Generating Functions for 1(n) and 2(n)
Das, Sabuj and Mohajan, Haradhan
Assistant Professor, Premier University, Chittagong, Bangladesh.
17 March 2014
Online at
https://mpra.ub.uni-muenchen.de/83046/
1
Generating Functions for
β
1(
n
) and
β
2(
n
)
Sabuj Das
Senior Lecturer, Department Of Mathematics Raozan University College, Bangladesh.
Email: [email protected]
Haradhan Kumar Mohajan2
Assistant Professor, Premier University, Chittagong, Bangladesh Email: [email protected]
ABSTRACT
This paper shows how to prove the two Theorems, which are related to the terms β1(n) and β2(n)
respectively Theorem: N(0,5,5n+1)= β1(n)+N (5,5,5n+1) and Theorem: N(1,5,5n+1)= β2(n)+
N(2,5,5n+2).
Keywords:Generating functions, Jecobi’s triple product.
1. INTRODUCTION
We give the definitions of
, Rank of partition, N m,n , N
m,t,n
, z,
x (zx)∞,
m nx , 1
n , 2
n ,
m kx
x ; 5 collected from Partitions Yesterday and Today (Garvan 1979), Generalizations of Dyson’s
rank (Garvan 1986), Ramanujan’s Lost Notebook (Andrews 1979). We generate the generating functions for 1
n , 2
n (Andrews 1979) and prove the Theorems N
0,5,5n1
1
n +
2,5,5n1
N and N
1,5,5n1
2
n + N
2,5,5n2
. Finally we give two examples, which arerelated to the Theorem 1 and Theorem 2 respectively when n =2.
2. DEFINITIONS
: A partition.
2
mnN , : The number of partitions of n with rank m.
mt n
N , , : The number of partition of n with rank congruent to m modulo t.
n0
: The number of partitions of n with unique smallest part and all other parts the double of the smallest part.
n
1
: The number of partitions of n with unique smallest part and all other parts one plus the double of the smallest part.z: The set of complex numbers.
x: The product of infinite factors is defined as follows:
x 1x
1x2
1x3
... .
zx : The product of infinite factors is defined as follows:
1
1 2
1 3
... zx zx
zx
zx .
m nx : The product of m factors is defined as follows:
1
2
1
1... 1
1
1
n n n nm
m n
x x
x x
x .
m kx
x ; 5 : The product of m factors is defined as follows:
; 5
1 k
1 k5
1 k10
...
1 km1 5
m
k x x x x x
x .
n1
: The number of partitions of ninto 1’s and parts congruent to 0 or –1 modulo 5 with the largest part 0mod55 times the number of 1’s the smallest part 1
mod5
. n
2
: The number of partitions of n into 2’s and parts congruent to 0 or – 2 modulo 5 with the largest part 0
mod5
5 times the number of 2’s the smallest part 2
mod5
.3. GENERATING FUNCTIONS (FROM RAMANUJAN’S LOST NOTE BOOK)
From Ramanujan’s Lost Note Book (Andrews 1979), Mock Theta Functions (2) (Watson 1937), G. E. Andrews and F. G. Garvan (Andrews and Garvan 1989), we quote the relations as follows:
... ) 5 2 cos 2 1 )( 5 2 cos 2 1 (
... ) 1 )( 1 )( 1 ( )
(
4 2
2
3 2
x n x x n x
x x x x
F
3 2 / 5 2 cos 2 1 1 ) ( x n x x x
f
... ) 5 2 cos 2 1 )( 5 2 cos 2 1
( 2 2 4
4 x n x x n x x
,n1or 2.
( ) 5 4 cos 2 ) ( 5 2 cos 4 ) ( ) ( 5 2 5 1 5 1 x C n x x B n x x A x
F ( )
5 2 cos 2 5 3 x D n
x . (1)
) ( 5 2 cos 2 ) ( ) ( 5 2 sin 4 ) ( ) ( 5 2 5 1 2 5 1 x C n x x B x x n x A x
f
x x n x D n
x . ( )
5 2 sin 4 ) ( 5 2 cos
2 5 2
3
. (2)
... 1 1 1 ... 1 2 6 2 4 2 9 3 2 x x x x x x x A ,
... 1 1 1 ... 1 1 1 6 4 15 10 5 x x x x x x x B ,
... 1 1 1 ... 1 1 1 7 3 2 15 10 5 x x x x x x x C ,
... 1 1 1 ... 1 2 7 2 3 2 2 7 4 x x x x x x x D ,
54 6
1 1 1 1 1 1 x x x x x x
1 1 1 1 ...
