Two truncated identities of Gauss

Two truncated identities of Gauss Victor J W Guo 1 and Jiang Zeng 2 1 Department of Mathematics, East China Normal University, Shanghai 200062, People s Republic of China jwguo@mathecnueducn, http://mathecnueducn/~jwguo 2 Université de Lyon; Université Lyon 1; Institut Camille Jordan, UMR 5208 du CNRS; 43, boulevard du 11 novembre 1918, F-69622 Villeurbanne Cedex, France zeng@mathuniv-lyon1fr, http://mathuniv-lyon1fr/~zeng Abstract Two new expansions for partial sums of Gauss triangular and suare numbers series are given As a conseuence, we derive a family of ineualities for the overpartition function p(n and for the partition function p 1 (n counting the partitions of n with distinct odd parts Some further ineualities for variations of partition function are proposed as conjectures Keywords: Partition function; Overpartition function; Gauss identities; Shanks identity AMS Subject Classifications: 11B65; 11P81; 05A17 1 Introduction The partition function p(n has the generating function p(n n = 1 1 n = 1 + + 22 + 3 3 + 5 4 + 7 5 + 11 6 + (11 Two classical results in the partition theory [1, p 11] are Euler s pentagonal number theorem ( 1 1 + ( 1 j ( j(3j 1/2 + j(3j+1/2 = 1, (12 1 n and Euler s recursive formula for computing p(n: p(n + ( 1 j (p(n j(3j 1/2 + p(n j(3j + 1/2 = 0, (13 where p(m = 0 for all negative m Recently, Merca [7] has stumbled upon the following ineuality: p(n p(n 1 p(n 2 + p(n 5 0, (14 1

and then, Andrews and Merca [3] proved more generally that, for k 1, ( 1 k 1 k 1 ( 1 j (p(n j(3j + 1/2 p(n (j + 1(3j + 2/2 0 (15 with strict ineuality if n k(3k + 1/2 The proof of (15 in [3] is based on the truncated formula of (12: where 1 k 1 ( 1 j j(3j+1/2 (1 2j+1 = 1 + ( 1 k 1 (; (a; = (1 a n, (a; M = (a;, (a M ; n=k and (k+1n+(k 2 (; n [ ] M = N [ ] n 1, (16 k 1 (; M (; N (; M N Motivated by Andrews and Merca s work [3], in this paper we shall prove new truncated forms of two identities of Gauss [1, p 23]: 1 + 2 ( 1 j j2 = (; ( ;, (17 ( 1 j j(2j+1 (1 2j+1 = (2 ; 2 ( ; 2, (18 and derive similar overpartition function and special partition function ineualities Theorem 1 For < 1 and k 1, there holds ( ( ; k 1 + 2 ( 1 j j2 = 1 + ( 1 k (; n=k+1 [ ] ( ; k ( 1; n k (k+1n n 1 (; n k (19 The overpartition function p(n, for n 1, denotes the number of ways of writing the integer n as a sum of positive integers in non-increasing order in which the first occurrence of an integer may be overlined or not, and p(0 = 1 (see Corteel and Lovejoy [5] It is easy to see that p(n n = Corollary 2 For n, k 1, there holds 1 + n 1 n = 1 + 2 + 42 + 8 3 + 14 4 + 24 5 + 40 6 + ( 1 k (p(n + 2 k ( 1 j p(n j 2 0 (110 2

with strict ineuality if n (k + 1 2 For example, p(n 2p(n 1 0, (111 p(n 2p(n 1 + 2p(n 4 0, p(n 2p(n 1 + 2p(n 4 2p(n 9 0 Theorem 3 For < 1 and k 1, there holds ( ; 2 ( 2 ; 2 k 1 ( 1 j j(2j+1 (1 2j+1 = 1 + ( 1 k 1 ( ; 2 k ( ; 2 n k 2(k+1n k ( 2 ; 2 n n=k [ ] n 1 (112 k 1 2 Let p 1 (n denote the number of partitions of n in which odd parts are not repeated It is easy to see that p 1 (n n = Corollary 4 For n, k 1, there holds ( 1 k 1 k 1 1 + 2n 1 1 2n = 1 + + 2 + 2 3 + 3 4 + 4 5 + 5 6 + 7 7 + ( 1 (p j 1 (n j(2j + 1 p 1 (n (j + 1(2j + 1 0 (113 with strict ineuality if n (2k + 1k For example, p 1 (n p 1 (n 1 p 1 (n 3 + p 1 (n 6 0, p 1 (n p 1 (n 1 p 1 (n 3 + p 1 (n 6 + p 1 (n 10 p 1 (n 15 0 A nice combinatorial proof of (13 was given by Bressoud and Zeilberger [4] It would be interesting to find a combinatorial proof of (15, (110 and (113 A combinatorial proof of (111 will be given in Section 4 2 Proof of Theorem 1 Generalizing Shanks work [8, 9], Andrews, Goulden, and Jackson [2, Theorem 1] established the following identity n (b; j (1 b 2j (b/a; j a j j2 (1 b(; j (a; j = (b; n (a; n When b = 1 and a = 1, the identity (21 reduces to 3 n (b/a; j a j (n+1j (21 (; j

