ON PARTITION FUNCTIONS OF ANDREWS AND STANLEY

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1 ON PARTITION FUNCTIONS OF ANDREWS AND STANLEY AE JA YEE Abstract. G. E. Andrews has established a refinement of the generating function for partitions π according to the numbers O(π) and O(π ) of odd parts in π and the conugate of π, respectively. In this paper, we derive a refined generating function for partitions into at most M parts less than or equal to N, which is a finite case of Andrew s refinement.. Introduction A partition π of n is a weakly decreasing sequence of positive integers whose sum equals n; π can be written as π ( f f 3 f 3 4 f4 ), where exactly f i part of π are equal to i (see, p.].) We denote the conugate of π by π and the number of odd parts of π by O(π). R. P. Stanley 5, 6] has shown that t(n) (p(n) + f(n)), (.) where t(n) is the number of partitions of n for which O(π) O(π ) (mod 4), p(n) is the number of partitions of n, and f(n) is defined by f(n)q n ( + q i ) ( q 4i )( + q 4i. n0 i Definition.. For any integers N, n, r, and s, let S N (n, r, s) be the number of partitions π of n, where each part of π is less than or equal to N, O(π) r, and O(π ) s. Andrews ] has examined S N (n, r, s). As a corollary, he has derived a refinement of the generating function for partitions according to O(π) and O(π ) and proved (.) in an analytic way. A. V. Sills 4] and C. E. Boulet 3] have independently given combinatorial proofs of the case when N tends to infinity. Definition.. For any integers N, M, n, r, and s, let S N,M (n, r, s) be the number of partitions π of n counted by S N (n, r, s) where π has at most M parts. In this paper, we examine the generating function for S N,M (n, r, s) and establish a combinatorial proof of an explicit expression for the generating function. The idea of the combinatorial proof will be generalized to provide combinatorial proofs of Andrews results. We explain what partitions f(n) counts in Section 3. Throughout this paper, we use boxes for Ferrers graphs instead of dots. P. A. MacMahon introduced modular partitions, p.3]. Let n and k be positive integers. Then there exist h 0 and 0 < k such that n kh +. The modular representation Research partially supported by a grant from the Number Theory Foundation.

2 AE JA YEE of a partition to the modulus k is a modification of the Ferrers graph so that n is represented by a row of h k s and one. Thus the representation of to the modulus is the following. Let. Main Results (a; q) 0 :, (a; q) n : ( a)( aq)( aq ( aq n ), n, and let the q-binomial coefficient be defined by n (q; q) n (.) k] (q; q) k (q; q) n k q for k 0,,..., n and a nonnegative n. Note that the q-binomial coefficient (.) is the generating function for partitions into at most k parts less than or equal to n k. Throughout this paper, for convenience, we define { n for k 0, k] 0 for k > 0, for any n < k. Theorem.. If S N,M (n, r, s) is as defined in Section, then S N,M (n, r, s) q n z r y s M k0 M k0 ] N + k (zq) k k 0 N (zq/y) S N,M+ (n, r, s) q n z r y s ] N + k (zq) k k 0 q (yq) () q 4 0 m 0 N (zq/y) (yq) () q 4 0 m 0 (yzq) m ] M + N k m3 N m q 4, (yzq) m M + k ] M + N k m3 N q 4, m M k + m m M k + m m

3 ON PARTITION FUNCTIONS OF ANDREWS AND STANLEY 3 and M k0 S N+,M+ (n, r, s) q n z r y s (zq) k N + k k ] N M k + m m (yq) () q 4 0 m 0 0 (zq/y) ( + yzq)(yzq 5 ) m M + k M k Proof. Let π be a partition counted by S N,M (n, r, s). We write π ( o e 3 o 4 e (N ) o N (N) e N ), m M + N k m3 N where exactly o i parts of π are equal to i and exactly e i parts of π are equal to i. Let Π be the modular representation of π to the modulus. Then Π has parts of size i a total of (o i + e i ) times for i,..., N. There are o i parts of size i whose last boxes have. For instance, Π of ( ) is We cut the odd parts of size i out o i / times from π so that the parts ending with in the modular representation Π of the resulting partition π are distinct. We form a modular partition Π using the parts cut out from π. Let k N i o i/, which is less than or equal to M. Since Π is a modular partition into k parts that are less than or equal to N, and end with, and the parts of Π have even multiplicity, Π is generated by ] N + k (zq) k. (.) k q 4 For instance, Π and Π of ( ) are as follows. Note that the parts of Π ending with are all distinct, and Π has at most M k parts. Moreover, the number of odd parts of π is equal to that of the conugate π of π, since

