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Munich Personal RePEc Archive

Generating Function for M(m, n)

Mohajan, Haradhan

Assistant Professor, Premer University, Chittagong, Bangladesh.

4 April 2014

Online at https://mpra.ub.uni-muenchen.de/83594/

MPRA Paper No. 83594, posted 03 Jan 2018 16:31 UTC

(2)

Generating Function for M(m, n)

Sabuj Das

Senior Lecturer, Department of Mathematics Raozan University College, Bangladesh

Email: sabujdas.ctg@gmail.com

Haradhan Kumar Mohajan

Premier University, Chittagong, Bangladesh Email: haradhan_km@yahoo.com

Abstract

This paper shows that the coefficient of x in the right hand side of the equation for M(m, n) for all n >1is an algebraic relation in terms of z. The exponent of z represents the crank of partitions of a positive integral value of n and also shows that the sum of weights of corresponding partitions of n is the sum of ordinary partitions of n and it is equal to the number of partitions of n with crank m. This paper shows how to prove the Theorem “The number of partitions π of n with crank C(π) = m is M(m, n) for all n >1.”

Keywords: Crank, j-times, vector partitions, weight, exponent.

1. Introduction

First we give definitions of P

 

n , the crank of partitions,

 

x ,

 

zx ,

 

x2;x and M

 

m,n . 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 n1). 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 M

 

m,n , weight 

 

and crank

 

. We prove the Theorem

“The number of partitions  of n with crank C

 

m is M

 

m,n for all n1.”

2. Definitions

Now we give some definitions following ([3], [4] and [5]).

(3)

 

n

P : Number of partitions of n, like 4, 3+1, 2+2, 2+1+1, 1+1+1+1. Therefore, P

 

4 5 and similarlyP

 

5 7etc.

Crank of partitions [2]: For a partition  , let l

 

 denotes the largest part of , 

 

 denote the number of 1’s in , and

 

denote the number of parts of  larger than

 

, the crank

 

c is given by;

     

     



 

. 0 if

;

0 if

;

lc

 

x

1x

 

1x2



1x3

...

 

zx

1zx

 

1zx2



1zx3

...

 

x2;x

1x2



1x3



1x4

...

 

mn

M , : The number of partitions of n with crank m.

2.1 Notations

For all integers n0 and all integers m, the number of n with crank equal to m is

 

1,1 1

M , like;

Partitions of 1

 

Largest part

 

l

Number of 1’s

 

 Number of parts larger than 

 

 

Crank

 

c

1 1 1 0 –1

 

1,1 1

M .

But we see that;

 

1,1 M

 

1,1 M

 

0,1 1

M .

Since, the coefficient of x in the right hand side of the equation;

   

   



x z zx x x

z n m

M m n

m n

1 0

,

is z1z1 i.e., z1z1z0 the exponent of z being the crank of partition.

Therefore, M

 

1,1 M

 

1,1 M

 

0,1 1.

(4)

3. The Generating Function for M

 

m,n

The generating function for M

 

m,n is given by [2];

   

n



n

n

n n m

m n zx z x

x x z n m

M 1

0 1 1 1

, 1



 

 

 

    

1

 

11 2

 

1...1 11



1... 1 2

...

3 2

x z x z zx

zx

x x

x

 

    

1

 

11 2

 

1...1 11



1... 1 2

...

3 2

x z x z zx

zx

x x

x

 

 

    

  



 

1 1 ...

...

1 1

1

2 1 1

3 2

x z x z

x x

zx x

 

     

 







 

0

1 1

j j

j j

x xz zx zx

x , by Andrews [1],

       

     

     

 







    

 

...

1 1

3 13 3 2

12 2 1

11 1

x xz zx x

xz zx x

xz zx zx

x

     

     

   



 

 

 

 

2

2 2 2 1

1 1

1 1 1

1 1 1

x x

z x zx zx x

xz zx zx

x

   

 

  

...

1 1 ...

1 ...

1 1

1

3 2

2 2 2

1

2 x zx

z x zx

xz zx

zx

x

    

1 1

 

1 1



1

...

