#A8 INTEGERS 12 (2012) PARTITION OF AN INTEGER INTO DISTINCT BOUNDED PARTS, IDENTITIES AND BOUNDS

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1 #A8 INTEGERS 1 (01) PARTITION OF AN INTEGER INTO DISTINCT BOUNDED PARTS, IDENTITIES AND BOUNDS Mohammadreza Bidar 1 Deartment of Mathematics, Sharif University of Technology, Tehran, Iran mrebidar@gmailcom Received: 3/15/11, Acceted: 1/10/1, Published: 1/13/1 Abstract The artition function Q(n), which denotes the number of artitions of a ositive integer n into distinct arts, has been the subject of a dozen aers In this aer, we study this kind of artition with the additional constraint that the arts are bounded by a fixed integer We denote the number of artitions of an integer n into distinct arts, each k, by Q k (n) We find a shar uer bound for Q k (n), and more, an infinite series lower bound for the artition function Q(n) In the last section, we exhibit a grou of interesting identities involving Q k (n) that arise from a combinatorial roblem 1 Introduction Let Q(n) be the number of ways of artitioning a ositive integer n into distinct summands The generating function for this kind of artition is Q(x) = Q(n)x n = (1 + x j ) (1) n=0 Euler noted that he could easily convert Q(x) to something else, which is in fact another generating function: Q(x) = (1 + x j ) = (1 xj ) (1 xj ) = 1 1 x j 1 The last roduct is the generating function for artitioning an integer into odd summands Consequently, he concluded that there was a bijection between the set of artitions of a ositive integer n into distinct arts, and set of artitions of n 1 The author is currently a guest researcher of the School of Mathematics at the Institute for Research in Fundamental Sciences (IPM)

2 INTEGERS: 1 (01) into odd arts To read a brief history of Euler work concerning this bijection see, [9] If σ o (n) denotes the odd divisor function, ie, the sum of odd divisors of n, then the artition function Q(n) satisfies the recurrence equation (see [1], 86) Q(n) = 1 n 1 Q(k)σ o (n k), n > 0 () n k=0 It is easily seen that σ o (n) = σ(n) 1 σ( n ) = σ(n)/(a(n)+1 1), where σ(n) is the sum of divisors of n, and a(n) is the ower of in the decomosition of n into rime factors Therefore, we are able to modify our recurrence equation as follows: Q(n) = 1 n n σ(k) Q(n k), n > 0 (3) a(k)+1 1 An investigation in the table of amounts of Q(n) for large numbers demonstrates that it has a considerably slower growth than the unrestricted artition function P (n) To have a comarison with P (n), it is worthwhile to mention Rademacher like series for Q(n) (see [5], [6] and [7]): where Q(n) = 1 { ( d A k 1 (n) [J dn 0 A k (n) = k h=1 (h,k)=1 i k 1, e i[s(h,k) s(h,k)] e ihn/k, s(h, k) = ( 1 n ) )]}, 4 n=n k 1 r=1 ( r hr k k hr 1 ) k Here s(h, k) is a Dedekind sum, and J 0 (x) is the zeroth order Bessel function of the first kind This series reresentation of Q(n) leads to an asymtotic formula for Q(n): 1 Q(n) 4 3 1/4 n e n 3/4 3, as n (4) Comaring this asymtotic formula with the one for P (n) demonstrates the slower growth of Q(n) (see also [4], ) In Sections and 3, we derive suitable uer and lower bounds for the artition function Q(n) It is well known that the general artition function P (n), n > 0, is convex (see [8]) The convexity for the amounts of Q(n) takes lace if n 4, which means that the inequality Q(n) 1 {Q(n + 1) + Q(n 1)} holds for n 4 A short roof of this fact is resented in Section 3 In this aer, we are mainly concerned to a restricted form of artition of an integer into distinct arts Let Q k (n) denote the number of artitions of a ositive