1 4 6 9 11
20 x x x x x x .
But we get;
( ) 5 4 cos 2 ) ( 5 2 cos 4 )
( 5 5 2 5
x C x x B x x
A ( )
5 2 cos
2 3 5
x D x
2 2 3 5
2 5 2 cos 2 5 4 cos 2 5 2 cos 4
1 x x x x ...
5 2 cos 2 5 2 cos
4x6 2 x8 x10
2 2 53 7
1 1 1 1 1 1 x x x x x x
1 1 1 1 ...
1 2 3 7 8 12
20 x x x x x x . Now,
1 1 1 1 1 ...
1 3 4 5
4
x A
x3 1 .
And,
1 1 1 1 1 ...
1 3 4 5
3 3
2 2
x x x
x x
x x x
x
x xD
x
3 .
We assume without loss of generality that n = 1. Let 5 2
exp
i
, then we may write the definitions of F(x) and ′(x) as;
x x
x x
F 1
and,
1
1 1
1 1 ... 1
1 1
2
n
n n
n
x x
x x
x x
f
1
1
2
n n n
n
x x
x
,
where we have used the relations;
a 0 1,
1
1 ... 1
1
n
n a ax ax
a , for n1
and,
1
1
1
n
n n
n a ax
Lim
a .
After replacing x by x5 we see that (1) and (2) are identities for F(x) and ′(x). We note that the numerators in the definitions of A(x) and D(x) are theta series in x and hence may be written as
infinite products using Jecobi’s triple product identity;
n
n
n
n
x x
z
zx
1 1 1
5
2 1
1
n n n n n
x
z (3)
... 1
... 2 1 2 3
x z zx z
x
z .
where z 0 and x< 1.
Replacing x by x5 and z by x3 we get from (3);
n
n
n
nx x
x5 3 5 2 5
1
1 1
1
... x11 x3 1 x2 x9 ...
1 x2 x3 x9 x11 ... .
Again replacing x by x5 and z by x3 equation (3) becomes;
n
n
n
nx x
x5 4 5 1 5
1
1 1
1
... x13 x4 1 x x7 ...
1 x x4 x7 x13 ... .
In fact we have;
5
4
2
5 1
2
5 2 5 3 51 1 1
1 1
1
n n
n n
n
n x x
x x
x x
A ,
5
4
5
5 1
1 1 1
1
n n n
n x x
x x
B ,
5
3
5 2
511 1
1
n n n
n x x
x x
C ,
5 4
5 3
5 1
5 2
5
1 1 1
1 1
1
n n n n n
n x x
x x
x x
D .
6
3.1 Rank of a Partition
The rank of a partition is defined as the largest part minus the number of parts. Thus the partition 6 + 5 + 2 + 1 + 1 + 1 + 1 of 17 has rank, 6–7 = –1 and the conjugated partition, 7 + 3 + 2 + 2 + 2 + 1 has rank, 7–6 = 1. i.e., the rank of a partition and that of the conjugate partition differ only in sign. The rank of a partition of 5 belongs to any one of the residues (mod 5) and we have exactly 5 residues. There is similar result for all partitions of 7 leading to (mod 7).
The generating function for the rank is of the form (Garvan 1986);
1
1 1
1 3 2 1
1 1
1 n
j j
n n m n n n
x x
x
1 0
1 2 3 2 1 2 3 2 1
1
n k
k m
n n m n n n
x k P x
x
... ...
.
0 2 3 2 5 2 6
1 m m m m
m
x x x
x x
n nx n m N
0
, .
The generating function for N
m,t,n
is of the form;
1
1 1
1 3
2 1
1 1
n n
j j tn
m t n mn n n n
x x
x x x
1
1 3 2 1
n n
m t n mn n n n
x x
x
0 2
... 1
k
k tn
tn
x k P x
x
nn
x n t m N , ,
0
;
which shows that all the coefficients of xn (where n is any positive integer) are zero.
Now we define the generating function;
d
7
where
nn a
a d r d t N a t tn d x
r , , ,
0
, and
d r
d t r d r dra,b a,b , a b .
nn
x d tn t b N d tn t a
N
0 , , , ,
.
The generating function
x is of the form;
54 6
1 1 1 1
1 1
x x x
x x
x
1 1 1 1 ...
1 4 6 9 11
20
x x x x x
x
,
1 1 x x2 ... x51 x x2 ...
1x4...
1x6...
