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1 Eat Tenneee State Univerity Digital Eat Tenneee State Univerity Electronic Thee and Diertation Student Work Vector Partition Jennifer French Eat Tenneee State Univerity Follow thi and additional work at: Part of the Dicrete Mathematic and Combinatoric Common, and the Number Theory Common Recommended Citation French, Jennifer, "Vector Partition" (208). Electronic Thee and Diertation. Paper Thi Thei - Open Acce i brought to you for free and open acce by the Student Work at Digital Eat Tenneee State Univerity. It ha been accepted for incluion in Electronic Thee and Diertation by an authorized adminitrator of Digital Eat Tenneee State Univerity. For more information, pleae contact digilib@etu.edu.

2 Vector Partition A thei preented to the faculty of the Department of Mathematic Eat Tenneee State Univerity In partial fulfillment of the requirement for the degree Mater of Science in Mathematical Science by Jennifer French May 208 Rodney Keaton, Ph.D., Chair Robert A. Beeler, Ph.D. Anant Godbole, Ph.D. Keyword: number theory, integer partition, vector partition.

3 ABSTRACT Vector Partition by Jennifer French Integer partition have been tudied by many mathematician over hundred of year. Many identitie exit between integer partition, uch a Euler dicovery that every number ha the ame amount of partition into ditinct part a into odd part. Thee identitie can be proven uing method uch a conjugation or generating function. Over the year, mathematician have worked to expand partition identitie to vector. In 963, M. S. Cheema proved that every vector ha the ame number of partition into ditinct vector a into vector with at leat one component odd. Thi parallel Euler reult for integer partition. The primary purpoe of thi paper i to ue generating function to prove other vector partition identitie that parallel reult of integer partition. 2

4 Copyright by Jennifer French 208 All Right Reerved 3

5 TABLE OF CONTENTS ABSTRACT LIST OF TABLES LIST OF FIGURES INTRODUCTION INTEGER PARTITIONS Background Generating Function Integer Partition Identitie VECTOR PARTITIONS Background Generating Function for Vector Partition Vector Partition Identitie FUTURE WORK BIBLIOGRAPHY VITA

6 LIST OF TABLES Example illutrating Euler Identity Example illutrating Theorem 2.4 for r= Example Illutrating Theorem 2.5 for r = Example Illutrating Theorem 2.5 for r = Example illutrating Theorem Example calculated by Sage illutrating Theorem Example illutrating Theorem 3.4 for r=

7 LIST OF FIGURES Example of a Ferrer graph Example of conjugation Sage code for the number of partition with at leat one odd component in each vector Sage code for the number of partition into ditinct part Sage code for the number of partition with at leat one component of each vector not diviible by r Sage code for the number of partition in which there are le than r copie of each vector Sage code for the number of partition in which every component i a multiple of r Sage code for the number of partition in which each vector i repeated a multiple of r time Sage code for partition enumerated by A r (n) Sage code for partition enumerated by B r (n)

8 INTRODUCTION The tudy of integer partition ha been of interet to many mathematician ince Leibniz aked Bernoulli about the number of partition of an integer n [6]. Over three hundred year of tudie conducted on thi ubject ha led to ome ignificant reult. In 748, Euler made everal ignificant dicoverie related to integer partition. One urpriing dicovery that Euler made wa that the number of partition of an integer into ditinct part i equal to the number of partition into odd part [2]. In 93, S. Ramanujan and G. H. Hardy contributed to the area of integer partition by proving everal ignificant reult [6]. An example of the modern approach to partition can be found in The Web of Modularity: Arithmetic of the Coefficient of Modular Form and q-erie by Ken Ono [5]. More recently, tudie have been conducted on vector partition. Vector partition function are a natural generalization of integer partition function that are cloely related to plane partition. Plane partition are two-dimenional partition where we conider row of integer. The row are left jutified, and there i a non-increae along row and column. For information on plane partition, ee Chapter 0 of Integer Partition by Andrew and Erikon [2]. In 963, Cheema provided a ignificant contribution to the area of vector partition in hi paper Vector Partition and Combinatorial Identitie [3]. In thi paper, he proved that the number of partition of a vector into ditinct part i equal to the number of partition in which each part ha at leat one odd component. Thi parallel Euler integer partition reult regarding ditinct part and odd part a a recurion relation for vector partition. Cheema ue generating function to prove hi reult. For a more detailed look at 7