3 INTEGERS: 1 (01) 3 integer n into distinct arts, each k This artition function has an interesting combinatorial interretation If X = {1,, 3,, k}, then Q k (n) is the number of subsets of X for which the sum of the members is n The artition function Q k (n) has the generating function Q k (x) = (1 + x j ) = θ Q k (n)x n, θ = n=0 k(k + 1) (5) We know that Q k (x) is a symmetric unimodal olynomial It means that its coefficients goes u to somewhere, (for Q k (x) the climax occurs at k(k+1) 4 ) then symmetrically goes down The symmetry of the coefficients is almost evident, but roving the unimodal roerty of Q k (x) is difficult To my knowledge there is not a known combinatorial roof for this fact, but there is a non-elementary roof based on semi-simle Lie Algebras The interested reader might have a look at [10] to see a roof of the unimodal roerty for Q k (x) (For further discussion on the unimodal roerty and Lie algebras see, [1]) Since Q(n) = Q k (n) for each k > n, the unimodal roerty of Q k (x) leads to the monotonicity of the artition function Q(n) Let P k (n) denote the number of artitions of an integer n into arts, each k, and let P k (x) be its generating function The relation between Q k (n) and P k (n) can be stated by means of the identity P k (x) = k 1 k (1 xj ) = (1 + xj ) k (1 xj ) = P k(x )Q k (x), which leads to a recurrence equation relating P k (n) to Q k (n): n P k (n) = Q k (n i)p k (i) (6) i=0 An Elementary Uer Bound for Q k (n) Pribitkin [11] has introduced a remarkable elementary method to obtain a shar uer bound for the artition function P k (n) With modification of his method, we are able to find a shar uer bound for Q k (n) As in [11], we emloy the dilogarithm function Li (x) = m=1 xm m, where x < 1 It is clear that Li (1) = 6 We also will need the simle fact ex e x > x for x > 0, that has aeared in [11] The main result of this section is stated in the next theorem Theorem 1 Let k,n be ositive integers, n k(k+1) 4 Then we have the following inequality: A(k, n) Q k (n) < e n/3 1 3nLi(e α/ 3n), n

4 INTEGERS: 1 (01) 4 where A(k, n) = n k +k 4n+ + Proof If 0 < x < 1, we have Q k (x) = 3, α = k/ After taking logarithm, we observe that log(q k (x)) < (1 + x j 1 ) < (1 x)(1 x 3 ) (1 x α 1 ) α log(1 x j 1 ) = Now we let x = e u, u > 0, to find that = = α m=1 m=1 m=1 1 m x (j 1)m α m x (j 1)m 1 x m m 1 x m (1 xαm ) log(q k (e u )) < m=1 e mu (1 e αmu ) m(1 e mu ) 1 e αmu = m(e mu e mu ) m=1 < 1 1 e αmu u m m=1 = 1 ( ) u 6 Li (e αu ) We exloit the unimodal and symmetry roerties of Q k (x) to obtain that for all 0 < x < 1, and n k(k+1) 4, Q k (x) Q k (n)(x n + x n x k(k+1) n ) = Q k (n)x n 1 x k(k+1) n+1 1 x Therefore, we realize that log(q k (n)) < nu + log(1 e u k(k+1) ( ) log(1 e + 1 ( ) u 6 Li (e αu ) k(k+1) ( < nu + log(u) log(1 e + 1 ( ) u 6 Li (e αu ) n+1)u ) n+1)u )

5 INTEGERS: 1 (01) 5 Here we have alied the simle estimation 1 e x < x, that is valid for x > 0 Now we let u = 1 λ 1, λ > 0, and estimate the best λ Substitute u by u = n λ n in the right hand side of the inequality to find that Q k (n) < e( 1 λ + λ 1 ) n λ n e 1 λ nli (e αu ) k(k+1) 1 ( n+1) (1 e λ n ) Calculate the best ossible λ to minimize the multile of n at the first exonential term; it turns out that λ = 3 and u = Hence we conclude that 3n Q k (n) < e n/3 3n e 1 3nLi(e α 3n ) (7) k(k+1) ( n+1) (1 e 3n ) Note that for x 0, 1 + x e x, or e x 1/(1 + x) Subtracting both sides from 1, gives us the estimate x 1 + x 1 e x ; the roof is now comlete when we aly this inequality to the right hand side of (7) for x = ( k(k+1) n + 1) 3n Fix n and let k ; since A(k, n) tends to, and 3 e 1 3nLi(e α/ 3n) tends to 1, we are able to determine a very nice uer bound for Q(n) Corollary Let Q(n) denote the number of unrestricted artitions into distinct arts Then, we have Q(n) < e n/3 3n Remark It is clear that for all feasible amounts of k, n, the value of e small enough to make Li (x) > x a good estimation; hence we conclude that α 3n is Q k (n) < A(k, n) e (/ 3 e 3n α 3/) n n 3 Simle Lower Bounds for Q(n) Analytic methods, like the saddle oint method (see [4], ) are excellent for asymtotic estimations or finding uer bounds, but they seem oor to derive lower bounds Likewise, the dilogarithm scheme is not alicable to find a lower bound for Q k (n) However, we are able to find a lower bound for Q(n) by alying other methods First, we take a detour and rove the convexity of Q(n)