1

3 2

3 3 3

2

x x

x

z x zx zx

zx

1 2



1 3



1 4

... ...

3

3

zx x

x

z x

         

 

1

1 1 - j 2

2 ... ;

1 1

1

j

j j j

zx x x

z x zx

zx

x (1)

 

 

1 z1 z 1 x z2 z2 x2 z3 z3 1 x3

1z2 z2 z4 z4

x4

1zz1z3z3z5 z5

x5

(5)

1zz1z2z2z3z3z4z4z6z6

x6...

We see that the exponent of z represents the crank of partitions of n (for n1). As for examples when n = 5 and 6,

For n = 5,

Partitions of 5

 

 Largest part

 

l

Number of 1’s

 

 Number of parts larger than

 



 

Crank

 

c 5

4+1 3+2 3+1+1 2+2+1 2+1+1+1 1+1+1+1+1

5 4 3 3 2 2 1

0 1 0 2 1 3 5

1 1 2 1 2 0 0

5 0 –1 3 –1 3 –5 For n = 6,

Partitions

 Largest part

 

l

Numbers of ones

 

 Number of parts

larger than 

 

 

Crank

 

c

6 6 0 1 6

5+1 5 1 1 0

4+2 4 0 2 4

4+1+1 4 2 1 –1

3+3 3 0 2 3

3+2+1 3 1 2 1

3+1+1+1 3 3 0 –3

2+2+2 2 0 3 2

2+2+1+1 2 2 0 –2

2+1+1+1+1 2 4 0 –4

1+1+1+1+1+1 1 6 0 –6

4. Vector Partitions of n

Let, VDPP, 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, VDPP. For

1,2,3

V , where  1 2 3

weight = 

   

  1# 1

, the crank

     

# 2 # 3 .

(6)

We have 41 vector partitions of 4 are given in the following table:

Vector partitions of 4 Weight

 

Crank

 

, ,4

1  

 

+1 –1

, ,3 1

2  

+1 –2

, ,2 2

3    

+1 –2

, ,2 1 1

4    

+1 –3

, ,1 1 1 1

5      

+1 –4

,1,3

6

 

+1 0

,1,2 1

7   

+1 –1

,1 1 1 1

8     

+1 –2

,2 2

9   

+1 0

,2,1 1

10   

+1 –1

,1 1,2

11  

+1 1

,1 1,1 1

12   

+1 0

,3,1

13

 

+1 0

,2 1,1

14  

+1 1

,1 1 1,1

15   

+1 2

 

16  ,4,

+1 1

 

17  ,31,

+1 2

 

18 ,22,

+1 2

 

19,211,

+1 3

20  ,1111,

+1 4

1, ,3

21

  –1 –1

1, ,2 1

22   

–1 –2

1, ,1 1 1

23   

–1 –3

 

1,1,2

24

–1 0

1,1,1 1

25  

–1 –1

 

1,2,1

26

–1 0

1 1,1,1

27  

–1 1

28 1,3, –1 1

29  1,21, –1 2

30 1,111, –1 3

2, ,2

31

  –1 –1

(7)

2, ,1 1

32   

–1 –2

2,1,1

33

–1 0

34 2,2, –1 1

35 2,11, –1 2

3, ,1

36

  –1 –1

2 1, ,1

37

  

+1 –1

38 3,1, –1 1

39  21,1,

+1 1

40  4, , –1 0

 

41 31, ,

+1 0

From the above table we have,

           

0,4 6 9 12 13 24 M

+ 

       

26 33 40 41

= 1+1+1+1–1–1–1–1+1 = 1

 

1,4

M

   

11 14 ...

 

39

= 1 + 1 + 1–1–1–1–1+1 = 0.

and

1,4

M

   

1 7 ...

 

37

= 1 + 1 + 1–1–1–1–1+1 = 0

 

2,4

M 1 + 1 + 1–1–1= 1

2,4

M 1 + 1 + 1–1–1= 1

 

3,4

M 1–1= 0

3,4

M 1–1= 0

 

4,4

M 1

4,4

M 1

(8)

    

M m,4 ;

i.e.,

   

 



m V m

m M

crank 4

4

, = 5

i.e.,

   

 



m V m

m M

crank 4

4

, = P

 

4 .

