O r (mod p) is distinct and higher than parts congruent to 0 (mod p).

O r (mod p) is distinct and higher than parts congruent to 0 (mod p). Our theorem is stated below. Theorem 3. Let O(n, p, r) be the PX-478 Autophagy number of Leukotriene D4 medchemexpress partitions of n in which parts are congruent to 0, r (mod p), components r (mod p) are distinct, and each integer congruent to r (mod p) smaller than the biggest part that is congruent to r (mod p) need to appear as a component. Then, D (n, p, r) = O(n, p, r). Proof. Setting p = 2, r = 1 in Theorem three gives rise to Theorem two. We give a preferred bijective proof. Let be enumerated by O(n, p, r). We’ve the decomposition = (1 , 2) exactly where 1 will be the subpartition of whose parts are r (mod p), and 2 would be the subpartition of whose parts are congruent to 0 (mod p). Then, the image is given by 1 2 , i.e., 1 2 . The inverse on the bijection is provided as follows: Let be a partition enumerated by D (n, p, r). Then, decompose as = ( , ) exactly where would be the subpartition with components congruent to r (mod p) and may be the subpartition with components congruent to 0 (mod p). Construct as = ( p – p r, p – 2p r, p – 3p r, . . . , r 2p, r p, r) where would be the number of parts in . Then the image of is given by [ – ].Example 1. Contemplate p = four, r = 1 and an O(190, four, 1)-partition = (32, 32, 21, 17, 16, 13, 9, eight, eight, eight, 8, 5, 4, four, four, 1). By our mapping, decomposes as follows: = ((21, 17, 13, 9, five, 1), (32, 32, 16, 8, 8, eight, eight, four, four, 4)). The image is then given by(21, 17, 13, 9, 5, 1, 0, 0, 0, 0) (32, 32, 16, eight, eight, eight, eight, four, 4, 4)(we append zeros to the subpartition with smaller sized length), and addition is componentwise within the order demonstrated. Thus (53, 49, 29, 17, 13, 9, 8, four, four, 4) which can be a partition enumerated by D (190, 4, 1). To invert the process, beginning with = (53, 49, 29, 17, 13, 9, eight, 4, four, four), enumerated by D (190, 4, 1), we’ve got the decomposition = ( , ) = ((53, 49, 29, 17, 13, 9), (eight, four, four, 4)) where = (53, 49, 29, 17, 9) and = (eight, 4, 4, four). Note that = 5 to ensure that = (17, 13, 9, 5, 1). Hence, the image isMathematics 2021, 9,four of [ – ] = (eight, four, 4, four) (21, 17, 13, 9, five, 1) [(53, 49, 29, 17, 13, 9) – (21, 17, 13, 9, 5, 1)]= (eight, 4, 4, 4) (21, 17, 13, 9, 5, 1) (32, 32, 16, 8, eight, 8) = (32, 32, 21, 17, 16, 13, 9, 8, 8, 8, 8, five, 4, four, 4, 1),that is enumerated by O(186, 4, 1) and the we began with. Corollary 1. The amount of partitions of n in which all components 0 (mod p) kind an arithmetic progression with popular distinction p as well as the smallest portion is less than p equals the number of partitions of n in which parts 0 (mod p) are distinct, possess the similar residue modulo p and are higher than components 0 (mod p). Proof. By Theorem 3, we’ve O(n, p, r) = D (n, p, r).r =1 r =1 p -1 p -3. Associated Variations In Theorem 2, if we reverse the roles of odd and also components by letting any positive even integer significantly less than the biggest even part appear as a portion and every single odd portion be greater than the biggest even component, we acquire the following theorem. Theorem four. Let r = 1, three in addition to a(n, r) denote the amount of partitions of n in which each and every even integer less than the biggest even aspect seems as a part and the smallest odd element is no less than r the largest even component. Then, A(n, r) is equal to the quantity of partitions of n with components r, two (mod 4). Proof.n =A(n, r)qn =q n ( n 1) 1 1 2 2 2jr) 2jr) j =0 (1 – q n =1 ( q ; q) n j = n (1 – q= = = = = = = =q n ( n 1) 1 (q2 ; q2)n n (1 – q2jr) j= n =q n ( n 1) 1 two ; q2) ( q2nr ; q2) n n =0 ( q q n ( n 1) ( q r ; q two) n 2 2 two n=0 ( q ; q)n ( q; q) 1 (q; q2) 1 (q; q2) q n ( n 1) r 2 (q ; q)n 2 two n =0 ( q ; q) nn =1 n =.