Plane Partitions and Divisors

: In this paper, we consider the sum of divisors d of n such that n / d is a power of 2 and derive a new decomposition for the number of plane partitions of n in terms of binomial coefﬁcients as a sum over partitions of n . In this context, we introduce a new combinatorial interpretation of the number of plane partitions of n


Introduction
Recall that a plane partition π of the positive integer n is a two-dimensional array π = (π i,j ) i,j 1 of non-negative integers π i,j such that n = ∑ i,j 1 π i,j , which is weakly decreasing in rows and columns: π i,j π i+1,j , π i,j π i,j+1 , for all i, j 1.
If we ignore the entries equal to zero in a plane partition, it can be considered as the filling of a Young diagram with positive integers with entries weakly decreasing in rows and columns and such that the sum of all entries is equal to n.On the other hand, there is a desirable way to represent a plane partition as a three-dimensional object: this is achieved by replacing each part of size k of the plane partition by a stack of k unit cubes (Figure 1).This is a natural generalization of the concept of classical partitions [1].Different configurations are counted as different plane partitions.As usual, we denote by PL(n) the number of plane partitions of n.For convenience, we define PL(0) = 1.Plane partitions were introduced by MacMahon [2] who proved the following highly non-trivial result: The expansion starts as ∞ ∏ n=1 1 (1 − q n ) n = 1 + q + 3 q 2 + 6 q 3 + 13 q 4 + 24 q 5 + 48 q 6 + 86 q 7 + • • • . ( An n-color partition of a positive integer m is a partition in which a part of size n can come in n different colors denoted by subscripts: n 1 , n 2 , . . ., n n .The parts satisfy the following order: They were introduced by A. K. Agarwal and G. E. Andrews [3] nearly a century after MacMahon introduced pane partitions.For example, there are thirteen n-color partitions of 4: (4 4 ), ( 43 ), ( 42 ), ( It was pointed out in [3] that the right-hand side of ( 1) is also a generating function for the number of n-color partitions.Thus, the following statement holds.
Theorem 1.The number of plane partitions of m equals the number of n-color partitions of m.
We also note that the set of plane partitions with strict decrease along columns (of the Young diagram) is in one-to-one correspondence with the set of symmetric matrices with non-negative integer entries ( [1], Corollary 11.6).Moreover, by the Knuth-Schensted correspondence ( [1], Theorem 11.4), in the set of pairs of plane partitions (π, π ) in which there is strict decrease along columns, each entry is at most k, and the corresponding rows of π and π s are of the same length are in bijection with the set of k × k matrices with non-negative integer entries.
There is a well-known connection between plane partitions and divisors.In [4], it is shown that In this article, we consider a restricted sum of divisors function and find connections with the sequence PL(n).
For a positive integer n, we denote by s n the sum of divisors d of n such that n/d is a power of 2. For example, the divisors of 12 are 1, 2, 3, 4, 6, 12.
The generating function for s n is given on the page for A129527.It can be derived as follows: where we have used the identity |q| < 1.
with q replaced by q 2 n .On the other hand, it is not difficult to prove that for n odd, n + s n/2 , for n even. ( Logarithmic differentiation of the generating Function (1) gives the following identity: In Section 3, we show that Then, logarithmic differentiation of the generating function (5) gives Equating the coefficients of q n−1 in the Equations ( 4) and (6), we obtain the following identity: On the other hand, by ( 4) and ( 6), we see that Therefore, we deduce the relation n = s n , for n odd, s n − s n/2 , for n even, which implies identity (3).
From (3), we see that Theorem 2 is trivial when n is odd.However, for n even, this theorem provides an interesting decomposition of σ 2 (n).For example, The case n = 6 of Theorem 2 reads as follows: For any positive integer m, we denote by PL (m) (n) the number of m-tuples of plane partitions of non-negative integers n 1 , n 2 , . . ., n m where For r ∈ {−1, 0, 1}, we define the numbers PL (m,r) (n) as follows: Recently, Merca and Radu [6] considered specializations of complete homogeneous symmetric functions and provided the following formula for PL (m,r) (n).
This formula provides a decomposition of PL (m,r) (n) as a sum over all the partitions of n in terms of binomial coefficients involving the multiplicities of the parts.
In this paper, we provide a new decomposition of PL (m,r) (n) as a sum over partitions of n in terms of binomial coefficients.This time, in addition to the multiplicities of part sizes, we also need the sequence s n .Theorem 4. For m 1, r ∈ {−1, 0, 1} and n 0, The case m = 1 and r = 0 of Theorem 4 reads as follows.
Corollary 1.For n 0, While the sum above is over all partitions of n, not all terms are non-zero.Due to the fact that ( s j t j ) = 0 when t j > s j , in this sum it suffices to consider the partitions of n in which, for each j ∈ {1, 2, . . ., n}, part j occurs at most s j times.For example, the partitions of four with this restriction can be rewritten as Therefore, the case n = 4 of Corollary 1 reads as follows: The case m = 2 and r = 0 of Theorem 4 gives the following identity: Considering the partitions of four with t 1 2, the case n = 4 of Corollary 1 reads as follows: On the other hand, according to (2) we can write By Corollary 2, we easily deduce the following congruence identity.
Corollary 3.For n 0, where PL(x) = 0 if x is not a non-negative integer.
As a consequence of Theorems 3 and 4, we remark the following identity.
Corollary 4. For m 1, r ∈ {−1, 0, 1} and n 0, The remainder of this paper is organized as follows.In Section 2, we provide an analytic proof of Theorem 4. In Section 3, we introduce a new combinatorial interpretation for PL(n).In Section 4, we make a connection to the Josephus problem.In Section 5, we give some concluding remarks.