Again we have 83 vector partitions of 5 are given in the following table:

Vector partitions of 5 Weight

 

Crank

 

, ,5

1  

 

+1 –1

, ,4 1

2   

+1 –2

, ,3 2

3    

+1 –2

, ,3 1 1

4    

+1 –3

, ,2 2 1

5    

+1 –3

, ,2 1 1 1

6      

+1 –4

, ,1 1 1 1 1

7       

+1 –5

 

8  5, , –1 0

 

9  ,5,

+1 1

 

10,41,

+1 2

 

11 41, ,

+1 0

12 4,1, –1 1

13 1,4, –1 1

,4,1

14

 

+1 0

,1,4

15

 

+1 0

1, ,4

16

  –1 –1

4, ,1

17

  –1 –1



18 32, ,

+1 0

 

19  ,32,

+1 2

20 3,2, –1 1

21 2,3, –1 1

,3,2

22

 

+1 0

,2,3

23

 

+1 0

(9)

3, ,2

24

  –1 –1

2, ,3

25

  –1 –1

 

26 ,311,

+1 3

27 31,1,

+1 1

28 1,31, –1 2

,3 1,1

29  

+1 1

,1,3 1

30  

+1 –1

3 1, ,1

31

  

+1 –1

1, ,3 1

32  

–1 –2

33 3,11, –1 2

,1 1,3

34  

+1 1

,3,1 1

35  

+1 –1

3, ,1 1

36  

–1 –2

37 ,221,

+1 3

38 1,22, –1 2

,2 2,1

39  

+1 1

,1,2 2

40  

+1 –1

1, ,2 2

41  

–1 –2

42 21,2,

+1 1

43 2,21, –1 2

,2,2 1

44   

+1 1

,2 1,2

45  

+1 1

2 1, ,2

46

  

+1 –1

2, ,2 1

47   

–1 –2

 

48 ,221,

+1 4

,2 1 1,1

49   

+1 2

,1,2 1 1

50   

+1 –2

51 1,211, –1 3

1, ,2 1 1

52   

–1 –3

53 21,11,

+1 2

,2 1,1 1

54   

+1 0

,1 1,2 1

55   

+1 0

2 1, ,1 1

56   

+1 –2

,1 1 1,2

57   

+1 2

(10)

,2,1 1 1

58   

+1 –2

59 2,111, –1 3

2, ,1 1 1

60   

–1 –3

 

61 ,11111,

+1 5

,1 1 1 1,1

62    

+1 3

,1,1 1 1 1

63    

+1 –3

1, ,1 1 1 1

64    

–1 –4

65 1,1111, –1 4

,1 1,1 1 1

66    

+1 –1

,1 1 1,1 1

67    

+1 1

1,1,1 1 1

68  

–1 –2

1,1 1 1,1

69  

–1 2

1,1 1,1 1

70  

–1 0

1,1 1,2

71 

–1 1

1,2,1 1

72 

–1 –1

2,1 1,1

73 

–1 1

2,1,1 1

74 

–1 –1

2,2,1

75

–1 0

2,1,2

76

–1 0

1,2,2

77

–1 0

 

3,1,1

78

–1 0

 

1,3,1

79

–1 0

 

1,1,3

80

–1 0

1 2,1,1

81 

+1 0

1,1 2,1

82  

–1 1

1,1,1 2

83 

–1 –1

From this table we have;

 

0,5

M

       

8 11 14 15 +

         

18 22 23 54 55

+

         

70 75 76 78 79

+

     

79 80 81

(11)

= –1+1+1+1+1+1+1+1+1–1–1–1–1–1–1–1–1+1 = 1.

