The reciprocal of $$(q;q)_n$$ ( q ; q ) n as sums over partitions
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The reciprocal of (q; q)n as sums over partitions Mircea Merca1,2 Received: 17 March 2020 / Accepted: 14 August 2020 © Springer Science+Business Media, LLC, part of Springer Nature 2020
Abstract In this paper we show that the reciprocal of the finite product (q; q)n can be expressed as a sum over all the partitions of n in two different ways. We derive similar results for the reciprocal of 1 − q n . Product identities involving sums over partitions are derived as consequences of these results. Keywords Partitions · Multinomial coefficients · Product identities Mathematics Subject Classification 11P81 · 11P83 · 05A17,
1 Introduction Let n be a positive integer. A partition of n is a sequence of positive integers λ = (λ1 , λ2 , . . . , λk ) such that λ1 ≥ λ2 ≥ · · · ≥ λk and λ1 + λ2 + · · · + λk = n. The positive integers λi in the sequence are called parts [1] and the k is called the number of parts of λ, denoted by (λ). If λ is a partition of n then we use the notation λ n. For example, the following are the partitions of 5: 5, 4 + 1, 3 + 2, 3 + 1 + 1, 2 + 2 + 1, 2 + 1 + 1 + 1, 1 + 1 + 1 + 1 + 1.
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Mircea Merca [email protected]
1
Department of Mathematics, University of Craiova, Craiova 200585, Romania
2
Academy of Romanian Scientists, Ilfov 3, Sector 5, Bucharest, Romania
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M. Merca
The multiplicity of the part i in λ, denoted by ti , is the number of parts of λ equal to i. Then the partition λ can be written as t1 + 2t2 + · · · + ntn = n
(1)
and the number of parts is given by t1 + t2 + · · · + tn . As usual, we denote by p(n) the number of the partitions of n. The number of partitions of n is thus seen to be equal to the number of solution of the equation (1). To choose an arbitrary partition λ of k, we can decide independently for each positive integer j how many times to include j as a part of λ. Each use of j as a part contributes j to the total size k. Thus the generating function for the choice of any number of repetitions of the part j is 1 + q j + q2 j + · · · =
1 . 1−qj
Multiplying for all j we get the generating function for p(k), i.e., ∞
p(k)q k =
k=0
∞ j=1
1 . 1−qj
If we want a particular coefficient p(k) we need only multiply out those factors involving q to a power k or less, and there finitely many of these. Thus the infinite product makes sense since only a finite number of the factors contribute to any given term. Let pn (k) be the number of the partitions of k whose parts are less than or equal to n. To count partitions whose parts are less than or equal to n, use only the factors for j = 1, 2, ..., n to get ∞ k=0
pn (k)q k =
1 , (q; q)n
where (a; q)n = (1 − a)(1 − aq)(1 − aq 2 ) · · · (1 − aq n−1 ) is the q-shifted factorial with (a; q)0 = 1. For each partition λ we consider the following multinomial coefficient: λ1 . mλ = λ1 − λ2 , λ2 − λ3 , . . . , λ(λ)−1 − λ(λ) , λ(λ)
If (λ) = 1 it is clear that m λ = 1. For (λ) > 1, we can write m λ as a product of binomial coefficients: λ(λ)−1 λ1 λ2 ··· . mλ = λ2 λ3 λ(λ)
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The recip
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