N =1 n=1 (1 – q4n-r)(1 q2n)(1 – q2n) (q; q2) (q2 ; q2) (1 – q4n-r)(1 – q4n)(1 – q4n-r)(1 q2n) (by (1))1 – qn(1 – q4nr)(1 – q4n2) .Mathematics 2021, 9,5 ofThe Bijective Proof Let be a D-Fructose-6-phosphate disodium salt site partition enumerated by A(n, r). Execute the following steps: 1. 2. Conjugate getting . If has no aspect with odd multiplicity, set := and go to step 4. Otherwise, decompose = (,) where could be the subpartition of consisting of all parts much less than or equal for the largest part which has odd multiplicity and may be the subpartition sub(,). Recall that may be written as m m m = 1 1 two two . . . m m exactly where 1 2 . . . m . We use this notation of in the next step. a. If m 1 r (mod 4), then update and as follows: := sub( , 2) , := 2 . 1 1 b. For j = two, three, . . . , m, if m j 0 (mod four), then update and as follows: := sub( , two) , := 2 . j j Now call the new updated and , and , respectively. Observe that = . Compute = L2 (( L2 )). Note that = is a partition into components r, 2 (mod four). Ahead of giving the inverse mapping, let us examine an instance. The inverse Let be a partition of n into parts r, 2 (mod four). Decompose as follows = 1 r where 1 will be the subpartition of with components 2 (mod 4) and r would be the subpartition with parts r (mod four). Compute- – h = L2 1 (-1 ( L2 1 (1))).3.4.Then = r h is usually a partition inside a(n, r). Instance Let r = 1 with = 232 171 113 81 63 41 24 A(124, 1). Then, = 152 112 102 72 63 36 26 . Hence = 152 112 102 72 and = 63 36 26 . Updating and yields: = 152 112 102 72 62 32 22 and = 61 34 24 . Now we’ve got L2 = 30, 22, 20, 14, 12, six, 4 in order that ( L2 ) = 152 112 72 54 36 14 . Thus = L2 (( L2 )) = 301 221 141 102 63 22 . Since = 92 51 13 , the image is = = 301 221 141 102 92 63 51 22 13 .Mathematics 2021, 9,six ofTo discover the inverse, take into account = 301 221 141 102 92 63 51 22 13 inside the Sobetirome Purity & Documentation example above (r = 1). Then, 1 = 301 221 141 102 63 22 and r = 92 51 13 . – – Now L2 1 (1) = 152 112 72 54 36 14 in order that -1 ( L2 1 (1)) = 30, 22, 20, 14, 12, 6, four . – Therefore h = L2 1 ( 30, 22, 20, 14, 12, 6, 4) = 152 112 102 72 62 32 22 and that two 121 103 81 63 41 24 . Hence, h = 14 = 92 51 13 142 121 103 81 63 41 24 = 232 171 113 81 63 41 24 .Theorem 5. Let C (n) be the number of partitions where if 2j happens, then all even integers significantly less than 2j take place as components and any part greater than 2j is odd. Then, C (n) 1 (mod two) if and only if j ( j 1) n = two for some j 0. Proof.n =C (n)qn =n =q246…2n(1 – q)(1 – q2) . . . (1 – q2n)(1 – q2n1)(1 – q2n3) . . .=qn n (q; q)2n (q2n1 ; q2) n =qn n (q; q2)n = (q; q)2n (q; q2) n == = =1 (q; q2) 1 (q; q2)qn n 2 two n =0 ( q ; q) nn =1 n =1 n =n =1 1 – q4n(1 q2n) (by (1),a = 0, q := q2)1 – qn(1 – q n)three qn(n1)/(mod two) (mod two).Remark 1. It is actually clearly observable from line six from the proof that C (n) is equal towards the quantity of partitions of n into parts not divisible by four. To prove this partition identity combinatorially, decompose C (n) into (o , e) exactly where o will be the subpartition consisting of odd components, and e is definitely the subpartition consisting of even parts. Then compute (o) and conjugate e . Split each a part of e into two identical parts, acquiring Then, ((o) can be a partition in which parts aren’t divisible by 4. This transformation is invertible. Theorem six. Let B(n) be the number of partitions of n in which either (a) all components are even and distinct or (b) 1 need to seem and odd parts seem without the need of gaps, even components are distinct and each is higher than or equal to 3 the largest odd component. Denote by Be (n) (resp. Bo (n)), the number of B(n)-partitions with an even (resp. od.
