Note on square-root partitions into distinct parts
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Note on square-root partitions into distinct parts Shane Chern1 Received: 10 April 2019 / Accepted: 26 June 2019 © Springer Science+Business Media, LLC, part of Springer Nature 2020
Abstract Let r D (n) be the number of square-root partitions of n into distinct parts. We will give the asymptotic formula of r D (n), r D (n) ∼ 2−7/6 3−1/3 π −1/2 ζ (3)1/6 n −2/3 4/3 3 ζ (3)1/3 2/3 ζ (2) ζ (2)2 1/3 × exp n − n + , 2 2 · 31/3 ζ (3)1/3 72ζ (3) by adjusting the well-known approach of Meinardus. Keywords Square-root partition · Distinct part · Asymptotic formula · Saddle point method · Mellin transform Mathematics Subject Classification 11P82 · 05A17
1 Introduction √ √ The square-root partition, which is a partition into parts with the order 1, 2, √ √ 3, . . ., k, . . . , was introduced by Balasubramanian and Luca [2]. For example, √ √ √ 1 has three square-root partitions: 1, 2, and 3. Let r (n) be the number of square-root partitions of n. It is not hard to see that r (n) has generating function: n≥0
B 1
r (n)q n =
k≥1
1 . (1 − q k )2k+1
Shane Chern [email protected] Department of Mathematics, Penn State University, University Park, PA 16802, USA
123
S. Chern
In [4], Luca and Ralaivaosaona studied the asymptotic behavior of r (n). They showed that, as n → ∞, r (n) = 1 + o(1) 25/18 3−1/2 π −1/2 ζ (3)7/18 n −8/9 3ζ (3)1/3 2/3 ζ (2) ζ (2)2 1/3 + 2ζ (−1) + ζ (0) , × exp n + 2/3 n − 21/3 2 ζ (3)1/3 24ζ (3) where as usual ζ (·) is the Riemann zeta function. In general, if we are given a prescribed ordered set of parts, then apart from partitions into parts in this set, we are often interested in partitions into distinct √ parts√as well. In√the square-root partition case, we will assume that, for instance, 1, 2, and 3 are different parts, although they have the same numerical value. Let r D (n) be the number of square-root partitions of n into distinct parts. One can see that the generating function of r D (n) is F(q) :=
n≥0
r D (n)q n =
(1 + q k )2k+1 =
k≥1
(1 − q 2k )2k+1 . (1 − q k )2k+1
(1.1)
k≥1
Similar to the asymptotic formula of r (n), we will prove the following result. Theorem 1 As n → ∞, we have that r D (n) = 1 + o(1) 2−7/6 3−1/3 π −1/2 ζ (3)1/6 n −2/3 4/3 3 ζ (3)1/3 2/3 ζ (2) ζ (2)2 1/3 n + . × exp n − 2 2 · 31/3 ζ (3)1/3 72ζ (3)
(1.2)
We remark that for a general infinite product k≥1
1 , (1 − q k )ak
where a1 , a2 , . . . is a “nice” sequence of non-negative integers, Meinardus’ Theorem [5] is a powerful tool used to study the asymptotic behavior of its Taylor coefficients. A delicate presentation of Meinardus’ approach is given in Chapter 6 of George Andrews’ book The theory of partitions [1]. However, Meinardus’ original theorem requires that the associated Dirichlet series of the sequence (ak )k≥1 , D(s) :=
ak k≥1
ks
,
has only one simple pole. But if D(s) has multiple singularities, Meinardus’ approach is still admissible, provided that we make suitable adjustments. A general result on such case was given by Granovsky a
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