A crank for bipartitions with designated summands
aa r X i v : . [ m a t h . C O ] F e b A crank for bipartitions with designated summands
Robert X. J. Hao , Erin Y. Y. Shen Department of Mathematics and Physics,Nanjing Institute of TechnologyNanjing 211167, P. R. China School of Science, Hohai UniversityNanjing 210098, P. R. China [email protected], [email protected] Abstract.
Andrews, Lewis and Lovejoy introduced the partition function
P D ( n ) as thenumber of partitions of n with designated summands. A bipartition of n is an ordered pairof partitions ( π , π ) with the sum of all of the parts being n . In this paper, we introducea generalized crank named the pd -crank for bipartitions with designated summands andgive some inequalities for the pd -crank of bipartitions with designated summands modulo2 and 3. We also define the pd -crank moments weighted by the parity of pd -cranks µ k,bd ( − , n ) and show the positivity of ( − n µ k,bd ( − , n ). Let M bd ( m, n ) denote thenumber of bipartitions of n with designated summands with pd -crank m . We prove amonotonicity property of pd -cranks of bipartitions with designated summands and findthat the sequence { M bd ( m, n ) } | m |≤ n is unimodal for n = 1 , , Keywords : bipartition with designated summands; pd -crank; moment; monotonicity AMS Classifications : 11P81, 11P83, 05A17, 05A20 Introduction
A partition λ of a positive integer n is a finite nonincreasing sequence of positive integers λ = ( λ , . . . , λ l ) such that | λ | = P li =1 λ i = n . Let p ( n ) denote the number of partitionsof n . There are two vital statistics in the theory of partitions named Dyson’s rank [8]and the Andrews–Garvan–Dyson crank [2]. The rank [8] of a partition λ , denoted by r ( λ ), is defined as the largest part of λ minus the number of the parts. The rank can beused to provide combinatorial interpretations for the following two Ramanujan’s famouscongruences p (5 n + 4) ≡ , (1.1) p (7 n + 5) ≡ . (1.2)1he crank [2] of a partition λ of n > c ( λ ) = ( λ , if n ( λ ) = 0 ,µ ( λ ) − n ( λ ) , if n ( λ ) > , where n ( λ ) denotes the number of parts equal to one in λ and µ ( λ ) denotes the num-ber of parts in λ larger than n ( λ ). The crank can be used to provide combinatorialinterpretations for congruences (1.1), (1.2) as well as p (11 n + 6) ≡ . (1.3)Let M ( m, n ) denote the number of partitions with crank m . For n ≤ m = ± , M ( m, n ) = 0. For n = 1 and m = ± ,
0, we define M ( − ,
1) = M (1 ,
1) = 1 , M (0 ,
1) = − . The generating function of M ( m, n ) is obtained by Andrews and Garvan [2, 10] as givenby C ( z ; q ) = ∞ X m = −∞ ∞ X n =0 M ( m, n ) z m q n = ( q ; q ) ∞ ( zq ; q ) ∞ ( z − q ; q ) ∞ , (1.4)where here and throughout this paper, ( a ; q ) ∞ stands for the q -shifted factorial( a ; q ) ∞ = ∞ Y n =1 (1 − aq n − ) , | q | < . Let M ( m, t, n ) denote the number of partitions of n with crank congruent to m modulo t . Setting z = − C ( − , q ) = ∞ X n =0 ( M (0 , , n ) − M (1 , , n )) q n = ( q ; q ) ∞ ( − q ; q ) ∞ . (1.5)Andrews and Lewis [3] proved that M (0 , , n ) > M (1 , , n ) , (1.6) M (1 , , n + 1) > M (0 , , n + 1) , (1.7)for n ≥