9 technique involving generating function, the intereted reader i referred to [8]. The following chapter will explore integer and vector partition. Chapter 2 will introduce integer partition and generating function for everal type of partition. Generating function will be ued to prove everal known reult related to integer partition. Chapter 3 will extend thee concept to vector partition. Generating function for vector partition will be ued to prove everal reult that parallel the reult of Chapter 2. Chapter 4 will dicu poible future direction for the tudy of vector partition. 8

10 2 INTEGER PARTITIONS Thi chapter will introduce integer partition. Unle otherwie noted, all material from thi chapter will reference Integer Partition by Andrew and Erikon [2]. 2. Background A partition i a way of writing a poitive integer a a um of poitive integer where the order of the ummand doe not matter. The partition tay the ame regardle of the order of the ummand, o we may chooe by convention to order the part from larget to mallet. For n N, we define the partition function, denoted p(n), to be the number of partition of n. Since the empty um i the only partition of zero, we have p(0) =. Example 2.. The following are all of the poible partition of the number four: 4 = 4, 3 +, 2 + 2, 2 + +, Therefore, we have p(4) = 5. Often, we are intereted in partition of a number n that atify ome condition. We denote the number of uch partition by p(n [condition] ). Example 2.2. The following are all of the poible partition of the number five: 5, 4 +, 3 + 2, 3 + +, , , So the partition of 5 with ditinct part are 5, 4 +, Therefore, p(5 ditinct part) = 3. 9

11 2.2 Generating Function Uing generating function, we can keep track of the number of partition that atify ome condition. Convergence will not be an iue ince each integer n can only be partitioned uing number from the et {,..., n}. Suppoe we want to illutrate all poible partition with ditinct element from S = {, 2, 3}. A generating function i a power erie whoe coefficient give a equence of number. Define p S (n) = p(n ditinct part in S). Note, p S (n) = 0 for n 7, and by convention p S (0) =. The following value can be eaily calculated: p S () =, p S (2) =, p S (3) = 2, p S (4) =, p S (5) =, and p S (6) =. Therefore, we have p S (n)q n = + q + q 2 + 2q 3 + q 4 + q 5 + q 6 + 0q 7 + 0q n 0 = + q + q 2 + q 2+ + q 3 + q 3+ + q q 3+2+ = ( + q + q 2 + q 2+ )( + q 3 ) = ( + q )( + q 2 )( + q 3 ) 3 = ( + q k ). k= The coefficient of q n in the firt line i the number of partition of n into ditinct part from S = {, 2, 3}. For intance, ince there are two uch partition of the number three, we have the term 2q 3. Note that the exponent in the econd line diplay all the partition of thi type. Thu, the generating function for thee partition i given by the following: 3 p(n ditinct part in {,2,3} )q n = ( + q k ). () n 0 k= 0

12 If we allow ditinct part from S = {k,..., k r }, we can extend the generating function given in () to the following: p(n ditinct part in S)q n = n 0 r i=( + q ki ) = k S ( + q k ). (2) It follow that if S = N, we have p(n ditinct part )q n = ( + q k ). (3) n 0 k= Suppoe that we wih to allow part to repeat up to a certain amount of time. For intance, let S = {, 2} and uppoe we allow part to repeat up to three time. Define p S (n; 3) = p(n part in S, no part repeated more than 3 time). The following value can be eaily calcuated: p S (0; 3) =, p S (; 3) =, p S (2; 3) = 2, p S (3; 3) = 2, p S (4; 3) = 2, p S (5; 3) = 2, p S (6; 3) = 2, p S (7; 3) = 2, p S (8; 3) =, p S (9; 3) =, and p S (n; 3) = 0 for n 0. Therefore we have p S (n; 3)q n = + q + 2q 2 + 2q 3 + 2q 4 + 2q 5 + 2q 6 + 2q 7 + q 8 + q 9 + 0q n 0 = + q 2 + q q q + q 2+ + q q q + + q q q q ++ + q q q =( + q + q + + q ++ )( + q 2 + q q ) 2 = ( + q k + q k+k + q k+k+k ) = k= 2 ( + q k + q 2k + q 3k ). k=

