The Cranks for 5-Core Partitions

It is well known that the number of 5-core partitions of 5kn + 5k − 1 is a multiple of 5k. In [1] a statistic called a crank was developed to sort the 5-core partitions of 5n + 4 and 25n + 24 into 5 and 25 classes of equal size, respectively. In this paper we will develop the cranks that can be used to sort the 5-core partitions of 5kn + 5k − 1 into 5k classes of equal size.


Introduction
A t-core partition of n is a partition of n that contains no hook numbers that are multiples of t [2, 2.7.40].The generating function for t-core partitions is given by 2 10 t t n b n n nZ q     where the vector 1 (1,1, ,1)  in t Z and (0,1, , 1)  bt [3].In [3] Garvan, Stanton, and Kim showed that the statistic 4n 0 + n 1 + n 3 + 4n 4 (mod 5), where the n i 's are the components of the vector in the generating function for 5-cores, can be used to sort the 5-cores of 5n + 4 into 5 classes of equal size.In a sequel to this paper [1] Garvan explicitly describes a crank for the 5-cores of 25n + 24.In this paper a crank for the 5-cores of 5 k n + 5 k − 1 will be given using techniques similar to those used by Garvan, Stanton, and Kim.

Description of the Crank
For ease of working with the vector n we will write it as (a, b, c, d, e).Using the fact that a + b + c + d + e = 0, the exponent on q in the generating function for the 5-core partitions can be expressed as Note that A, B, C, and D are integers since where Theorem 1.1 The 5-core partitions of 5n + 4 corresponding to the vectors (A, B, C, D) can be sorted into 5 classes of equal size by looking at the values of B modulo 5.

An Illustration of the Crank
The following series of Tables 1-3  Table 1.Solutions corresponding to 5-cores of 14.

Solutions of
cores of integers of the form 5n + 4 are associated with the values of a, b, c, and d satisfying .Evaluating G(a, b, c, d) with a = A − C − 2D, b = −2A + B − C + D, c = −B + 4C − D, d = 2A − B − C + 2D + 1, we get an expression in A, B, C, and D which we will label as H(A, B, C, D).

Theorem 1 . 2
8) Note that A, B, C, and D are integers and for each of these changes of variable H(A, B, C, D) becomes 5G(A, B, C, D) + 4. Hence G(A, B, C, D) = n and for each solution of this equation we have 5 solutions (A, B, C, D) of H(A, B, C, D) = 5n + 4, one with (mod 5) Bm  for each choice of m = 0, 1, 2, 3, 4, which can be transformed to a solution (a, b, c, d) of G(a, b, c, d) = 5n + 4.This completes the proof of the theorem.The 5-core partitions of 5 k n + 5 k − 1 can be sorted into 5 k classes of equal size.From the proof of Theorem 1.1 we can transform a solution of G(a, b, c, d) = n into 5 solutions of G(a, b, c, d) = 5n + 4. Each solution of G(a, b, c, d) = 5n + 4 can be transformed into 5 solutions of G(a, b, c, d) = 25n + 24.Iterating this process k times we easily see that a solution of G(a, b, c, d) = n can be transformed into 5 k solutions of G(a, b, c, d) = 5 k n + 5 k − 1.At each stage in the transformation process we can keep track of the congruence class modulo 5 of B to get a k-tuple of values m (mod 5) associated with each solution of G(a, b, c, d) = 5 k n + 5 k − 1.These k-tuples can be used to sort the solutions of G(a, b, c, d) = 5 k n + 5 k − 1 into 5 k classes of equal size.
show the 2 solutions of G(a, b, c, d) = 2 transformed into 250 solutions of G(a, b, c, d) = 374.The intermediate solutions of H(A, B, C, D) are shown in order to easily see the classes of B (mod 5) which can be used to sort these 250 solutions into 125 classes of equal size.