...
x x2 x3 x4 2x5 2x6 2x7 2x8 ...
nn
x n N n
N1,5,5 2,5,5 0
0
2 , 1
r
.The generating function A(x) is defined as;
... 1
1 1
... 1
2 6 2 4 2
9 3 2
x x
x
x x x x
A
1 x2 x3 x9 ... 1 2x 3x2 ...
12x43x8...
...
1 2x 2x2 x3 2x4 ...
0
5 , 5 , 2 5
, 5 , 0 1
n
n N
n
N
25 , 5 , 2 2 5 , 5 ,
1 n N n x
N
n
n
x n N
n
N 0,5,5 2,5,5
1
0
nn
x n N n
N1,5,5 2,5,5 2
0
0 2
01r0,2 r1,2
8 The generating function is of the form;
5 4
5 1
5
1 1 1
1
n n n
n x x
x
1 1 ... 1 ...
1 5 4
5 5
1
n n
n n
x x
x
1 0 3 2x 12 11x2 x3 2x4 ...
nn
x n N
n N
0
1 5 , 5 , 2 1 5 , 5 , 0
12 , 0
r
.
The generating function is of the form;
5 3
5 2
5
1 1 1
1
n n n
n x x
x
1 5
1 5 3 10 6 ...
1
n n
n n
x x
x
1 0 3 3 x 16 15 x2 ...
n n
x n N
n
N 0,5,5 2 2,5,5 2
0
...
1
x
5n 2x
10n 4
2 2 , 1r
.
The generating function (x) is of the form;
2 2 53 7
1 1 1 1
1 1
x x x
x x
x
1 1 1 1 ...
1 2 3 7 8 12
20
x x x x x
x
1 ...
1 ...
1 2 4 5 2
x x x x
1 3 6...
1 7...
...x x
x
9 Hence,
... 2
2 7 8 9
6 5 4 3
x x x x x x x x x
x
nn
x n N n
N 2,5,5 3 0,5,5 3
0
3 0 , 2r
and,
x x r0,2 3 .
The generating function D(x) is of the form;
... 1
1 1
... 1
2 7 2 3 2 2
7 4
x x
x
x x x x
D
1 x x4 x7 ... 1 2x2 3x4 ...
12x3...
12x7 ...
...
1 2 2 0. 3 ...
x x x
0
3 5 , 5 , 1 3 5 , 5 , 0 n
n N n
N
nx n N n
N 0,5,5 3 2,5,5 3
0
3 5 , 5 , 1 3 5 , 5 , 0
n
n
x n N n
N
0
3 5 , 5 , 2 3 5 , 5 , 0
n
n
x n N n
N
3 0,2
3 1,
0 r
r
4. THE GENERATING FUNCTIONS FOR 1
n AND 2
n
First we shall establish the following identity, which is used in proving the Theorems. If a and t are both real numbers with a < 1 and t < 1, we have;
... 1
1 1
... 1
1 1
2 2
tx tx t
atx atx
at t
10
1
at
1
atx
...
1
t
t
2...
1txt2x2 ...
1t4x4...
...
1 t 1 x x2 ... a1 x x2 ...
t
2
1
x
2
x
2
2
x
3
...
a
1
2
x
3
x
2
...
22 32 4...
...2
x x x x a
1 1 a t1 x x2 ...
1a 1ax
t2
1x2x22x3...
...
2 2
1 1
1 1 1
1 1
x x
t ax a
x t
a
... 1
1 1
1 1 1
3 2
3 2
x x x
t ax ax a
0
n n
n n x
t a
i.e.,
0
n n
n n x
t a t
at
. (4)
The generating function for 1
n is defined as;
0
5 4 5 5 5
;
;
n
n n
n
x
x
x
x
x
... 1
1 1
1
14 9
4
x x
x
1
1
1
...1
14 9
5
x x
x
1 5
1 10
1 14
...... 2x x
x
x
1 x x2 x3 2x4 x5 2x6 3x8 ...
n nx n 0
1
, (5)
were we have assumed 1
0 1.The generating function for 2
n is defined as;
0
5 3 5 5 5
2
;
;
n
n n
n
x
x
x
x
11
... 1 1 1
1 13 8
3
x x
x
1 5
1 8
1 13
... 2
x x x
x
1 5
1 10
1 13
... ...4
x x
x
x
1 x2 x3 x4 2x6 x7 2x8 ...
n nx n
0 1
, (6)
were we have assumed 1
0 1.Here we give two Theorems, which are related to the terms 1
n and 2
n respectively.Theorem 1:
N
0
,
5
,
5
n
1
1
n
+N
2
,
5
,
5
n
1
,where 1
n is the number of partitions of ninto 1’s and parts congruent to 0 or –1 modulo 5 with the largest part 0
mod5
5 times the number of 1’s the smallest part 1
mod5
.Proof: From (4) by replacing
z1x for a and z for t we have;
x z z
x 1
0 1 1
1
n n
n n x
z x z x
z ,
where z 1 but z0
...
