Some results concerning partitions with designated summands
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Some results concerning partitions with designated summands Shane Chern1 · Michael D. Hirschhorn2 Received: 1 April 2019 / Accepted: 26 July 2019 © Springer Science+Business Media, LLC, part of Springer Nature 2020
Abstract Let P D(n) and P D O(n) count, respectively, the number of partitions of n with designated summands and the number of partitions of n with designated summands where all parts are odd, and let P Dt (n) and P D Ot (n) count, respectively, the number of tags (that is, designated summands) in the partitions enumerated by P D(n) and P D O(n). We give elementary proofs of congruences for these partition functions. Keywords Partitions with designated summands · Tagged parts · Congruences Mathematics Subject Classification 11P83 · 05A17
1 Introduction The notion of partitions with designated summands was introduced by Andrews et al. [2]. These partitions are constructed by taking ordinary partitions and tagging (that is, designating) exactly one of each part size. Thus, for example, there are 15 partitions of 5 with designated summands. 5 , 4 + 1 , 3 + 2 , 3 + 1 + 1, 3 + 1 + 1 , 2 + 2 + 1 , 2 + 2 + 1 , 2 + 1 + 1 + 1, 2 + 1 + 1 + 1, 2 + 1 + 1 + 1 , 1 + 1 + 1 + 1 + 1, 1 + 1 + 1 + 1 + 1, 1 + 1 + 1 + 1 + 1, 1 + 1 + 1 + 1 + 1, 1 + 1 + 1 + 1 + 1 .
B
Michael D. Hirschhorn [email protected] Shane Chern [email protected]
1
Department of Mathematics, The Pennsylvania State University, University Park, PA 16802, USA
2
School of Mathematics and Statistics, UNSW Sydney, NSW 2052, Australia
123
S. Chern, M. D. Hirschhorn
The number of partitions of n with designated summands is denoted by P D(n). Thus P D(5) = 15. The authors of [2] showed that pd(q) =
f6 f1 f2 f3
(1.1)
(1 − q nk ).
(1.2)
P D(n)q n =
n≥0
where, as usual,
f k = (q k ; q k )∞ =
n≥1
They also considered P D O(n), the number of partitions of n with designated summands, and all parts odd, and showed that pdo(q) =
P D O(n)q n =
n≥0
f 4 f 62 . f 1 f 3 f 12
(1.3)
Also in [2, Corollary 7], it was shown that P D(3n + 2) ≡ 0
(mod 3),
(1.4)
f 42 f 65
(1.5)
[2, Theorem 20] that
P D(2n)q n =
n≥0
2 f 13 f 2 f 33 f 12
P D(2n + 1)q n =
n≥0
,
2 f 25 f 12
f 15 f 3 f 42 f 6
,
(1.6)
[2, Theorem 21] that
f 42 f 64 , 2 2 f 1 f 32 f 12
P D O(2n)q n =
n≥0
P D O(2n + 1)q n =
2 f 26 f 12
(1.7) ,
(1.8)
(1 − q n )−a(n) ,
(1.9)
n≥0
f 14 f 42 f 62
[2, Theorem 23] that n≥0
P D(3n)q n =
n≥1
with a(6n + 1) = 5, a(6n + 3) = 2, a(6n + 5) = 5,
123
(1.10)
Some results concerning partitions with designated summands
and [2, Theorem 22] that
P D O(3n)q n =
n≥0
P D O(3n + 1)q n =
f 22 f 64 2 f 14 f 12
,
(1.11)
f 24 f 33 f 12
(1.12)
f 15 f 4 f 62
n≥0
and
f 23 f 6 f 12
P D O(3n + 2)q n = 2
f 14 f 4
n≥0
.
(1.13)
They also prove a number of congruences of the form P D(an + b) ≡ 0 (mod m) and P D O(an + b) ≡ 0 (mod m). Chen et al. [3] took up the same topic, gave explicit formulas for P D(3n)q n , n≥0 P D(3n + 1)q n and P D(3n + 2)q n , the last of w
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