Keywords: crank, jtimes, vector partitions, weight, exponent
Turkish Journal of Analysis and Number Theory, 2014 2 (4),
pp 125129.
DOI: 10.12691/tjant244
Copyright © 2013 Science and Education Publishing. All Rights Reserved.
1. Introduction
First we give definitions of , the crank of partitions, , , and . We generate some generating functions related to the crank and show the coefficient of x is the algebraic relations in terms of various powers of z, the exponent of z represent the crank of partitions of n (for all ). We show the results with the help of examples when n = 5 and 6 respectively. We introduce the special term weight related to the vector partitions V and show the relations in terms of , weight and crank . We prove the Theorem “The number of partitions of n with crank is for all .”
2. Definitions
Now we give some definitions following (^{[3, 4, 5]}).
: Number of partitions of n, like 4, 3+1, 2+2, 2+1+1, 1+1+1+1. Therefore, and similarlyetc.
Crank of partitions ^{[2]}: For a partition , let denotes the largest part of , denote the number of 1’s in , and denote the number of parts of larger than , the crank is given by;
: The number of partitions of n with crank m.
2.1. NotationsFor all integers and all integers m, the number of n with crank equal to m is , like;
But we see that;
Since, the coefficient of x in the right hand side of the equation;
is i.e., the exponent of z being the crank of partition.
Therefore,
3. The Generating Function for
The generating function for is given by ^{[2]};
 (1) 
We see that the exponent of z represents the crank of partitions of n (for ). As for examples when n = 5 and 6,
For n = 5,
For n = 6,
4. Vector Partitions of n
Let, , where D denotes the set of partitions into distinct parts and P denotes the set of partitions. The set of vector partitions V is defined by the Cartesian product, .
For , where weight = , the crank .
We have 41 vector partitions of 4 are given in the following table:
From the above table we have,
and
1–1= 0
i.e., =
i.e.,
Again we have 83 vector partitions of 5 are given in the following table:
From this table we have;
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+1–1–1–1 =1
1+1–1–1–1–1+1+1+1–1= 0
1+1–1–1–1–1+1+1+1–1= 0
1+1–1–1+1= 1
1+1–1–1+1= 1
1–1= 0
1–1= 0
1
1
;
i.e., = 7
i.e., = .
From above discussion we get;
Theorem: The number of partitions of n with crank is for all .
Proof: The generating function for is given by;
 (2) 
Now we distribute the function into two parts where first one represents the crank with and second one represents the crank with .
The first function is;
Counts (for ) the number of partitions with no 1’s and the exponent on z being the largest part of the partition where , like;
Here n = 4, the 5^{th} term is .
Again second partition is,
which counts the number of partitions with and the exponent on z is clearly , since , like;
Here n = 4, the 5^{th} term is i.e., .
Thus in the double series expansion of , we see that the coefficient of is the number of partitions of n in which . Equating the coefficient of from both sides in (2) we get the number of partitions of n with is for all . Hence the Theorem.
5. Conclusion
We have verified that the coefficient of x in the right hand side of the generating function for is an explanation of z, the exponents of z represent the crank of partitions, it is already shown with examples for n = 5 and 6. We have satisfied the result = , it is already shown when n = 4 and 5 respectively. For any positive integer of n we can verify the corresponding Theorem. We have already satisfied the Theorem for n = 4 and 5.
Acknowledgment
It is a great pleasure to express our sincerest gratitude to our respected Professor Md. Fazlee Hossain, Department of Mathematics, University of Chittagong, Bangladesh. We will remain ever grateful to our respected Late Professor Dr. Jamal Nazrul Islam, RCMPS, University of Chittagong, Bangladesh.
References
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