4 4 AE JA YEE we cut out from π the odd parts of size i a total of o i / times. We add up the numbers inside the boxes in rows and of the graph of Π and put the sums into the boxes in rows deleting rows to obtain a Ferrers graph fitting inside the rectangle of size (M k) N. For instance, Π of ( ) becomes Let m i be the number of columns whose last boxes have i for i,, 3, 4 in the resulting graph. Then, m + m + + m 4 N. The columns with in the last boxes contribute twice to s, and the columns with 3 in the last boxes contribute once to each of r and s. Thus those columns are generated by ] (yq) m M k + m m ] (yq) (m + ) M k + m m ] (zyq 3 ) q 4(m3 m q 4 3 q 4 (zq/y) q 4(m3 q 4, ] M k q 4, (.3) since the columns with 3 in their last boxes are of different length. On the other hand, the remaining columns have either or 4 in their last boxes. The columns with are of different length and they contribute once to r and s at the same time. Thus those columns are generated by (zyq) m ] M k m q 4 q 4m 4 ] M k + m4 m 4 q 4. (.4) Since m + m + + m 4 N, let m + N for some and m + m 4. The sum of (.3) all over m and with m + N becomes 0 (yq) () (zq/y) ] M k + N m3 N and the sum of (.4) all over m and m 4 with m + m 4 becomes since m 0 m i0 (zyq) m q 4m 4 by identity (3.3.9) in ]. m ] M k + m m ] ] M k + m4 M k + m m 4 m q 4 (.5) q 4, (.6)

5 ON PARTITION FUNCTIONS OF ANDREWS AND STANLEY 5 Summing identities (.), (.5), and (.6) over all k M and N, we obtain the first identity of the theorem M k0 S N,M (n, r, s)q n z r y s ] N + k (zq) k k 0 N (zq/y) (yq) () q 4 0 m 0 (zyq) m ] M k + N m3 N m q 4. M k + m m Similarly, we can prove the other two identities. We omit the proofs. Andrews ] has found special cases of the generating function for S N,M (n, r, s) when M tends to infinity. We prove Andrews results in the following theorem. Corollary.. If S N (n, r, s) is as defined in Section, then S N (n, r, s)q n z r y s ( N ] ) ) N ( zyq; q 4 ) ( zy q; q 4 ) (yq) N / ((q 4 ; q 4 ) N (z q ; q 4 ) N, q 4 and 0 S N+ (n, r, s)q n z r y s ( N ] ) ) N ( zyq; q 4 ) + ( zy q; q 4 ) (yq) N / ((q 4 ; q 4 ) N (z q ; q 4 ) N+. q 4 0 Proof. We take M to infinity in the first formula in Theorem.. Then, we have lim M S N (n, r, s)q n z r y s lim 0 k0 M k0 (zq) k N + k k (zq/y) ] N + k (zq) k k M ] M k N S N,M (n, r, s) q n z r y s (yq) () q 4 0 m 0 N (yq) () q 4 0 m 0 (yzq) m M + N k m3 N (yzq) m (q 4 ; q 4 ) m (q 4 ; q 4 ) m m 0 M k + m m (zq/y) (q 4 ; q 4 ) m3 (q 4 ; q 4 ) m3