13 The exponent of the polynomial beginning on the econd line diplay all partition with element from S = {, 2} where the element are allowed to repeat up to three time, and the coefficient of q n in the firt line i the number of uch partition of n. Therefore, we have the following generating function: p(n part in {,2}, no part repeated more than 3 time )q n n 0 = (4) 2 ( + q k + q 2k + q 3k ). k= For the et S = {k,..., k r } with part allowed to repeat up to d time, the generating erie in (4) i extended to the following: p(n part in S, none repeated more than d time )q n n 0 = r ( + q k i + q 2k i + + q dk i ) i= (5) = k S( + q k + q 2k + + q dk ). It follow that if S = N, we have p(n no part repeated more than d time )q n n 0 = ( + q k + + q dk ). k= (6) By uing the formula of a geometric erie, we obtain the following generating function: 2

14 p(n part in S)q n = n 0 = r ( + q k i + q 2k i + q 3k i +... ) i= r i= = k S q k i q k. (7) If we take the generating function given in (7) and let S = N, we obtain the tandard generating function a follow: p(n)q n = q. k (8) n 0 k=0 The preceding generating function are ueful for proving certain reult related to integer partition. 2.3 Integer Partition Identitie When every number ha the ame amount of integer partition of one type a another type, we ay thi i a partition identity. Many partition identitie exit. One uch identity ay that every number ha the ame amount of integer partition into ditinct part a into odd part. Thi identity wa firt proved by Leonhard Euler in 748 [2]. Table illutrate thi identity for poitive integer up to ix. Notice for each poitive integer n in thi table, there are alway the ame amount of partition in each column. 3

15 Table : Example illutrating Euler Identity. n Odd Part Ditinct Part Theorem 2.3 (Euler Partition Identity [2]). The number of partition of n into odd part i equal to the number of partition of n into ditinct part. Proof. Conider the following generating function which reult from the equation given in (3) and (7), repectively: p(n part ditinct)q n = ( + q k ), n=0 k= p(n part all odd)q n = n=0 k odd q k. 4

16 Then we have ( + q k ) =( + q)( + q 2 )( + q 3 )( + q 4 )( + q 5 )( + q 6 )... k= =( q2 q )( q4 q = ( q)( q 3 )( q 5 )... = q. k k odd q6 q8 q0 q2 )( )( )( )( 2 q3 q4 q5 q )... 6 Therefore, the generating function are equal. Hence, for every poitive integer n, we have p(n part ditinct) = p(n part all odd). Euler Partition Identity can be generalized to give u another partition identity. For any integer r 2, every number ha the ame amount of partition in which no part i diviible by r a partition in which there are le than r copie of each part. Notice, if we let r = 2, then we have every number ha the ame amount of partition in which no part i diviible by two a partition in which there are le than two copie of each part. Thi tatement i equivalent to Euler Theorem. Table 2 illutrate thi identity for integer up to ix and r = 3. 5

17 Table 2: Example illutrating Theorem 2.4 for r=3 n No Part Diviible by 3 Le Than 3 Copie of Each Part Theorem 2.4. [2] The number of partition of n in which no part i diviible by r i equal to the number of partition of n in which there are le than r copie of each part. Proof. The proof i imilar to that of Euler Partition Identity. Conider the following generating function which reult from the equation given in (6) and (7), repectively: p(n le than r copie of each part)q n = ( + q k + q 2k + + q (r )k ), n=0 k= 6

18 p(n no part diviible by r)q n = n=0 k not diviible by r q k. Then for r 2 we have that ( + q k + q 2k + + q (r )k ) i equal to k= ( + q + + q r )( + q q 2(r ) )( + q q 3(r ) )... ( ) ( ) ( ) ( ) q r q 2r q 3r q 4r =... q q 2 q 3 q 4 = q. k k not diviible by r Therefore, the generating function are equal. Hence, for every poitive integer n, we have p(n no part diviible by r) = p(n le than r copie of each part). Another partition identity give u that every number ha the ame amount of partition into even part a into part that are repeated an even number of time. Table 3 illutrate thi identity for poitive even integer up to ix. Note for odd integer, there are no uch partition of either kind. Thi identity can alo be generalized to partition in which all part are multiple of a poitive integer r and partition in which all part are repeated a multiple of r time. Table 4 illutrate thi for integer up to nine and r = 3. A with even part, note that for number that are not a multiple of 3, there are no uch partition of either kind. 7

