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/
MPRA Paper No. 83046, posted 02 Jan 2018 23:06 UTC
1
Generating Functions for β
1( n ) and β
2( n )
Sabuj Das
Senior Lecturer, Department Of Mathematics Raozan University College, Bangladesh.
Email: sabujdas.ctg@gmail.com
Haradhan Kumar Mohajan2
Assistant Professor, Premier University, Chittagong, Bangladesh Email: haradhan_km@yahoo.com
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)∞,
mxn , 1
n , 2
n ,
m k xx ; 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 are related to the Theorem 1 and Theorem 2 respectively when n =2.2. DEFINITIONS
: A partition.
Rank of partition: The largest part of a partition minus the number of parts of .
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:
zx 1zx
1zx2
1zx3
... .
mxn : The product of m factors is defined as follows:
m 1 n
1 n1
1 n2
... 1 nm1
n x x x x
x .
m k xx ; 5 : The product of m factors is defined as follows:
; 5
m1 k
1 k5
1 k10
...
1 km1 5
k x x x x x
x .
n1 : The number of partitions of n into 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 ) cos2 2 1 5 )(
cos2 2 1 (
...
) 1 )(
1 )(
1 ) (
(
4 2
2
3 2
n x x n x
x
x x x x
F
3
2 /
5 cos2 2 1 1 ) (
n x x x x
f ...
5 ) cos2 2 1 5 )(
cos2 2 1
( 2 2 4
4
n x x n x
x
x
,n1or 2.
( )
5 cos4 2 ) 5 ( cos2 4 ) ( )
( 5
2 5
1 5
1
x n C x x n B x x A x
F
) 5 ( cos2 2 5
3
x n D
x . (1)
) 5 ( cos2 2 ) ( ) 5 ( sin 2 4 ) ( )
( 5
2 5
1 5 2
1
x n C x x B x n x
x A x
f
x
x x n
n D
x ( )
5 . sin 2 4 ) 5 ( cos2
2 5 2
3 . (2)
...
1 1
1
...
1
6 2 4 2
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 ,
112
21 3
21...7
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 cos4 2 ) 5 ( cos2 4 )
(x5 x B x5 x2 C x5
A
) 5 ( cos2
2x3 D x5
2 2 3 2 5
5 cos2 5 2
cos4 5 2
cos 2 4
1 x x x x
...
5 cos2 5 2
cos 2
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
5 3
2 3
x x x
x x
x x x
x
4
x A x
3 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 exp25
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,
a n 1a
1ax
...
1axn1
, 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 1 1
1
5
1 21
n n n n n
x
z (3)
... z2x z1 1 zx z2x3 ... . where z 0 and x< 1.
Replacing x by x5 and z by x3 we get from (3);
n
n
n
n
x 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
n
x 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
25 2 5 3 5
1 1 1
1 1
1
n n
n n
n
n x x
x x
x x
A ,
5 4
5 1
5
1 1 1
1
n n n
n x x
x x
B ,
5 3
5 2
5
11 1
1
n n n
n x x
x x
C ,
5 3
5 2
5 1 5 4 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 23
1 1 1
1
n
j j
n n m n n
n x x x
1 0
1 2 23 1 2 23
1 1
n k
m k n n m n n
n x x P k x
xm1 0.xm2 xm3 ... x2m5 x2m6 ...
n
n
x n m
N
0
, .
The generating function for N
m,t,n
is of the form;
1
1 1
1
23 1
1 1
n n
j tn j
m t n n mn
n
n x
x x x x
1
1 23
1
n n
m t n n mn
n
nx x x
0
2 ...
1
k
k tn
tn x P k 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;
dra for N
a,t,tnd
7
where
na n
a d r d t N a t tn d x
r , , ,
0
, and
d r
d t r d r d ra,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 4 1 6 1 9 1 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
02 ,
r1
.
The generating function A(x) is defined as;
11
2
1 4
21...6
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 N
1,5,5n
2N
2,5,5n
x2
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
0 1r0,2 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 5 1 5 4 ... 1 5 1 ...
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 ,
r0
.
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
1x5n2x10n4...
22 ,
r1
.
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 2
1 3
1 7
1 8
1 12
...20
x x x x x
x
1 ...
1 ...
1 2 4 5 2
x x x x
1x3x6...
1x7...
...
x2 x4 x6 x7 2x8 x9 2x10 ... .
9 Hence,
x x3 x4 x5 x6 2x7 x8 2x9 ...
x
x
nn
x n N n
N 2,5,5 3 0,5,5 3
0
3
0 ,
r2
and,
x r0,2 3 x .
The generating function D(x) is of the form;
112
21 3
21...7
2...7 4
x x
x
x x x x
D
1 x x4 x7 ... 1 2x2 3x4 ...
12x3...
12x7 ...
...
1 x 2x2 0.x3 ...
0
3 5 , 5 , 1 3 5 , 5 , 0
n
n N n
N N
0,5,5n3
N 2,5,5n3
xn
0
3 5 , 5 , 1 3 5 , 5 , 0
n
xn
n N n
N
0
3 5 , 5 , 2 3 5 , 5 , 0
n
xn
n N n
N
3 0,2
31 ,
0 r
r
4. THE GENERATING FUNCTIONS FOR 1 n AND 2
nFirst 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
at
10
1 at 1 atx ... 1 t t2 ...
1txt2x2 ...
1t4x4...
...
1 t 1 x x2 ... a1 x x2 ... t2
1x2x22x3...
a12x3x2...
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 11 ...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
1x5
1x91
1x14
...
1 5
1 10
1 14
......2
x x
x
x
1 x x2 x3 2x4 x5 2x6 3x8 ...
nn
x 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