0. The above two inequalities imply that the signs of the coefficients of ( q ; q ) ∞ ( − q ; q ) ∞ are alternating.Later, to study the higher-order spt-function spt k ( n ), Garvan [11] introduced the k -thsymmetrized moments µ k ( n ) of cranks of partitions of n . He defined the k -th symmetrizedmoments µ k ( n ) of cranks of partitions of n as µ k ( n ) = ∞ X m = −∞ (cid:18) m + ⌊ k − ⌋ k (cid:19) M ( m, n ) . (1.8)2ince the fact that M ( m, n ) = M ( − m, n ) [2, (1.9)], it is easy to see that µ k +1 ( n ) = 0.In a recent work, Ji and Zhao [13] introduced the crank moments weighted by the parityof cranks as given by µ k ( − , n ) = ∞ X m = −∞ (cid:18) m + k − k (cid:19) ( − m M ( m, n ) . (1.9)Let k = 0 in (1.9), we get µ ( − , n ) = M (0 , , n ) − M (1 , , n ) . Ji and Zhao [13] proved the following positivity property of ( − n µ k ( − , n ). Theorem 1.1. ([13, Theorem 1.2])
For all n ≥ k ≥ , we have ( − n µ k ( − , n ) > . (1.10)This property implies the two inequalities (1.6) and (1.7) of Andrews and Lewis [3].Recently, Ji and Zang [14] proved the following monotonicity property of cranks ofpartitions. Theorem 1.2. ([14, Theorem 1.7])
For all n ≥ and ≤ m ≤ n − , we have M ( m − , n ) ≥ M ( m, n ) . (1.11)Considering Theorem 1.2 and the symmetry M ( m, n ) = M ( − m, n ), the followingcorollary is true. Corollary 1.3. ([14, Corollary 1.8])
For n ≥ , we have M (1 − n, n ) ≤ · · · ≤ M ( − , n ) ≤ M (0 , n ) ≥ M (1 , n ) ≥ · · · ≥ M ( n − , n ) . This means the sequence { M ( m, n ) } | m |≤ n − is unimodal for n ≥ P D ( n ) denote the number of partitions of n with designated summands. There areten partitions of 4 with designated summands:4 ′ , ′ + 1 ′ , ′ + 2 , ′ , ′ + 1 ′ + 1 , ′ + 1 + 1 ′ , ′ + 1 + 1 + 1 , ′ + 1 + 1 , ′ + 1 , ′ . Hence we have
P D (4) = 10. Andrews, Lewis and Lovejoy [4] derived the generatingfunction of
P D ( n ) as given by ∞ X n =0 P D ( n ) q n = ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ . (1.12)3ndrews, Lewis and Lovejoy [4] obtained a Ramanujan-type congruence for the partitionfunction P D ( n ) as P D (3 n + 2) ≡ . (1.13)By introducing the pd -rank for partitions with designated summands, Chen, Ji, Jin andthe second author [7] gave a combinatorial interpretation of the congruence (1.13).A bipartition π of n is an ordered pair of partitions ( π , π ) with | π | + | π | = n . Let p − ( n ) denote the number of bipartitions of n . The generating function of p − ( n ) is ∞ X n =0 p − ( n ) q n = 1( q ; q ) ∞ . A bipartition with designated summands means a bipartition π = ( π , π ) for which π and π are both partitions with designated summands. Here π and π are allowed tohave one part size tagged in common. Let P D − ( n ) denote the number of bipartitions of n with designated summands. We get the generating function of P D − ( n ) as given by ∞ X n =0 P D − ( n ) q n = ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ . (1.14)Recently, arithmetic properties of bipartitions with designated summands have drawn anumber of interest, see, for example [12, 15].The main objective of this paper is to investigate bipartitions with designated sum-mands from three aspects. First, we introduce a generalized crank named the pd -crank forbipartitions with designated summands and establish some inequalities for the pd -crankof bipartitions with designated summands modulo 2 and 3. Second, we define the pd -crank moments weighted by the parity of pd -cranks µ k,bd ( − , n ) and show the positivityproperty of ( − n µ k,bd ( − , n ). Finally, we prove a monotonicity property of pd -cranks ofbipartitions with designated summands. The pd -crank and its inequalities In this section, we first introduce a generalized crank which called pd -crank for bipartitionswith designated summands. The definition of the pd -crank relies on the construction ofthe following bijection ∆ which Chen, Ji, Jin and the second author established in [7]. Theorem 2.1. ([7, Theorem 3.1])