19 Table 3: Example Illutrating Theorem 2.5 for r = 2 n All Part Even All Part Repeated an Even Number of Time Table 4: Example Illutrating Theorem 2.5 for r = 3 n All Part a Multiple of 3 All Part Repeated a Multiple of 3 Time Theorem 2.5. [2] The number of partition of n in which all part are a multiple of r i equal to the number of partition of n in which all part are repeated a multiple of r time. Proof. Conider the following generating function which reult from the equation given in (6) and (7), repectively: p(n part repeated a multiple of r time)q n = (+q rk +q 2rk +q 3rk +q 4rk +... ), n=0 k= 8

20 Then we have p(n all part a multiple of r)q n = n=0 k= k a multiple r ( + q rk + q 2rk + q 3rk + q 4rk +... ) = q rk = k= j a multiple r q k. q j. Therefore, the generating function are equal. Hence, for every poitive integer n, we have p(n part repeated a multiple of r time) = p(n all part a multiple of r). In order to prove the next partition identity, we mut introduce Ferrer graph and conjugate partition. A Ferrer graph i a way of repreenting a partition of a number. Each part in the partition i repreented by a row of dot. The Ferrer graph i arranged o that the part are ordered from larget to mallet, and the left ide of the graph i aligned vertically. For example, Figure how the Ferrer graph for the partition = Figure : Example of a Ferrer graph. Suppoe we take a Ferrer graph and rearrange it by taking it row and making them into column. Thi proce i called conjugation. Figure 2 how the Ferrer graph from Figure along with it conjugate partition. The reulting partition i 9

21 = Since conjugation preerve the number of dot in a Ferrer graph, it i ueful in proving certain partition identitie. If two type of partition are conjugate, then there i a partition identity between them. Figure 2: Example of conjugation. For more information on Ferrer graph and conjugate partition, refer to Chapter 3 of Integer Partition by Andrew [2]. The following theorem wa proved by MacMahon in the r = cae [4], and by Andrew for general r []. Theorem 2.6. [] Let A r (n) denote the number of partition of n of the form n = b +b 2 + +b, where b i b i+, all odd part are greater than or equal to 2r+, and if b i b i+ i odd then b i b i+ 2r +. Let B r (n) be the number of partition of n into part which are even or ele congruent to 2r + (mod 4r + 2). Then A r (n) = B r (n). Proof. The partition which are conjugate to thoe enumerated by A r (n) are jut thoe partition of n in which any part appearing an odd number of time appear at 20

22 leat 2r + time. Therefore A r (n)q n = ( + q 2k + q 4k + + q (2r 2)k + q 2rk + q (2r+)k + q (2r+2)k +... ) n=0 k= ( ) q 2k(r+) = + q(2r+)k q 2k q k k= ( ) q (2r+2)k = + q(2r+)k + q (2r+2)k q 2k q 2k k= ( ) + q (2r+)k =. q 2k k= Note by Euler Theorem, we have ( + q (2r+)k ) = k= ( k= q (2r+)(2k ) ). Thu, ( ) + q A r (n)q n (2r+)k = q 2k n=0 k= ( ) ( ) = q (2r+)(2k ) q 2k k= ( ) ( ) = q (4r 2)k (2r+) q 2k = k= B r (n)q n. n=0 2

23 3 VECTOR PARTITIONS The previou chapter wa an introduction to partition of poitive integer. The focu of thi chapter will be to extend thi concept to vector of integer. Sage will be ued to define function to verify all reult from thi chapter [7]. 3. Background A vector partition i a way of writing a vector a a um of other vector where the order of ummand doe not matter. Define N to be the nonnegative integer. Unle otherwie noted, all vector will be of the form n=. n n, where n,... n N with at leat one nonzero n i. We define the partition function, denoted p(n), to be 0 the number of partition of vector n. By convention, for the zero vector, 0= 0., 0 we define p(0) =. If we are intereted in the number of partition of a vector that atify ome condition, we denote thi by p(n [condition]). Example 3.. The following are all of the poible partition of the vector [ ] 2 : 2 [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] , +, +, + +, +, [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] 0 0 +, + +, + +, [ ] [ ] [ ] [ ] ([ ]) [ ] 2 2 Therefore p = 9. Since the only partition of with all component even [ ] [ 2 ] [ ] ([ ] are and +, we have p all component even) =