6 6 AE JA YEE (zq ; q 4 ) N N 0 (zq ; q 4 ) N (q 4 ; q 4 ) N (zq ; q 4 ) N (q 4 ; q 4 ) N (yq) () m 0 N ] N 0 (yzq) m (q 4 ; q 4 ) m (q 4 ; q 4 ) m q 4 (yq) () m 0 0 (zq/y) (q 4 ; q 4 ) m3 (q 4 ; q 4 ) m3 ] N (yzq) m] m q 4(m q 4 0 N ] N (yq) () ( zyq; q 4 ) (zq/y; q 4 ) q 4 0 by identities (3.3.7) and (3.3.6) in ]. We obtain the same result by taking the limit in the second formula in Theorem.. Here we present a combinatorial proof of the generating function for S N (n, r, s), which can be derived from the method for the finite cases. We rewrite the first identity as S N (n, r, s)q n z r y s (z q ; q 4 ) N N k0 Let π be a partition counted by S N (n, r, s). We write π ( o e 3 o 4 e (N ) o N (N) e N ), ( zyq; q 4 ) N k ( zy q; q 4 ) k (y q k (q 4 ; q 4 ) N k (q 4 ; q 4 ) k. where exactly o i parts of π are equal to i and exactly e i parts of π are equal to i. We consider the modular representation of π. We cut out part i ending with a total of o i / times. These parts form a partition π generated by (z q ; q 4 ) N. (.7) Let the resulting partition be π. Note that the modular representation Π of π has boxes with at the corners and the number of odd parts of the conugate π of π is equal to the number of odd parts of the conugate π of π. We add up the numbers inside the boxes in rows and of the graph of Π and put the sums into the boxes in rows deleting rows. Suppose that there are k columns whose last boxes have either or 3. The columns with contribute twice to s, and the columns with 3 contribute once to r and s. Thus those columns are generated by (y q + zyq 3 )(y q + zyq 7 ) (y q + zyq 4k ) (q 4 ; q 4 ) k, (.8) since the columns with 3 in their last boxes are of different length. On the other hand, the remaining columns have either or 4 in their last boxes. The columns with are of different length and they contribute once to r and s at the same time. Thus those columns are generated by ( + zyq)( + zyq 5 ) ( + zyq 4(N k) 3 ) (q 4 ; q 4 ) N k. (.9) Combining the two cases above, we obtain the first identity of the theorem. q 4 (zq/y) q 4(m3

7 ON PARTITION FUNCTIONS OF ANDREWS AND STANLEY 7 The generating function for S N+ (n, r, s) can be obtained in a similar way. We omit the proof. One of the referees has pointed out that we can get identity (4.) in Andrews paper ] by letting N tend to infinity after replacing by N in the first identity in Theorem.. By letting N tend to infinity in the formulas in Corollary., we have the following refinement of the generating function for partitions obtained by Andrews ]. A combinatorial proof of Corollary.3 can be derived by generalizing the idea we employed in the combinatorial proofs of the finite cases. Thus we omit the proof. Corollary.3. Let S (n, r, s) be the number of partitions π of n such that π has r odd parts and the conugate π of π has s odd parts. Then S (n, r, s)q n z r y s ( + zyq ) ( q 4 )( z q 4 ( y q 4. (.0) 3. Remarks As Andrews pointed out ], we can easily see from the generating function (.0) that O(π) O(π ) (mod ). Recall that t(n) is the number of partitions of n for which O(π) O(π ) (mod 4). Andrews showed that where f(n)q n n0 t(n) (p(n) + f(n)), (3.) i ( + q i ) ( q 4i )( + q 4i. To see that t(n) is equal to (p(n)+f(n))/, we examine the generating function (.0). In (.0), the factors ( + zyq ) and /( q 4 ) have nothing to do with the parities of O(π) and O(π ). However, by changing the negative sign to positive in the other factors /( z q 4 and /( y q 4, we see that π has weight ( ) O(π) O(π ) /. Thus f(n) is equal to the number of partitions π of n where O(π) O(π ) (mod 4) minus the number of partitions of π of n where O(π) O(π ) (mod 4). Therefore, t(n) is equal to half of (p(n) + f(n)). Acknowledgment. The author thanks B. C. Berndt for his suggestion of these problems and encouragement, and also thanks D. Stanton for his comments. The author is grateful to anonymous referees for very helpful comments. References ] G. E. Andrews, The Theory of Partitions, Addison Wesley, Reading, MA, 976; reissued by Cambridge University Press, Cambridge, 984. ] G. E. Andrews, On a partition function of Richard Stanley, E. J. Combinatorics, to appear. 3] C. E. Boulet, A four-parameter partition identity, preprint.

8 8 AE JA YEE 4] A. V. Sills, A combinatorial proof of a partition identity of Andrews and Stanley, submitted for publication. 5] R. P. Stanley, Problem 0969, Amer. Math. Monthly 09 (00), ] R. P. Stanley, Some remarks on sign-balanced and ma-balanced possets, Advances in Applied Math., to appear. Department of Mathematics, The Pennsylvania State University, University Park, PA 680, USA address: yee@math.psu.edu