There is a bijection ∆ between the set of partitions of n with designated summands and the set of vector partitions ( α, β ) with | α | + | β | = n ,where α is an ordinary partition and β is a partition into parts
6≡ ± . Chen, Ji, Jin and the second author gave a combinatorial proof of the above theoremwhich illustrates the construction of the bijection ∆, see [7, Theorem 3.1]. Under themap ∆, Chen, Ji, Jin and the second author defined the pd -rank of a partition λ withdesignated summands in terms of the pair of partitions ( α, β ).4 efinition 2.2. ([7, Definition 3.2]) Let λ be a partition with designated summands andlet ( α, β ) = ∆( λ ) . Then the pd -rank of λ , denoted r d ( λ ) , is defined by r d ( λ ) = l e ( α ) − l e ( β ) , (2.1) where l e ( α ) is the number of even parts of α and and l e ( β ) is the number of even parts of β . We are now ready to give the definition of the pd -crank for a bipartition ( λ , λ ) withdesignated summands under the map ∆. Definition 2.3.
Let ( λ , λ ) be a bipartition with designated summands and let (( α , β ) , ( α , β )) = (∆( λ ) , ∆( λ )) . Then the pd -crank of ( λ , λ ) , denoted r bd ( λ ) , is defined by r bd ( λ ) = l ( α ) − l ( α ) , (2.2) where l ( α ) is the number of parts of α . Let M bd ( m, n ) denote the number of bipartitions of n with designated summands with pd -crank m , and M bd ( k, t, n ) denote the number of bipartitions with designated summandsof n with pd -crank congruent to k modulo t . According to the Definition 2.3, it is clearthat M bd ( m, n ) = M bd ( − m, n ) . (2.3)Following the results of Andrews and Lewis [3], we investigate the inequalities of the pd -crank of bipartitions with designated summands modulo 2 and 3. By the definition of the pd -crank of bipartitions with designated summands, we have ∞ X m = −∞ ∞ X n =0 M bd ( m, n ) z m q n = ( q ; q ) ∞ ( zq ; q ) ∞ ( z − q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ . (2.4)The following theorem shows that the sequence M bd (0 , , n ) − M bd (1 , , n ) alternates insign. Theorem 2.4.
For n ≥ , we have M bd (0 , , n ) > M bd (1 , , n ) , (2.5) M bd (1 , , n + 1) > M bd (0 , , n + 1) . (2.6) Proof.
Setting z = − ∞ X n =0 ( M bd (0 , , n ) − M bd (1 , , n )) q n = ( q ; q ) ∞ ( − q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ = ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ = f f f f , (2.7)5here here and throughout this paper, f k is defined by f k = ( q k ; q k ) ∞ . By the following 2-dissection([17, Lemma 2.6]) f f = f f f f f f − q f f f f f f f f , we can deduce that f f = f f f f f f + q f f f f f f f f − q f f f f f f . (2.8)Applying (2.8) into (2.7), we get ∞ X n =0 ( M bd (0 , , n ) − M bd (1 , , n )) q n = f f (cid:18) f f f f f f + q f f f f f f f f − q f f f f f f (cid:19) , which implies the coefficient of q n in (2.7) is positive(negative) when n is even(odd). Thiscompletes the proof.Considering the pd -crank of bipartitions with designated summands modulo 3, we havethe following results. Theorem 2.5.