24 3.2 Generating Function for Vector Partition Much like we did with integer, we will ue generating function to keep track of the number of partition of a vector that atify ome condition. For example, uppoe we want to illutrate all poible partition with ditinct vector from the et {[ ] [ ] [ ]} [ ] 0 n S =,,. Let n= and define p 0 n S (n) = p(n ditinct vector in S). ([ ]) 2 0 By convention p S =. The following value can be calculated by writing out ([ 0 ]) ([ ]) ([ ]) ([ ]) 0 the partition: p S =, p S =, p 0 S = 2, p S =, ([ ]) ([ ]) p S =, p S =, otherwie p 2 S (n) = 0. Therefore, we have ps (n)x n y n 2 = + x 0 y + x y 0 + 2x y + x y 2 + x 2 y + x 2 y 2 = + x 0 y + x y 0 + (x y 0 )(x 0 y ) + x y + (x y )(x 0 y ) + (x y )(x y 0 ) + (x y )(x y 0 )(x 0 y ) [ ] = + x 0 y + x y 0 + (x y 0 )(x 0 y ) ][ + x y =( + x 0 y )( + x y 0 )( + x y ). Note, the exponent in the polynomial tarting on the econd line illutrate all poible {[ ] [ ] [ ]} 0 partition uing ditinct vector from S =,,, and the coefficient of 0 [ ] x n y n n 2 in the firt line give the number of uch partition of. Thu, the generating function for partition into ditinct vector from S i given by n 2 where k= [ k k 2 ]. p(n ditinct vector in S)x n y n 2 = ( + x k y k 2 ), (9) 23 k S

25 If we allow our et S to be an arbitrary et of vector of the form k=., we can extend the generating function given in (9) to the following: k k p(n ditinct vector in S)x n x n x n = ( + x k x k x k ). (0) k S Now uppoe that we wih to allow each vector to repeat up to a certain amount of {[ ] [ ]} 2 4 time. For example, ay S =,, and we wih to allow vector to repeat up 2 to two time. Define p S (n; 2) = p(n vector in S, none repeated more than 2 time) [ ] ([ ] ) n 0 where n=. By convention, p n S ; 2 =. The following value can be 2 0 ([ ] ) ([ ] ) 2 4 calculated by writing out the partition: p S ; 2 =, p S ; 2 = 2, ([ ] ) ([ ] ) ([ ] ) ([ 2 ] ) p S ; 2 =, p 3 S ; 2 = 2, p 4 S ; 2 =, p 5 S ; 2 =, 6 p S (n; 2) = 0 otherwie. Therefore, we have ps (n; 2)x n y n 2 = + x 2 y + 2x 4 y 2 + x 6 y 3 + 2x 8 y 4 + x 0 y 5 + x 2 y 6 = + x 2 y + (x 2 y )(x 2 y ) + x 4 y 2 + (x 4 y 2 )(x 2 y ) + (x 4 y 2 )(x 2 y )(x 2 y ) + (x 4 y 2 )(x 4 y 2 ) + (x 4 y 2 )(x 4 y 2 )(x 2 y ) + (x 4 y 2 )(x 4 y 2 )(x 2 y )(x 2 y ) [ ][ ] + x 2 y + (x 2 y )(x 2 y ) + x 4 y 2 + (x 4 y 2 )(x 4 y 2 ). Thu, the generating function for partition into vector from S = {[ ] [ ]} 2 4, where 2 24

