The ith biggest part of is equal to the sum of your ith largest

The ith biggest part of is equal to the sum of your ith largest parts in and In getting the sum the partition with smaller length need to have zeros appended to it in order to match in length with the other partition. Related guidelines apply to computing – Suppose is actually a subpartition of . We define a brand new partition sub(, to become a partition obtained by deleting from . For PK 11195 Cancer example sub((eight, 72 , 63 , 23), (7, 63 , 2)) = (eight, 7, 22). Further, Lk will be the partition obtained by multiplying k to each part of whose multiplicity is Siramesine Biological Activity divisible by k and dividing its multiplicity by k. Around the other hand, L-1 is obtained by k dividing by k every component divisible by k and multiplying its multiplicity by k. For q-series, we make use of the following regular notation:n -Publisher’s Note: MDPI stays neutral with regard to jurisdictional claims in published maps and institutional affiliations.Copyright: 2021 by the authors. Licensee MDPI, Basel, Switzerland. This short article is definitely an open access write-up distributed beneath the terms and conditions of the Inventive Commons Attribution (CC BY) license (licenses/by/ four.0/).( a; q)n =i =(1 – aqi),( a; q) = lim ( a; q)n ,n( a; q)n =( a; q) . ( aqn ; q)Mathematics 2021, 9, 2693. ten.3390/mathmdpi/journal/mathematicsMathematics 2021, 9,2 ofSome q-identities that will be helpful are recalled as follows:( a; q)n n(n1) q two = (1 – aq2n-1)(1 qn), (q; q)n n =0 n =(1)q2n (1 q8n-3)(1 q8n-5)(1 – q8n) , = (q; q)2n 1 – q2n n =0 n =(2)( a; q)n (b; q)n n (b; q) ( az; q) z = (q; q)n (c; q)n (c; q) (z; q) n =(c/b; q)n (z; q)n n b , |z| 1, |b| 1, |q| 1. (3) (q; q)n ( az; q)n n =For proof of your above identities, see [2,4,5], respectively. Euler found the following theorem. Theorem 1 (Euler, [2]). The number of partitions of n into odd parts is equal to the number of partitions of n into distinct parts. This theorem has an interesting bijective proof supplied by J .W. L Glaisher (see [6]). We shall denote Glaisher’s map by . In actual fact converts a partition into odd parts to a partition into disctinct components. m m m Let = (1 1 , 2 two , . . . , r r) be a partition of n whose parts are odd. Note that the notation for implies 1 2 . . . are parts with multiplicities m1 , m2 , . . ., respectively. Now, write mi ‘s in k-ary expansion, i.e., mi =m lj =aij 2jliwhere 0 aij 1.We map i i to ji=0 (two j i) aij , where now two j i can be a element with multiplicity aij . The image of which we shall denote by , is given byr li(two j i) aij .i =1 j =Clearly, this is a partition of n with distinct parts. f f Around the other hand, assume that = (1 , two , . . .) is actually a partition of n into ditinct parts. Write = 2ri ai exactly where 2 ai then map i to ( ai)2 i f i for every i, exactly where now ai is really a part with multiplicity 2ri f i . The inverse of is then given by -1 =i 1 fr( a i)two i f i .rIn the resulting partition, it is also clear that the parts are odd. We also recall the following notation from [3]. pod (n): the number of partitions of n in which odd components are distinct and higher than eu even components. Od (n): the number of partitions of n in which the odd components are distinct and every single odd integer smaller sized than the largest odd part need to seem as a aspect. Theorem 2 of [3] is restated below. Theorem 2 (Andrews, [3]). For n 0, we’ve got pod (n) = Od (n). eu In this paper, we generalise Theorem 2 and have a look at various variations.Mathematics 2021, 9,3 of2. A Generalisation of Theorem two Define D (n, p, r) to be the amount of partitions of n in which components are congruent to 0, r (mod p), and each and every portion congruent t.