For n ≥ , we have M bd (0 , , n ) > M bd (1 , , n ) , (2.9) M bd (0 , , n + 1) < M bd (1 , , n + 1) , (2.10) M bd (0 , , n + 2) > M bd (1 , , n + 2) , (2.11) M bd (0 , , n + 3) > M bd (1 , , n + 3) , (2.12) M bd (0 , , n + 4) = M bd (1 , , n + 4) = M bd (2 , , n + 4) , (2.13) M bd (0 , , n + 5) < M bd (1 , , n + 5) . (2.14) Proof.
Substituting z = ζ = e πi into (2.4), we find that ∞ X m = −∞ ∞ X n =0 M bd ( m, n ) ζ m q n = ∞ X n =0 2 X k =0 M bd ( k, , n ) ζ k q n = ∞ X n =0 ( M bd (0 , , n ) − M bd (1 , , n )) q n = ( q ; q ) ∞ ( ζ q ; q ) ∞ ( ζ − q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ = ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ . (2.15)6sing the 2-dissection of ( q ; q ) ∞ ( q ; q ) ∞ [17, Lemma 2.5], which is( q ; q ) ∞ ( q ; q ) ∞ = ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ − q ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ , (2.16)we get ∞ X n =0 ( M bd (0 , , n ) − M bd (1 , , n )) q n = ( q ; q ) ∞ ( q ; q ) ∞ (cid:18) ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ − q ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ (cid:19) . (2.17)Therefore we have ∞ X n =0 ( M bd (0 , , n ) − M bd (1 , , n )) q n = ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ = ( q ; q ) ∞ ( q ; q ) ∞ ψ ( q ) . (2.18)By the identity [5, p. 49] ψ ( q ) = f ( q , q ) + qψ ( q ) , we deduce that ∞ X n =0 ( M bd (0 , , n ) − M bd (1 , , n )) q n = ( q ; q ) ∞ ( q ; q ) ∞ (cid:0) f ( q , q ) + qψ ( q ) (cid:1) . (2.19)Since the coefficients of q n and q n +1 in (2.19) are both positive, (2.9) and (2.11) are true.Noting that the coefficient of q n +2 in (2.19) is zero, hence we have (2.13).According to (2.17), we find that ∞ X n =0 ( M bd (1 , , n + 1) − M bd (0 , , n + 1)) q n = ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ = 1 ψ ( q ) · ( q ; q ) ∞ ( q ; q ) ∞ . (2.20)Let A ( q ) = f f f f , by the identity [12, Lem. 2.2]1 ψ ( q ) = ψ ( q ) ψ ( q ) (cid:0) A ( q ) − qA ( q ) ψ ( q ) + q ψ ( q ) (cid:1) , (2.21)7e may obtain that ∞ X n =0 ( M bd (1 , , n + 1) − M bd (0 , , n + 1)) q n = ( q ; q ) ∞ ( q ; q ) ∞ · ψ ( q ) ψ ( q ) (cid:0) A ( q ) − qA ( q ) ψ ( q ) + q ψ ( q ) (cid:1) . (2.22)It is clear that the coefficients of q n and q n +2 in (2.22) are both positive and the coefficientof q n +1 is negative. These lead to (2.10), (2.14) and (2.12) respectively. This completesthe proof.Theorem 2.5 shows that the pd -crank we defined can be used to divide the set ofbipartitions of 6 n + 4 with designated summands into three equinumerous classes. Hencewe provide a combinatorial interpretation for the congruence P D − (6 n + 4) ≡ The pd -crank moments weighted by the parity of pd -cranks As a natural analog to the 2 k -th crank moments µ k ( − , n ) weighted by the parity ofcranks due to Ji and Zhao [13], we define the 2 k -th pd -crank moments of bipartitionswith designated summands weighted by the parity of pd -cranks as given by µ k,bd ( − , n ) = ∞ X m = −∞ (cid:18) m + k − k (cid:19) ( − m M bd ( m, n ) . (3.1)In this section, we aim to prove the following positivity property of ( − n µ k,bd ( − , n ). Theorem 3.1.