6 INTEGERS: 1 (01) 6 Lemma 3 If n > 3, then Q(n) 1 {Q(n + 1) + Q(n 1)} Assuming n > 3, we need to show that Q(n + 1) Q(n) Q(n) Q(n 1) Consider a artition of n into distinct arts and increase the greatest summand by 1; we obtain a artition of n + 1 into distinct arts in which the two greatest summands differ by at least Conversely, we can delete a 1 from the greatest summand of such artition and obtain a artition of n with distinct arts (for single art artitions of n, n + 1 there is a similar corresondence) Therefore, there is a bijection between the entire set of artitions of n into distinct summands, and set of distinct art artitions of n + 1 for which the greatest summand is at least more than the revious one (this set includes the single art artition of n + 1) Hence, we find that Q(n + 1) Q(n) is the cardinality of the set of all artitions of n + 1 into (more than 1) distinct summands with the greatest summand exactly 1 more than the revious summand Denote this set by Y, and the analogous set ertaining to n by X Decomose X into two disjoint sets, one consisting of those artitions that contain 1, say X 1, and the other one including all artitions without 1 in their summands, say X Partition Y in a similar way, and assume that (1, λ 1,, k 1 1, k 1 ) X 1, (λ 1,, k 1, k ) X (note that since n > 3, k 1, k > ) Define the two maings σ 1, σ in the following way: σ 1 : X 1 Y, σ 1 [(1, λ 1,, k 1 1, k 1 )] = (λ 1,, k 1, k 1 + 1), σ : X Y 1, σ [(λ 1,, k 1, k )] = (1, λ 1,, k 1, k ) It is quite straightforward to see that σ 1 is an injection from X 1 into Y, and σ is a bijection between X, Y 1 Therefore, X 1 Y, X = Y 1, and we conclude that X Y The recurrence equation (3) together with the convexity of Q(n) leads us to the following lower bound Theorem 4 If n > 0, then Q(n) satisfies the following inequality: Q(n) > e 7 (7/1) k ( ) n + k 1 1 k! n Proof Starting with the equation (3), we divide the right hand sum into arts, each consisting of four consecutive terms with the first one index in the form 4t + 1 If k = 4t + 1 > 1, then we have 3 σ o (k + j)q(n k j) (k + 1)Q(n k) + k + Q(n k 1) 3 + (k + 3)Q(n k ) + Q(n k 3) 7 3 (k + j)q(n k j) 1

7 INTEGERS: 1 (01) 7 To acquire the last inequality, we have alied the monotonicity of Q(n), n 0 (the last inequality also holds for the last art which may have less than 4 terms) For k = 1, we could write that 4 σ o (j)q(n j) = Q(n 1) + Q(n ) + 4Q(n 3) + Q(n 4) > jq(n j) + 1 Q(n 1) 3 Here, we have exloited the monotonicity of Q(n) and the fact Q(n 1)+Q(n 3) > Q(n ), valid for n > 5 Thus, we conclude that Q(n) 1 7 Q(n 1) + 3n 1n n kq(n k), n > 5 Now, we define the function t(n) by the recurrence equation t(n) = 7 1n n kt(n k), t(0) = 1 A direct comutation shows that Q(i) t(i), 1 i 5 Hence, Q(i) t(i), i 0 Let T (x) be the generating function of t(n) It is easily seen that T (x) satisfies the equation T (x) (i + 1)x i = 1 7 T (x) i=0 After solving this differential equation, it turns out that 7 (7/1) k T (x) = T 0 e 1 1x = T0 (1 x) k k! Since t(0) = 1, the constant T 0 is equal to e 7 1 Thus, we have the following formula for t(n), n > 0: t(n) = e 7 (7/1) k ( ) n + k 1 1, k! n now the roof is comlete Let q k (n) denote the number of artitions of an integer n into exactly k distinct arts (note that q k (n) is quite different from Q k 1 (n k)) Clearly, Q(n) = a q k(n), a = 1 ( 1 + 8n + 1) ( It is easily verified that q k (n) = k n ( k ) ) Since k (n) 1 ( ) n 1 k! k 1 k=0