1 + 2 n ( 1 j j2 = n ( 1 j ( 1; j(; n (n+1j (; j ( ; n (22 By (22 and the -binomial theorem (see [1, Theorem 21], we have ( ( ; k 1 + 2 ( 1 j j2 = ( ; k ( 1 j ( 1; j(; k (k+1j (; (; (; j ( ; k = = k ( 1 j ( 1; j( k+1 ; (k+1j (; j ( k+1 ; k ( 1 j i=0 ( 1; j ( 1; i (k+1(i+j (; j (; i (23 By induction on k, it is easy to see that, for n 1, k ( 1 j ( 1; j( 1; n j = ( 1 k ( ; k ( 1; n k (; j (; n j (1 n (; n k 1 (; k Hence, letting i + j = n, the right-hand side of (23 can be written as as desired k ( 1 j ( 1; j( 1; n j (k+1n (; j (; n j = 1 + ( 1 k ( ; k ( 1; n k (k+1n (1 n (; n k 1 (; k [ ] = 1 + ( 1 k ( ; k ( 1; n k (k+1n n 1 (; n k n=k+1 3 Proof of Corollary 2 By Theorem 1, we see that the generating function for the seuence ( } k {( 1 k p(n + 2 ( 1 j p(n j 2 is given by ( 1 k + n=k+1 [ ] ( ; k ( 1; n k (k+1n n 1 (; n k which clearly has nonnegative coefficients of m for m 1 and has positive coefficients of m for m (k + 1 2 This completes the proof 4,,

4 A combinatorial proof of (111 Let P n denote the set of all overpartitions of n We now construct a mapping φ: P n P n 1 as follows: For any λ = (λ 1,, λ k P n, let (λ 1,, λ k 1, if λ k = 1, (λ φ(λ = 1,, λ k 1, λ k 1, if λ k 1, 1, (λ 1,, λ k 2, ˆ1,, 1, if λ }{{} k = 1, λ k 1 1 s where ˆ1 = 1 if λ k 1 is overlined and ˆ1 = 1 otherwise It is easy to see that φ 1 (µ 2 for any µ P n 1 For example, for n = 4, the mapping φ gives 4 3, 4 3, (3, 1 3, (3, 1 (1, 1, 1, (3, 1 (1, 1, 1, (3, 1 3, (2, 2 (2, 1, (2, 2 (2, 1, (2, 1, 1 (2, 1, (2, 1, 1 (2, 1, (2, 1, 1 (2, 1, (2, 1, 1 (2, 1, (1, 1, 1, 1 (1, 1, 1, (1, 1, 1, 1 (1, 1, 1 This proves that p(n 2p(n 1 5 Proof of Theorem 3 and Corollary 4 Shanks [8, 9] proved that n 1 + ( j(2j 1 + j(2j+1 = or euivalently, n 1 j(2j+1 (1 + 2j+1 = n 1 n (; 2 j ( 2 ; 2 n j(2n+1 ( 2 ; 2 j (; 2 n, (; 2 j ( 2 ; 2 n j(2n+1 ( 2 ; 2 j (; 2 n (51 By (51 (with replaced by and the -binomial theorem (see [1, Theorem 21], we have ( ; 2 ( 2 ; 2 k 1 ( 1 j j(2j+1 (1 2j+1 = ( ; 2 ( 2 ; 2 k 1 ( 1 j ( ; 2 j ( 2 ; 2 k (2k+1j ( 2 ; 2 j ( ; 2 k k 1 = ( 1 j ( ; 2 j ( 2k+1 ; (2k+1j ( 2 ; 2 j ( 2k+2 ; k 1 = ( 1 j i=0 ( ; 2 j ( 1 ; 2 i (2k+1(i+j+i ( 2 ; 2 j ( 2 ; 2 i (52 5