For n ≥ k ≥ , we have ( − n µ k,bd ( − , n ) > . (3.2)It is worth mentioning that Theorem 3.1 reduces to Theorem 2.4 when k = 0 in (3.2).With the aid of the proof of Theorem 1.3 of Ji and Zhao [13] and applying Andrews’ j -fold generalization of q -Whipple’s theorem [1], we obtain the generating function of µ k,bd ( − , n ) as ∞ X n =0 µ k,bd ( − , n ) q n = ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( − q ; q ) ∞ X n k ≥ n k − ···≥ n ≥ ( − k q n + n + ··· + n k (1 + q n ) (1 + q n ) · · · (1 + q n k ) . (3.3)8onsidering the proof of Theorem 1.4 of Ji and Zhao [13], we find that the generatingfunction (3.3) is equivalent to ∞ X n =0 µ k,bd ( − , n ) q n = ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( − q ; q ) ∞ X m k >m k − > ··· >m ≥ ( − m k m ( m − m ) · · · ( m k − m k − ) q m k (1 − q m )(1 − q m ) · · · (1 − q m k ) . (3.4)We are now stand at the point to prove Theorem 3.1. Proof of Theorem3.1.
Replacing q by − q in (3.4), we get ∞ X n =0 ( − n µ k,bd ( − , n ) q n = ( q ; q ) ∞ ( q ; q ) ∞ ( − q ; − q ) ∞ ( q ; − q ) ∞ X m k >m k − > ··· >m ≥ m ( m − m ) · · · ( m k − m k − ) q m k (1 − ( − q ) m )(1 − ( − q ) m ) · · · (1 − ( − q ) m k ) . (3.5)Using the facts that 1( q ; − q ) ∞ = ( − q ; q ) ∞ , ( − q ; − q ) ∞ = ( − q ; q ) ∞ ( q ; q ) ∞ , (3.5) turns out to be ∞ X n =0 ( − n µ k,bd ( − , n ) q n = ( q ; q ) ∞ ( − q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( − q ; q ) ∞ X m k >m k − > ··· >m ≥ ( − q ; q ) ∞ · m ( m − m ) · · · ( m k − m k − ) q m k (1 − ( − q ) m )(1 − ( − q ) m ) · · · (1 − ( − q ) m k )= ( q ; q ) ∞ ( − q ; q ) ∞ ( − q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( − q ; q ) ∞ X m k >m k − > ··· >m ≥ ( − q ; q ) ∞ · m ( m − m ) · · · ( m k − m k − ) q m k (1 − ( − q ) m )(1 − ( − q ) m ) · · · (1 − ( − q ) m k )(3.6)Multiplying the right hand side of (3.6) by( q ; q ) ∞ ( − q ; q ) ∞ ( − q ; q ) ∞ ( q ; q ) ∞ ( − q ; q ) ∞ ( − q ; q ) ∞ ,
9e get ∞ X n =0 ( − n µ k,bd ( − , n ) q n = ( − q ; q ) ∞ ( − q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ × X m k >m k − > ··· >m ≥ ( − q ; q ) ∞ · m ( m − m ) · · · ( m k − m k − ) q m k (1 − ( − q ) m )(1 − ( − q ) m ) · · · (1 − ( − q ) m k ) . (3.7)Applying the Jacobi triple product identity [6, (1.3.10)]( − z ; q ) ∞ ( − q/z ; q ) ∞ ( q ; q ) ∞ = ∞ X n = −∞ z n q ( n )into (3.7) with q replaced by q and z replaced by q , we deduce that ∞ X n =0 ( − n µ k,bd ( − , n ) q n = P ∞ n = −∞ q n (3 n − / ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ × X m k >m k − > ··· >m ≥ ( − q ; q ) ∞ · m ( m − m ) · · · ( m k − m k − ) q m k (1 − ( − q ) m )(1 − ( − q ) m ) · · · (1 − ( − q ) m k ) (3.8)Given m k > m k − > · · · > m ≥