1 1
1 1 1 1 1 ... 1 1
1
2 2 2 1 1 1
2 1 1
x x
z x z x z x
z x z x
z x z
... 1
1 ... 1
1
1
2 1 2
1 1
x z x
z x
z x
z
1
1 2
1 1 3
......2
x z x x
z
... 1
1 ... 1 1
... 1 1
2 1 1
2
x z x z zx
z
x x
Replacing x by x5 and z by x, we obtain;
1
1
...
1
1
1
...1
14 9 5 9
4
x x x
x x
x
1 5
1 10
1 14
......2
x x
x
12
... 1 1 ... 1 1
... 1 1
9 4 6
10 5
x x x
x
x x
.
Hence,
5 4 55 5
0
5 4 5 5 5
; ;
; ;
; x x x x
x x x
x x x
x
n
n n
n
0,2
10
1 n x r
n
n
, by above;
nn
x n N n
N 0,5,5 1 2,5,5 1
0
.
Equating the coefficient of xnon both sides, we get;
0,5,5 1
2,5,5 1
1 n N n N n .
Hence the Theorem.
Theorem 2: N
1,5,5n1
2 n + N
2,5,5n2
, where 2 n is the number of partitions of n into 2’sand parts congruent to 0 or – 2 modulo 5 with the largest part 0
mod5
5 times the number of 2’s the smallest part 2
mod5
.Proof: From (4) by replacing
z1x for a, and z for t we have;
x z z
x 1
0 1 1
1
n n
n n x
z x z x
z ,
where z 1 but z0
...
1 1
1 1 1 1 1 ... 1 1
1
2 2 2 1 1 1
2 1
1 x x
z x z x z x
z x z x
13
... 1
1 ... 1
1
1
2 1 2
1 1
x z x
z x
z x
z
1
1 2
1 1 3
......2
x z x x
z
... 1
1 ... 1 1
... 1 1
2 1 1
2
x z x z zx
z
x
x .
After replacing x by x5, and z by x2, we get;
1 1 ...
1 1 1
... 1
14 9 5 9
4
x x x
x x
x
1 5
1 10
1 14
...... 2x x
x
x
.
We get by replacing x by x5, and z by x2;
1 1 ...
1 1 1
... 113 8 5
2 8
3
x x x
x x
x
1 5
1 10
1 13
...... 4x x
x
x
... 1 1 ... 1 1
... 1 1 1
8 3 7
2
15 10
5
x x x
x
x x x
.
Hence,
2 5 3 55 5
0
5 3 5 5 5
2
;
;
;
;
;
x
x
x
x
x
x
x
x
x
x
x
n
n n
n
1,2
20
2 n x r n
n
, by above;
nn
x n N
n
N1,5,5 2 2,5,5 2
0
.
Equating the coefficient of xn on both sides, we get;
1,5,5 2
2,5,5 2
2 n N n N n
1,5,5n2
2
n N
2,5,5n2
N .
Hence the Theorem.
14
Example 1: N(0, 5, 11) = 12, N (2, 5, 11) = 11, 1
2 1, with the relevant partition is 1 + 1.N(0, 5, 11) = 1
2 + N(2, 5, 11).Example 2: N(1, 5, 12) = 16, N(2, 5, 12) = 15, 2
2 1 , with the relevant partition is 2.N(1, 5, 12) = 2
2 + N(2, 5, 12).5. CONCLUSION
We have verified for any positive integer of n, the two Theorems related to the terms 1
n and
n2
respectively. We have also verified the Theorems for n = 2.
6. 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 Professor Late Dr. Jamal Nazrul Islam, JNIRCMPS, University of Chittagong, Bangladesh.
REFERENCES
1. Andrews, G.E. and Garvan, F.G. (1989). Ramanujan’s “Lost” Notebook VI: The Mock Theta Conjectures, Advances in Mathematics, 73: 242–255.
2. Andrews, G.E. (1979). An Introduction to Ramanujan’s “Lost” Notebook, American Mathematical Monthly, 86: 89–108.
3. Garvan, F.G. (1986). Generalizations of Dyson’s Rank, Ph. D. Thesis, Pennsylvania State University.
4. Garvan, F.G. (1979). Partitions Yesterday and Today, New Zealand Mathematical Society, Wellington.