FORMULAS FOR THE NUMBER OF k-COLORED PARTITIONS AND THE NUMBER OF PLANE PARTITIONS OF n IN TERMS OF THE BELL POLYNOMIALS

SUMIT KUMAR JHA

Abstract. We derive closed formulas for the number of k-coloured partitions and the number of plane partitions of n in terms of the Bell polynomials.

1. Introduction Definition 1 (k-colored partitions of n [1]). A partition of a positive integer n is a finite Pr weakly decreasing sequence of positive integers λ1 ≥ λ2 ≥ · · · ≥ λr > 0 such that i=1 λi = n. The λi’s are called the parts of the partition. Let p(n) denote the number of partitions of n. A partition is said to be k-coloured if each part can occur as k colours. Let pk(n) denote the number of k coloured partitions of n. For example, there are five 2-coloured partitions of 2: 1 + 1, 10 + 1, 10 + 10, 2, 20

so p2(2) = 5.

The for pk(n) is given by ∞ ∞ X Y 1 p (n)qn = (|q| < 1). k (1 − qj)k n=0 j=1 The right hand side of the above equation is also the generating function for the multi- partitions which appear in many papers on the representations theory of Lie algebras, and some with applications in physics [2]. The unrestricted partition function p(n) = p1(n) can be computed using Hardy–Ramanujan– Rademacher formula with soft complexity O(n1/2+o(1)) and very little overhead [3]. In section 2, we obtain a formula for the number of k-colored partitions of n in terms of the partial Bell polynomials which allows computing them efficiently. For example, using our formula we can calculate

p30(200) = 23945275792616100703623332622769220026826156718318470749445535353589 which are the number of thirty colored partitions of 200.

2010 Subject Classification. 05A15. Key words and phrases. k-coloured partition; plane partitions. 1 2 SUMIT KUMAR JHA

Definition 2 (Plane partitions of n). A plane partition of n is a two-dimensional array of integers ni,j that are non-increasing in both indices, that is,

ni,j ≥ ni,j+1 ni,j ≥ ni+1,j and X n = ni,j. i,j Let PL(n) denote the number of plane partitions of n. For example, there are six plane partitions of 3: 1 1 1 2 1 1 1 1 2 1 3 1 1 1 so PL(3) = 6. Macmohan [4] obtained the generating function of PL(n) as ∞ ∞ X X 1 PL(n) qn = (|q| < 1). (1 − qj)j n=0 j=1 In section 3, we obtain an expression for PL(n) in terms of the complete Bell polynomials and sum of squares of divisors of n. This also gives a expression for PL(n). For example, we can calculate PL(700) = 1542248695905922088013690041381656661664744761954709483748320717869. Definition 3 (Partial Bell polynomials). For n and k non-negative integers, the partial (n, k)th partial Bell polynomials in the variables x1, x2, . . . , xn−k+1 denoted by Bn,k ≡ Bn,k(x1, x2, . . . , xn−k+1) [5, p. 206] can be defined by n−k+1 ` X n! Y xi  i B (x , x , . . . , x ) = . n,k 1 2 n−k+1 Qn−k+1 `i! i! 1≤i≤n,`i∈N i=1 i=1 Pn i=1 i`i=n Pn i=1 `i=k The partial Bell polynomials have the following generating function: ∞ !k ∞ 1 X tj X tn x = B (x , . . . , x ) , k = 0, 1, 2, . . .. k! j j! n,k 1 n−k+1 n! j=1 n=k The partial Bell polynomials can also be computed efficiently by a [6]: n−k+1 X n − 1 B = x B , n,k i − 1 i n−i,k−1 i=1 where

B0,0 = 1 COLORED AND PLANE PARTITIONS 3

Bn,0 = 0 for n ≥ 1;

B0,k = 0 for k ≥ 1. Cvijovi´c[7] gives the following formula for calculating these polynomials

k n−1 α1−1 αk−1−1 z  }|  { 1 X X X n α1 αk−1 Bn,k+1 = ··· ··· (k + 1)! α1 α2 αk α1 =k α2 =k−1 αk =1 | {z } k