1, we define X m ≥ h m ,m ,...,m k ( m ) q m = ( − q ; q ) ∞ (1 − ( − q ) m )(1 − ( − q ) m ) · · · (1 − ( − q ) m k ) . (3.9)In (3.9), when m i is odd, the corresponding term (1 − ( − q ) m i ) in denominator can becanceled by ( − q ; q ) ∞ since m i differs from each other. When m i is even, the corre-sponding term 1 / (1 − ( − q ) m i ) does not have negative coefficients. Therefore we find that h m ,m ,...,m k ( m ) ≥ h m ,m ,...,m k (0) = 1. This implies that ( − n µ k,bd ( − , n ) > n ≥ k . This completes the proof. A monotonicity property of M bd ( m, n ) In this section, we investigate the following monotonicity property of M bd ( m, n ) whichleads to the unimodality of the sequence { M bd ( m, n ) } | m |≤ n with n = 1 , , heorem 4.1. For m ≥ and n ≥ , M bd ( m − , n ) ≥ M bd ( m, n ) , (4.1) except for ( m, n ) = (1 , , (1 , or (1 , . Fu and Tang [9] studied a generalized crank named k -crank for k -colored partitions.Let M k ( m, n ) denote the number of k -colored partitions of n with k -crank m . The gen-erating function of M k ( m, n ) is given by ∞ X m = −∞ ∞ X n =0 M k ( m, n ) z m q n = ( q ; q ) − k ∞ ( zq ; q ) ∞ ( z − q ; q ) ∞ (4.2)for k ∈ N .Now we provide a proof for Theorem 4.1. Proof of Theorem 4.1.
Setting k = 2 in (4.2) and applying it to (2.4), we get ∞ X m = −∞ ∞ X n =0 M bd ( m, n ) z m q n = ( q ; q ) ∞ ( zq ; q ) ∞ ( z − q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ = ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ∞ X m = −∞ ∞ X n =0 M ( m, n ) z m q n . It is clear that ∞ X n =0 ( M bd ( m − , n ) − M bd ( m, n )) q n = ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ∞ X n =0 ( M ( m − , n ) − M ( m, n )) q n . (4.3)According to [16, Theorem 1.4] and [16, (6.3)], Zang and Zhang proved that M k ( m − , n ) ≥ M k ( m, n ) for m ≥ k ≥ n ≥
0. Hence by (4.3), we find that M bd ( m − , n ) ≥ M bd ( m, n ) when m ≥ n ≥ m = 1, we have ∞ X n =0 ( M bd (0 , n ) − M bd (1 , n )) q n = ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ∞ X n =0 ( M (0 , n ) − M (1 , n )) q n . (4.4)Let { c n } ∞ n =0 be a sequence of nonnegative integers. By [16, (6.5)] and a simple calculation,we can derive that ∞ X n =0 ( M (0 , n ) − M (1 , n )) q n = 1 − q + q + ∞ X n =6 c n q n . (4.5)11oting that ( q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ = ( q ; q ) ∞ ( − q ; q ) ∞ ( q ; q ) ∞ ( q ; q ) ∞ ( − q ; q ) ∞ = ∞ Y i =1 q i − q i = ∞ Y i =1 (cid:18) q i − q i (cid:19) , (4.6)and substituting (4.5), (4.6) into (4.4), we have ∞ X n =0 ( M bd (0 , n ) − M bd (1 , n )) q n = − q + q + ∞ X n =6 c n q n ! ∞ Y i =1 (cid:18) q i − q i (cid:19) . (4.7)Next, we aim to show the coefficients of(1 − q + q ) ∞ Y i =1 (cid:18) q i − q i (cid:19) (4.8)are nonnegative except for n = 1 , , P ( n ) denote the number of partitions of n with each part exists at least twice.Let P ( n ) denote the number of partitions of n counted by P ( n ) with 1 occurs zero, twoor four times. We deduce that(1 − q + q ) ∞ Y i =1 (cid:18) q i − q i (cid:19) = ∞ Y i =1 (cid:18) q i − q i (cid:19) − q (1 − q ) (cid:18) q − q (cid:19) ∞ Y i =2 (cid:18) q i − q i (cid:19) = ∞ Y i =1 (cid:18) q i − q i (cid:19) − q (1 + q + q ) ∞ Y i =2 (cid:18) q i − q i (cid:19) = 1 + ∞ X n =1 ( P ( n ) − P ( n − q n . (4.9)After a simple calculation, we find that(1 − q + q ) ∞ Y i =1 (cid:18) q i − q i (cid:19) = 1 − q + q + 2 q − q + 4 q − q + 6 q + 8 q + 15 q + ∞ X n =14 a n q n . (4.10)In light of (4.9) and (4.10), we wish to construct an injection Ω from the set of partitions α counted by P ( n −
1) to the set of partitions β enumerated by P ( n ) to prove thecoefficients a n of (4.10) are nonnegative for n ≥ α counted by P ( n −
1) has 1 as its part, 1 may occur twice or fourtimes. By adding a 1 to α as a part, we get a partition β = Ω( α ) enumerated by P ( n ).Hence β has three or five 1s. For instance, let α = (4 , , , , ,
1) and it belongs to the setof partitions counted by P (16). Under the map Ω, the corresponding partition β = Ω( α )should be (4 , , , , , ,
1) and it belongs to the set of partitions counted by P (17).Suppose α is a partition enumerated by P ( n −
1) with no part equals to 1. Let thelargest part size of the partition α be m and the second largest part size be α . We havethe following cases. • If m ≥ m = 3 and m appears more than twice in α , we obtain β by rewritingone m in α plus one more 1 as m + 1 1s. For example, let α = ( m, m, . . . , m, | {z } l, l> α , α , . . . , α n , α n ) , the corresponding partition β = Ω( α ) should be β = ( m, . . . , m, | {z } l − α , α , . . . , α n , α n , , , . . . , | {z } m +1 ) . • If m ≥ m = 3 and m appears exactly twice in α , we obtain β by rewriting two m in α plus one more 1 as 2 m + 1 1s. For example, let α = ( m, m, α , α , . . . , α n , α n ) , the corresponding partition β = Ω( α ) should be β = ( α , α , . . . , α n , α n , , , . . . , | {z } m +1 ) . • If m = 4 and 4 appears more than three times or exactly twice in α , we obtain β by rewriting two 4s in α plus one more 1 as nine 1s. For example, let α = (4 , , . . . , , | {z } l, l =2 or l> α , α , . . . , α n , α n ) , the corresponding partition β = Ω( α ) should be β = (4 , , . . . , , | {z } l − α , α , . . . , α n , α n , , , . . . , | {z } ) . • If m = 4 and 4 appears exactly three times in α , we obtain β by rewriting the three4s in α plus one more 1 as 3 , , , ,
2. For example, let α = (4 , , , α , . . . , α , α , α , . . . ) , the corresponding partition β = Ω( α ) should be β = (3 , , , α , . . . , α , , , α , α , . . . ) . If m = 2 and 2 appears more than or equal to seven times in α , we obtain β byrewriting five 2s in α plus one more 1 as 3 , , , ,
1. For example, let α = (2 , , . . . , , | {z } l, l ≥ ) , the corresponding partition β = Ω( α ) should be β = (3 , , , , , . . . , , | {z } l − , . Therefore the map Ω is an injection. Hence the coefficients a n of (4.10) are nonnegativefor n ≥