8 INTEGERS: 1 (01) 8 (see [], 56-57), we obtain a finite sum lower bound for Q(n): Q(n) = a k ( n ( k ) ) a ( ( 1 n k ) ) 1 k! k 1 (8) Remark To imrove the lower bound series in Section 3, one may sort terms of the recurrence identity concerning Q(n), modulo 8 or even 16; also a similar argument could be done to derive a lower bound for the artition function P (n) The first lower bound series is a quickly convergent satisfying lower bound The second lower bound sum, although not as straightforward as the first one, is shar In fact, emirical evidence shows that if n is greater than , then the amount of the lower bound series is greater than e 084 n/3 /n 3/4, and for n > 1500, the amount of the second lower bound is greater than e 093 n/3 /n 3/4 4 Identities Involving Prime Factors of the Bound Integer In this section, we find a grou of interesting identities which arise from a combinatorial roblem The key idea here is the uniqueness of a basis reresentation for the cyclotomic field Q(ζ ), when you look at it as a Q-vector sace We consider ( x; x) k = (1 + x j ) = k(k+1) n=0 Q k (n)x n, and consider as an odd rime factor of k Let ζ be the rimitive -th root of unity, ie, ζ = e i/ Let Q(ζ) be the field extension of ζ over Q First, we calculate the amount of ( ζ; ζ) k in Q(ζ) We have ( ζ; ζ) k = 1 (1 + ζ j ) = (1 + ζ j ) Since ζ is a rimitive root of unity we have Let z = 1 in this equation to find that 1 P (z) = z 1 = (z ζ j ) 1 (1 + ζ j ) = k/

9 INTEGERS: 1 (01) 9 Therefore, we have ( ζ; ζ) k = k/ The olynomial P (z) z 1 = 1 + z + z + + z 1 is a minimal olynomial for ζ over Q So we conclude that the set A = { 1, ζ, ζ, ζ 3,, ζ } constitutes a Q-basis for Q(ζ) as a Q-vector sace Thus, ( ζ; ζ) k = k/ = k/ ζ + 0 ζ ζ is the basis reresentation of ( ζ; ζ) k in Q(ζ) On the other hand, let us assume that ( ζ; ζ) k = (1 + ζ j ) = a 0 + a 1 ζ + a ζ + + a 1 ζ 1 In fact, we have exanded the roduct and reduced the owers modulo It is clear that α i a i = Q k (j + i), where α i = k(k + 1)/ i/ (9) The exansion above could be written in the form (a 0 a 1 ) 1 + (a 1 a 1 ) ζ + (a a 1 ) ζ + + (a a 1 ) ζ, as a reresentation over the basis A Since the reresentation over a basis is unique, we have the following system of equations: a 0 a 1 = k/, a i a 1 = 0, 1 i and which leads to the solution a 0 = k/ + k k/ So we have the following result:, a i = k k/ 1 a i = k, i=0 for 1 i 1 Theorem 5 Let k be a ositive integer and an odd rime factor of it Then, we have the following identities: α 0 Q k (j) = k/ + k k/, where α i = k(k + 1)/ i/ α i Q k (j + i) = k k/ for 1 i <,

10 INTEGERS: 1 (01) Alication Case 1 Let X = {1,, 3,, k}, and consider as an odd rime factor of k We are interested in the number of subsets of X for which the sum of the members is congruent to i modulo In fact, Q k (i), Q k ( + i), Q k ( + i), are equal to the numbers of subsets of X for which the sum of the members are i, + i, + i,, resectively Looking at equation (9) makes it clear that each a i in the exansion of ( ζ; ζ) k = (1 + ζ j ) = a 0 + a 1 ζ + a ζ + + a 1 ζ 1 describes the number of subsets of X for which the sum of the members is congruent to i modulo So if we denote the sum of the members of S X by σ(s), then we have { k/ + k k/, i = 0 #{S X : σ(s) i (mod )} = k k/, i 0 Case In the case X = {1,, 3,, k}, k = t + 1, there would be a similar argument for the coefficients a i of ( ζ; ζ) k But in this case we have t 1 ( ζ; ζ) k = (1 + ζ j ) (1 + ζ) = (k 1)/ + (k 1)/ ζ So we have the following system of equations: with the solutions a 0 a 1 = (k 1)/, a 1 a 1 = (k 1)/, a i a 1 = 0, i, and 1 a i = k, i=0 a 0 = a 1 = k + ( ) (k 1)/, a i = k + (k+ 1)/ Hence, we conclude that #{S X : σ(s) i (mod )} = for i 1 k +( ) (k 1)/, i = 0, 1 k + (k+ 1)/, i 0, 1 Case 3 In the case X = {1,, 3,, k}, k = t 1, we have t ( ζ; ζ) k = (1 + ζ j ) = (1 + ζ j ) (1 + ζ j ) = (k +1)/