· xn−α1 xα1−α2 ··· xαk−1−αk xαk (n ≥ k + 1, k = 1, 2,...) (1) The nth complete Bell polynomials are defined as n X Bn(x1, x2, ··· , xn) = Bn,k(x1, x2, . . . , xn−k+1). k=0 The complete Bell polynomials can be recurrently defined as n X n B (x , . . . , x ) = B (x , . . . , x )x n+1 1 n+1 i n−i 1 n−i i+1 i=0 with the initial value B0 = 1. We would frequently use Fa`adi Bruno’s formula [5, p. 134] which is n dn X f(g(q)) = f (l)(g(q)) · B g0(q), g00(q), . . . , g(n−l+1)(q) . (2) dqn n,l l=1

2. Formula for k-coloured partitions of n Theorem 1. For all positive integers n, k n 1 X p (n) = (−1)l (k)(l) B (λ , λ , ··· , λ ) k n! n,l 1 2 n−l+1 l=0 where  (−1)m i! if i = m(3m+1) for some positive integer m  2 m m(3m−1) λi = (−1) i! if i = 2 for some positive integer m , 0 otherwise where (k)(l) = k (k + 1) ··· (k + l − 1) represents the rising . Proof. We recall the Euler’s pentagonal number theorem [9, Equation 7.8] ∞ ∞ 2 Y j X n 3n +n g(q) := (1 − q ) = (−1) q 2 . j=1 n=∞ 4 SUMIT KUMAR JHA

Let f(q) = q−k. Then the Fa`adi Bruno’s formula (2) gives

n dn X (k)(l) g(q)−k = (−1)l B g0(q), g00(q), . . . , g(n−l+1)(q) . dqn (g(q))l+k n,l l=1

Letting q → 0 in the above we obtain our result readily. 

3. Another expression for pk(n) Theorem 2. For all positive integers n, k

B (k σ(1), 1! k σ(2), ··· , k (n − 1)! σ(n)) p (n) = n , k n! P where σ(n) = d|n d. Proof. Let ∞ Y 1 h(q) := . (1 − qj)k j=1 Then we have

∞ X log h(q) = − k log(1 − qj) j=1 ∞ ∞ X X qlj = k l j=1 l=1 ∞ X X = k qn d−1. n=1 d|n ∞ X qn k X = d. n n=1 d|n

Let f(q) = eq and g(q) = log h(q) in Fa`adi Bruno’s formula (2) gives

n dn X h(q) = h(q) B g0(q), g00(q), . . . , g(n−l+1)(q) . dqn n,l l=1

Letting q → 0 in the above gives us our result readily.  COLORED AND PLANE PARTITIONS 5

The determinant expression for the complete Bell polynomials [8, Theorem 2.1] gives us k σ(1) k σ(2) σ(3) k σ(4) ······ k σ(n)       −1 k σ(1) k σ(2) k σ(3) ······ k σ(n − 1)        0 −2 k σ(1) k σ(2) ······ k σ(n − 2)     1  0 0 −3 k σ(1) ······ k σ(n − 3) pk(n) = det   . n!        0 0 0 −4 ······ k σ(n − 4)      ......   ......      0 0 0 0 · · · −(n − 1) k σ(1) Thus 1  2 6 p (2) = det = 5. 2 2 −1 2 4. Formula for Plane partitions of n Theorem 3. For all positive integers n B (σ (1), 1! σ (2), ··· , (n − 1)! σ (n)) PL(n) = n 2 2 2 , n! P 2 where σ2(d) = d|n d . Proof. Let ∞ Y 1 h(q) := . (1 − qj)j j=1 Then we have ∞ X log h(q) = − j log(1 − qj) j=1 ∞ ∞ X X qlj = j l j=1 l=1 ∞ X X = n qn d−2. n=1 d|n ∞ X qn X = d2. n n=1 d|n 6 SUMIT KUMAR JHA

Let f(q) = eq and g(q) = log h(q) in Fa`adi Bruno’s formula (2) gives n dn X h(q) = h(q) B g0(q), g00(q), . . . , g(n−l+1)(q) . dqn n,l l=1 Letting q → 0 in the above gives us our result readily.  We immediately have the following determinant formula for PL(n) [8, Theorem 2.1]   σ2(1) σ2(2) σ2(3) σ2(4) ······ σ2(n)      −1 σ2(1) σ2(2) σ2(3) ······ σ2(n − 1)        0 −2 σ2(1) σ2(2) ······ σ2(n − 2)     1   PL(n) = det  0 0 −3 σ2(1) ······ σ2(n − 3) . n!        0 0 0 −4 ······ σ2(n − 4)      ......   ......      0 0 0 0 · · · −(n − 1) σ2(1) For example, using the above we have  1 5 10 1 PL(3) = det −1 1 5  = 6. 6 0 −2 1

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