14. Applying (4.10) into (4.8), we have(1 − q + q ) ∞ Y i =1 (cid:18) q i − q i (cid:19) = − q + q + 2 q − q + 4 q − q + 6 q + 8 q + 15 q + ∞ X n =14 a n q n ! ∞ Y i =1 (cid:18) q i − q i (cid:19) = (1 + q + q )(1 − q + q ) ∞ Y i =1 (cid:18) q i − q i (cid:19) + q + 3 q + 5 q + 7 q + 15 q + ∞ X n =14 a n q n ! ∞ Y i =1 (cid:18) q i − q i (cid:19) . (4.11)Substituting (4.10) into (4.11), we deduce that(1 − q + q ) ∞ Y i =1 (cid:18) q i − q i (cid:19) = (1 + q + q ) − q + q + 2 q − q + 4 q − q + 6 q + 8 q + 15 q + ∞ X n =14 a n q n ! + q + 3 q + 5 q + 7 q + 15 q + ∞ X n =14 a n q n ! ∞ Y i =1 (cid:18) q i − q i (cid:19) = (1 + q + q )( − q − q − q ) + (1 + q + q )(1 + q + 2 q + 4 q + 6 q + 8 q + 15 q + ∞ X n =14 a n q n ) + q ∞ Y i =1 (cid:18) q i − q i (cid:19) + q + 5 q + 7 q + 15 q + ∞ X n =14 a n q n ! ∞ Y i =1 (cid:18) q i − q i (cid:19) = G ( q ) + G ( q ) + G ( q ) + G ( q ) , (4.12)14here G ( q ) = (1 + q + q )( − q − q − q ) = − q − q − q − q − q − q , (4.13) G ( q ) = (1 + q + q )(1 + q + 2 q + 4 q + 6 q + 8 q + 15 q + ∞ X n =14 a n q n ) , (4.14) G ( q ) = q ∞ Y i =1 (cid:18) q i − q i (cid:19) , (4.15) G ( q ) = q + 5 q + 7 q + 15 q + ∞ X n =14 a n q n ! ∞ Y i =1 (cid:18) q i − q i (cid:19) . (4.16)It is clear that the coefficients of the functions G ( q ), G ( q ) and G ( q ) are all nonnegative.After a simple calculation, we get G ( q ) = q ∞ Y i =1 (cid:18) q i − q i (cid:19) = q (1 + q + q + 2 q + q + 4 q + 2 q + 6 q + 5 q + 9 q + 7 q + ∞ X n =12 b n q n ) , = q + q + q + 2 q + q + 4 q + 2 q + 6 q + 5 q + 9 q + 7 q + ∞ X n =12 b n q n +2 , (4.17)where b n are nonnegative for n ≥
12. Using (4.13) and (4.17), we deduce that G ( q ) + G ( q )= − q + q + q − q + 2 q − q + 4 q + q + 6 q + 3 q + 9 q + 6 q + ∞ X n =12 b n q n +2 . (4.18)Applying (4.18) into (4.12), we have(1 − q + q ) ∞ Y i =1 (cid:18) q i − q i (cid:19) = G ( q ) + G ( q ) + G ( q ) + G ( q )= − q + q + q − q + 2 q − q + 4 q + q + 6 q + 3 q + 9 q + 6 q + ∞ X n =12 b n q n +2 + G ( q ) + G ( q ) . (4.19)Since the coefficients of q , q and q in G ( q ) + G ( q ) are all zero, the coefficients of (4.8)are nonnegative except for n = 1 , ,
7. This completes the proof.15ith the aid of Theorem 4.1 and (2.3), we arrive at the following corollary.
Corollary 4.2.
For n = 1 , , , we have M bd ( n, n ) ≤ · · · ≤ M bd ( − , n ) ≤ M bd (0 , n ) ≥ M bd (1 , n ) ≥ · · · ≥ M bd ( n, n ) . This means the sequence { M bd ( m, n ) } | m |≤ n is unimodal for n = 1 , , Acknowledgments.
The first author was supported by the Scientific Research Founda-tion of Nanjing Institute of Technology (No. YKJ201627). The second author was sup-ported by the Natural Science Foundation of Jiangsu Province of China (No. BK20160855)and the National Natural Science Foundation of China (No. 11801139).
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