Proof of Theorem 4
Elementary techniques in the theory of partitions [1] give the following generating function: In order to prove our theorem, we consider the identity which can be rewritten as Then, by (10), with q replaced by q n , we obtain For |q| < 1, considering (10) and (11), the generating function of PL (m,r) (n) can be rewritten as follows: where we have used Cauchy multiplication of power series.
Using elementary techniques [1], we obtain the following generating function for Q s n (m): On the other hand, by (12) with m = 1 and r = 0, we obtain a new expression of the generating function of PL(n): Thus, we deduce the following result for which we give a combinatorial proof.
Theorem 5.The number of n-color partitions of m equals the number of s n -color partitions of m into distinct parts.
Proof.Given an integer n, we denote by n o the largest odd divisor of n.Then, n = 2 k n o for some non-negative integer k and Since 1 n o n, it follows that n s n 2n − 1.Note that, for odd n, we have s n = n.
Denote by P n (m) the set of n-color partitions of m.We define a bijection ϕ : P n (m) → Q s n (m).
Start with λ ∈ P n (m).For each part k j (size k, color j with 1 j k) that occurs more than once, we replace two parts equal to k j by a single part (2k) 2k+j (part of size 2k, color 2k + j).Since 1 j k, we have 2k + 1 2k + j 3k.Since s 2k = 4k − k o and k o k, the obtained partition is an s n -partition.We repeat the process until parts are distinct and obtain a partition µ ∈ Q s n (m).We define ϕ(λ) = µ.
To determine the inverse ϕ −1 , start with µ ∈ Q s n (m).Note that if k j is a part of µ, and k is odd then 1 j k.For each part k j with j > k, it follows that k is even and we replace k j by two parts (k/2) j−k .Note that if k/2 is odd, then k o = k/2 and s k = 2k − k/2.Then, 1 j 2k − k/2 and 1 j − k k/2.We continue the process until there are no parts k j with j > k to obtain a partition λ ∈ P n (m).Then, ϕ −1 (µ) = λ.
We remark the following consequence of Theorems 1 and 5.
Corollary 5.The number of plane partitions of m equals the number of s n -color partitions of m into distinct parts.