 

1,5

M 1–1–1–1–1+1+1+1+1+1+1+1–1–1–1 =1

1,5

M 1–1–1–1–1+1+1+1+1+1+1+1–1–1–1 =1

 

2,5

M 1+1–1–1–1–1+1+1+1–1= 0

2,5

M 1+1–1–1–1–1+1+1+1–1= 0

 

3,5

M 1+1–1–1+1= 1

3,5

M 1+1–1–1+1= 1

 

4,5

M 1–1= 0

4,5

M 1–1= 0

 

5,5

M 1

5,5

M 1

    

M m,5 ;

i.e.,

   

 



m V m

m M

crank 5

5

, = 7

i.e.,

   

 



m V m

m M

crank 5

5

, = P

 

5 .

From above discussion we get;

   

 



m n V m

n m M

crank

, = P

 

n .

(12)

Theorem: The number of partitions  of n with crank c

 

m is M

 

m,n for all n1.

Proof: The generating function for M

 

m,n is given by;

   

n



n

n n

n m

m n zx z x

x x z n m

M 1

0 1 1 1

, 1



 

 

 

(2)

 

   

   

 

1

1 1 - j 2

2 ... ;

1 1

1

j

j j j

zx x x

z x zx

zx

x .

Now we distribute the function into two parts where first one represents the crank with

   

l

c  and second one represents the crank with c

     

   . The first function is;

 

1zx

 

11zxx2



1zx3

...

1 z 1 x z2x2 z3x3 z2 z4 x4

z3z5

x5 ...

Counts (for n1) the number of partitions with no 1’s and the exponent on z being the largest part of the partition where c

   

 l  , like;

Partitions of 4

 

 Largest part

 

l

Number of 1’s

 

 Number of parts larger

than 

 

 

Crank

 

c

4 2+2

4 2

0 0

1 2

4 2 Here n = 4, the 5th term is

z2z4

x4.

Again second partition is,

   

1

1 1 - j 2,

j

j j j

zx x x

z

x



  



...

1 1

1 ...

1

1 2 3 4

2 2 3

2 1

zx zx

x

z x zx

zx

xz

1 2



1 3



1 4



1 5

... ...

3

3

zx zx

x x

z x

1 3

 

3 1 2 4

4 ...

2 2

1       

z x z x z x z z x

(13)

which counts the number of partitions with 

 

j and the exponent on z is clearly

     

  

c , since i0, like;

Partitions of 4

 

Largest part

 

l

Number of 1’s

 

Number of parts larger than 

 

 

Crank

 

c

3+1 2+1+1 1+1+1+1

3 2 1

1 2 4

1 0 0

0 –2 –4

Here n = 4, the 5th term is

1z2 z4

x4 i.e.,

z0z2z4

x4.

Thus in the double series expansion of

 

       

 

1

1 1 - j 2

2 ... ,

1 1

1

j

j j j

zx x x

z x zx

zx

x , we see that the coefficient of zmxn

n1

is the number of partitions of n in which c

 

m. Equating the coefficient of zmxn from both sides in (2) we get the number of partitions of n with c

 

m is M

 

m,n for all n1. Hence the Theorem.

5. Conclusion

We have verified that the coefficient of x in the right hand side of the generating function for

 

mn

M , 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

   

 



m n V m

n m M

crank

, =

 

n

P , 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.

(14)

References

[1] Andrews, G.E., The Theory of Partitions, Encyclopedia of Mathematics and its Application, vol. 2 (G-c, Rotaed) Addison-Wesley, Reading, mass, 1976 (Reissued, Cambridge University, Press, London and New York 1985). 1985.

[2] Andrews, G.E. and Garvan, F.G., Dyson’s Crank of a Partition, Bulletin (New series) of the American Mathematical Society, 18(2): 167–171. 1988.

[3] Atkin, A.O.L. and Swinnerton-Dyer, P., Some Properties of Partitions, Proc. London Math.

Soc. 3(4): 84–106. 1954.

[4] Garvan, F.G., Ramanujan Revisited, Proceeding of the Centenary Conference, University of Illinois, Urban-Champion. 1988.

[5] Garvan, F.G.. Dyson’s Rank Function and Andrews’ spt-function, University of Florida, Seminar Paper Presented in the University of Newcastle on 20 August 2013.2013.

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