By induction on k, it is easy to see that, for n 1, k 1 ( 1 j ( ; 2 j ( 1 ; 2 n j n j = ( 1 k 1 ( ; 2 k ( ; 2 n k n k ( 2 ; 2 j ( 2 ; 2 n j (1 2n ( 2 ; 2 n k ( 2 ; 2 k 1 Hence, letting i + j = n, the right-hand side of (52 can be written as k 1 ( 1 j ( ; 2 j ( 1 ; 2 n j (2k+1n+n j ( 2 ; 2 j ( 2 ; 2 n j = 1 + ( 1 k 1 ( ; 2 k ( ; 2 n k 2(k+1n k (1 2n ( 2 ; 2 n k ( 2 ; 2 k 1 = 1 + ( 1 k 1 ( ; 2 k ( ; 2 n k 2(k+1n k ( 2 ; 2 n n=k [ ] n 1 k 1 2 This proves Theorem 3 The proof of Corollary 4 is similar to that of Corollary 2 and is omitted here 6 Open problems In this section, we propose a common generalization of (15, (110 and (113 Let m, r be positive integers with 1 r m/2 Consider the generalized partition function J m,r (n defined by It is easy to see that J m,r (n n = 1 ( r ; m ( m r ; m ( m ; m (61 J 2,1 (n = p(n, J 3,1 (n = p(n, J 4,1 (n = p 1 (n Moreover, if r < m/2, then J m,r (n can be understood as the number of partitions of n into parts congruent to 0, ±r modulo m Now, Jacobi s triple product identity implies (see [6, p 375] that ( r ; m ( m r ; m ( m ; m = 1 + ( 1 j ( j(mj+m 2r/2 + j(mj+m+2r/2 (62 It follows from (61 and (62 that J m,r (n satisfies the recurrence formula: J m,r (n + ( 1 j (J m,r (n j(mj m + 2r/2 + J m,r (n j(mj + m 2r/2 = 0, where J m,r (s = 0 for all negative s 6 (63

Conjecture 5 For m, n, k, r 1 with r m/2, there holds ( 1 k 1 k 1 ( 1 (J j m,r (n j(mj + m 2r/2 J m,r (n (j + 1(mj + 2r/2 0 with strict ineuality if n k(mk + m 2r/2 (64 In fact, when m = 2 and r = 1, the conjectural ineuality (64 is euivalent to k 1 ( 1 (p(n k 1 + 2 ( 1 j p(n j 2 p(n k 2 (65 with strict ineuality if n k 2 It is clear that (65 is stronger than the proved ineuality (110 (with k replaced by k 1 Finally, along the same line of thinking, we consider the seuence {t(n} n 0 (see A000716 in Sloane s database of integer seuences [10] defined by t(n n 1 = (66 (; 3 = 1 + 3 + 9 2 + 22 3 + 51 4 + 108 5 + 221 6 + 429 7 + 810 8 + 1479 9 + Clearly, the number t(n counts partitions of n into 3 kinds of parts Now, invoking the identity of Jacobi [6, p 377]: (; 3 = ( 1 j (2j + 1 j(j+1/2, we derive the recurrence formula: ( 1 j (2j + 1t(n j(j + 1/2 = 0, (67 where t(m = 0 for all negative m We end the paper with the following conjecture: Conjecture 6 For n, k 1, there holds ( 1 k k ( 1 j (2j + 1t(n j(j + 1/2 0 with strict ineuality if n (k + 1(k + 2/2 For example, t(n 3t(n 1 0, t(n 3t(n 1 + 5t(n 3 0, t(n 3t(n 1 + 5t(n 3 7t(n 6 0 7

References [1] GE Andrews, The Theory of Partitions, Cambridge University Press, Cambridge, 1998 [2] GE Andrews, IP Goulden, and DM Jackson, Shanks convergence acceleration transform, Padé approximants and partitions, J Combin Theory Ser A 43 (1986, 70 84 [3] GE Andrews and M Merca, The truncated pentagonal number theorem, preprint, 2012 [4] DM Bressoud and D Zeilberger, Bijecting Euler s partitions-recurrence, Amer Math Monthly 92 (1985, 54 55 [5] S Corteel and J Lovejoy, Overpartitions, Trans Amer Math Soc 356 (2004, 1623 1635 [6] GH Hardy and EM Wright, An Introduction to the Theory of Numbers, 6th Ed, Oxford University Press, Oxford, 2008 [7] M Merca, Fast algorithm for generating ascending compositions, J Math Modelling and Algorithms 11 (2012, 89 104 [8] D Shanks, A short proof of an identity of Euler, Proc Amer Math Soc 2 (19 51, 747 749 [9] D Shanks, Two theorems of Gauss, Pacific J Math 8 (1958, 609 612 [10] NJA Sloane, On-Line Encyclopedia of Integer Seuences, http://oeisorg/ 8