A Connection with Josephus Problem
The Josephus problem is a math puzzle with a grim description for which we refer the reader to [7].Here, we give a friendlier adaptation of the problem: n rocks, labeled 1 to n, are placed in a circle.An person walks along the circle and, starting from the rock labeled 1, removes every k-th rock.As the process goes on, the circle becomes smaller and smaller, until only one rock remains.
We are interested in the case k = 2 of the Josephus problem.For k = 2, we denote by J n the order in which the first rock is removed.For example, if there are n = 7 rocks to begin with, they are removed in the following order: 2, 4, 6, 1, 5, 3, 7.
Therefore, the rock labeled 1 is eliminated at the fourth removal.Therefore, J 7 = 4.
The sequence The sequence (J n ) n 1 can be defined as follows: J n = (n + 1)/2, for n odd, n/2 + J n/2 , for n even.( 15) By ( 3) and ( 15), we easily deduce that for any positive integer n.
It is clear that our results can be expressed in terms of J n .For example, we remark the following version of Corollary 1: Corollary 6.For n 0, In this context, we denote by P s (n) the set of partitions of n with t j 2J j − 1, for each j ∈ {1, 2, . . ., n}, and define p s (n We also consider the set J defined as Note that, if n is odd, then nJ n = n(n+1) 2 , a triangular number.Thus, if m, n are both odd and m = n, then mJ m = nJ n .
QJ e (n) to be the number of partitions of n into an even number of distinct parts from J ; 2.
QJ o (n) to be the number of partitions of n into an odd number of distinct parts from J ; 3.
In certain conditions, p s (n) satisfies Euler's pentagonal number recurrence.
for n odd, QJ eo (n/2), for n even.
Analytic Proof.The generating function for p s (n) is given by: Assuming Conjecture 1, elementary techniques in the theory of partitions [1] give the following generating function: Thus, we can write This concludes the analytic proof.
We also provide a combinatorial proof of Theorem 6.First, we introduce some notation.We denote by P (n) the set of all partitions of n and set P := ∪ n 0 P (n).Given λ ∈ P, we denote by (λ) the number of parts in λ and by |λ| the sum of parts of λ (also referred to as the size of λ).For a pair of partitions (λ, µ), we write (λ, µ) n to mean |λ| + |µ| = n (and similarly for a triple of partitions).In general, given a set A(n) of partitions of n (or pairs of partitions with sizes adding up to n), we set A := ∪ n 0 A(n).We also write A e (n) (respectively, A e (n)) for the subset of λ ∈ A(n) with (λ) even (respectively, odd).
Combinatorial Proof of Theorem 6.Let Q(n) be the set of distinct partitions of n.As explained for example in [1], Franklin defined a sign-reversing involution ϕ F on a subset of the set of distinct partitions of n to prove combinatorially that the generating function for We define B(n Hence, the left-hand side of ( 16) is the generating function for We set 2J := {2nJ n | n ∈ N} and define and prove combinatorially that To do this, we define an involution ψ on the subset C * (n) of pairs (α, Next, we define a Glaisher-type bijection ϕ G between C(n) \ C * (n) and P s (n).Let (α, ∅) ∈ C(n) \ C * (n).For each part j is α with t j 2J j , replace 2J j parts of size j by a part of size 2jJ j .We repeat the process until we obtain a partition ξ ∈ P s (n), i.e., each part j of ξ satisfies t j 2J j − 1. Set ϕ G (α, ∅) := ξ.
If the mapping j → jJ j is injective (i.e., if Conjecture 1 holds), the transformation ϕ G is invertible.Starting with a partition ξ ∈ P s (n) if ξ has a part equal to 2jJ j for some j, we replace part 2jJ j into 2J j parts equal to j.We repeat the process until we obtain a partition α with no parts in 2J .

Concluding Remarks and Open Problems
In this section, we introduce some conjectures on the non-negativity of certain truncated theta series involving sequences studied in this article.
In [8], Andrews and Merca considered the truncation of the theta series arising from Euler's pentagonal number theorem.They considered the number M k (n) of partitions of n in which k is the smallest integer that does not occur as a part and there are more parts > k than there are < k.For example, we have M 3 (18) = 3 because the three partitions in question are (5, 5, 5, 2, 1), (6, 5, 4, 2, 1), (7, 4, 4, 2, 1).
As shown in [8], for every k 1, M k (n) is the coefficient of q n in the series There is a substantial amount of numerical evidence to state the following conjecture.Conjecture 2. For k 1, all the coefficients of the series are non-negative.The coefficient of q n is positive if and only if n k(3k + 1)/2.
Considering the generating functions of p s (n) and QJ eo (n), Conjecture 2 can be reformulated in its combinatorial form: Conjecture 3.For k 1,

1.
For n odd, we have with strict inequality if and only if n k(3k + 1)/2.
As shown in [25], for every k 1, P k (n) is the coefficient of q n in the series (−1) k−1 1 − 1 (q; q) ∞ k ∑ n=−k (−1) k q n(3n−1)/2 .Based on numerical evidence, we make the following conjecture which is analogous to Conjecture 2. Conjecture 4. For k 1, all the coefficients of the series are non-negative.The coefficient of q n is positive if and only if n (k + 1)(3k + 2)/2.
The combinatorial interpretation of this conjecture reads as follows.

2.
For n even, we have

→Figure 1 .
Figure 1.Representation of a plane partition of 32.

Conjecture 1 .
Let m, n be positive integers.If m = n, then m J m = n J n .