11 INTEGERS: 1 (01) 11 which by a similar argument finally leads us to the following answer: (k +1)/ + k (k +1)/, i = 0 #{S X : σ(s) i (mod )} = k (k +1)/, i 0 Remark The roblem can be solved for the cases k = t ± ( 1)/ in a similar way 5 Concluding Remarks Real Integral form of Q k (n) Consider Q k (z) az a comlex variable generating function of Q k (n) The fact that it has no singularities at z = 1 makes it ossible to find a real integral form of Q k (n) Since Q k (z) is analytic over C, we find that Q k (n) = 1 i Substituting z by e iθ, it follows that Q k (n) = k 1 0 z =1 [ ] 1 cos 4 (k + k 4n)θ Q k (z) dz (10) zn+1 cos jθ dθ (11) This formula enables us to study the behavior of Q k (n) for various amounts of k, n, on the background of basic calculus Let q k (n, r) denote the number of artitions of n with exactly r distinct arts, each k If k (n, r) denotes the number of artitions ( ( of n with exactly r arts, each k, it is easily seen that q k (n, r) = k r+1 n r ) ), r, and k (n, r) = (k 1, r, n r), where (k, r, n) is the coefficient of q n in the Gaussian olynomial [ ] k + r r (see [], 33-36) Since Q k (n) is the sum of q k (n, r) s, having a lower bound on Gaussian olynomials coefficients leads to a finite sum lower bound for Q k (n) In Section 4, if we had considered a factor m of k that was not necessarily rime, then we would have had to deal with the cyclotomic olynomial Φ m (X) = (X ζ), (1) ζ U m where U m is the subset of rimitive m-th roots of unity in the set of comlex numbers (to learn more about cyclotomic fields see [3, ] In this case, there are difficulties with a basis reresentation; also we have more variables than equations However, it still is ossible to obtain some new identities q

12 INTEGERS: 1 (01) 1 References [1] M Abramowitz and I A Stegun, Handbook of Mathematical Functions with Formulas, Grahs, and Mathematical Tables, 9th rinting, New York: Dover, 197 [] G E Andrews, The Theory of Partitions, Advanced Book Program Reading, Massachusetts: Addison-Wesley, 1976 [3] H Cohen, Number theory Volume I: Tools and Diohantine Equations, New York: Sringer- Verlag, 007 [4] P Flajolet and R Sedgewick, Analytic Combinatorics, Cambridge: Cambridge University Press, 009 [5] P Hagis, Jr, Partitions into Odd and Unequal Parts, Trans Amer Math Soc 86, 1964a, [6] P Hagis, Jr, On a Class of Partitions with Distinct Summands, Trans Amer Math Soc 11, 1964b, [7] P Hagis, Jr, A Correction of Some Theorems on Partitions, Trans Amer Math Soc 118, 1965, 550 [8] R Honsberger, Morsel46 On the Partition Function (n), More mathematical Morsels, The Mathematical Association of America, vol 10, 1991, [9] B Hokins and R Wilson, Euler s Science of Combinations, Studies in the History and Philosohy of Mathematics, Vol 5, 007, [10] J W B Hughes, Partitions of integers and Lie algebras, Grou Theoretical Methods in Physics: Sixth International Colloquium Tbingen, Editor: P Kramer, A Rieckers, Lecture Notes in Physics, vol 79, 1977, [11] W A Pribitkin, Simle uer bounds for artition functions, Ramanujan J, vol 18, 009, [1] R P Stanley, Unimodality and Lie Sueralgebras, Studies in Alied Mathematics, vol 7, 1985, 63-81