llog t b 7 t b (k+3)l b (k+3)l ) t ( ) t e b 9llog t = b 7 t. b (k+3)l ) t We will use the fact that h t to absorb h into (b 5 t) l(t ) in the following way: t h + (b 5 t) l(t ) t 3 (b 5 t) 3l(t ) /(e t b t ) et b t (b 5 t) 3(t ) Letting b 8 = 8., we may take b (k+3)l = e t ( ) t eb b 3 5t 3 ( eb ) t e t b 3 5t N = + b 8t 3 (b 5 t) 3l(t ) e t b t ( ) t e b 9llog t = b 7 t + b 8t 3 (b 5 t) 3l(t ) e t b t. We conclude the proof by restructuring the constants as follows: a = b 7 a = b a 3 = e b 9 a 4 = b a 5 = e b 8 a 6 = e b a 7 = b 5.

21 4 symptotic results Observe that as k, log t k log k, when l remains fixed. This implies that if l >, then a llog t 3, as k, a (k+)l 4 and so in this situation, the second term dominates in the expression for N(k,l) in Theorem.. If l = or, then a llog t 3 = e λ(k), a (k+)l 4 where λ (k) k log k, and (a 7 t) 6l = e λ (k), where λ (k) k log k. In this case the first term dominates in the expression for N(k,l). Thus we have proved the following corollary to Theorem.: Corollary 4.. Let l N be fixed. Then as k, ( F(k,l) = O t ) t a a llog t 3, if l =, ; a (k+3)l 4 ( F(k,l) = O t ( 3 a 6 (a 7 t) 6l) ) t, if l >. Theorem. follows from Corollary 4. simply by taking logarithms. 5 cknowledgements The author is supported by an NSERC CGS D research grant and wishes to acknowledge the Natural Sciences and Engineering Research Council for their generous funding, as well as the support of his supervisor Dr. Izabella Laba. References [] G. H. Hardy, and S. Ramanujan, symptotic formulae for the distribution of integers of various types, Proc. London Math. Soc., Ser., 6 (96), 3. [] P. Erdős, On an elementary proof of some asymptotic formulas in the theory of partitions, nn. Math., 43 (94), [3] C. G. Haselgrove, and H. N. V. Temperley, symptotic formulae in the theory of partitions, Proc. Cambridge Philos. Soc., 5 Part (954), 5 4.

22 [4] O. P. Gupta, and S. Luthra, Partitions into primes, Proc. Nat. Inst. Sci. India, (955), [5] P. T. ateman, and P. Erdős, Monotonicity of partition functions, Mathematika, 3 (956), 4. [6] P. T. ateman, and P. Erdős, Partitions into primes, Publ. Math. Debrecen, 4 (956), 98. [7] Takayoshi Mitsui, On the partitions of a number into the powers of prime numbers, J. Math. Soc. Japan, 9 (957), [8] E. Grosswald, Partitions into prime powers, Michigan Math. J., 7 (96), 97. [9] S. M. Kerawala, On the asymptotic values of lnp (n) and lnp ( d)(n) with as the set of primes, J. Natur. Sci. and Math., 9 (969), 9 6. [] L.. Richmond, symptotic relations for partitions, J. Number Theory, 7 (975), [] L.. Richmond, symptotic results for partitions (I) and the distribution of certain integers, J. Number Theory, 8 (976), [] L.. Richmond, symptotic results for partitions (II) and a conjecture of ateman and Erdős, J. Number Theory, 8 (976), [3] L.. Richmond, symptotic relations for partitions, Trans. mer. Math. Soc., 9 (976), Mathematics Subject Classification: Primary 57; Secondary P8. Keywords: partition function, difference function. (Concerned with seuences 67 and 9677.) Received pril 6; revised version received November 8 6. Published in Journal of Integer Seuences, December 9 6. Return to Journal of Integer Seuences home page.