26 each vector i allowed to repeat up to two time i given by: p(n vector in S, none repeated more than two time )x n y n 2 = k S [ ] + x k y k 2 + (x k y k 2 )(x k y k 2 ) () where k= [ k k 2 ( + x k y k 2 + x 2k y 2k 2 ), k S ]. Suppoe we allow vector to repeat up to d time. If we allow our et S to be an k arbitrary et of vector of the form k=., we can extend the generating function given in () to the following: k p(n vector in S, none repeated more than d time )x n x n x n = k S( + x k x k x k + x 2k x 2k x 2k + + x dk x dk x dk ). (2) Now uppoe we allow part to repeat arbitrarily many time. We can require x x 2... x < ince we only ue x, x 2,..., x to keep track of the partition and not for their value. Then, by uing the formula of a geometric erie, we obtain the following generating function: p(n vector in S)x n x n x n = k S( + x k x k x k + x 2k x 2k x 2k + x 3k x 3k x 3k +... ) = k S[ + x k x k x k + (x k x k x k ) 2 + (x k x k x k ) ] (3) = k S x k x k x k. 25

27 If we take the generating function given in (3) and let S be the et of all vector of k the form k=., we obtain the tandard generating function for vector partition a follow: k p(n)x n x n x n = k i 0 x k x k x k. (4) The preceding generating function are ueful for proving certain reult related to vector partition. 3.3 Vector Partition Identitie A vector partition identity exit when every vector ha the ame amount of partition of one type a another type. One uch identity ay that every vector ha the ame amount of partition into ditinct vector a into vector with at leat one component odd. Thi parallel Euler Theorem for integer partition and wa proven by Cheema[3]. Table 5 give one example of thi identity. The Sage code in Figure 3 [ ] x and Figure 4 can be ued to verify thi identity for other vector of the form n =. Thi code can eaily be extended to vector with more than two component. x 2 26

28 def atleatoneodd(x,y): E=[] L=VectorPartition([x,y]) for p in L: if all(v[0]%2!=0 or v[]%2!=0 for v in p): E.append(p) return(e) Figure 3: Sage code for the number of partition with at leat one odd component in each vector. def ditinct(x,y): L=VectorPartition([x,y]) i=0 for p in L: l=lit(p) if all(l.count(part)<2 for part in l): i+= return i Figure 4: Sage code for the number of partition into ditinct part. 27

29 Table 5: Example illutrating Theorem 3.2. n Ditinct Vector Vector With At Leat One Component Odd [ ] 2 [ ] 2 [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] Theorem 3.2 (Cheema Theorem [3]). The number of partition of n=. n n into vector with at leat one component odd i equal to the number of partition of n into ditinct part. Proof. By the generating function given in (0), we have p(n ditinct vector )x n x n x n = ( + x k x k x k ) k i 0 = k i 0 = k i 0 (x k x k x k ) 2 x k x k x k x k x k x k where at leat one k i i odd. Thi prove the reult, ince by the generating function 28,

30 given in (3) we have p(n vector with at leat one component odd)x n x n x n = k i 0 x k x k x k where at leat one k i i odd., A generalization of Cheema Theorem give u another vector partition identity. Every vector ha the ame amount of partition in which there are le than r copie of each vector a partition into vector in which at leat one component i not diviible by r. Note, if r = 2, then thi i equivalent to Cheema Theorem. Thi parallel Theorem 2.4. The Sage code in Figure 5 and Figure 6 can be ued to verify thi [ ] x identity for vector of the form n =. It can alo be extended to vector with x 2 more than two component. Although, there are too many partition of each type to lit, Table 6 give ome value of thee Sage function. 29

31 def notdiviiblebyr(x,y,r): E=[] L=VectorPartition([x,y]) for p in L: if all(v[0]%r!=0 or v[]%r!=0 for v in p): E.append(p) return(e) Figure 5: Sage code for the number of partition with at leat one component of each vector not diviible by r. def lethanrcopie(x,y,r): L=VectorPartition([x,y]) i=0 for p in L: l=lit(p) if all(l.count(part)<r for part in l): i+= return i Figure 6: Sage code for the number of partition in which there are le than r copie of each vector. 30

32 Table 6: Example calculated by Sage illutrating Theorem 3.3. n,r Le Than r Copie At Leat One Component Not Diviible by r [ ] 3, r = [ ] 4, r = [ ] 4, r = [ ] 5, r = Theorem 3.3. The number of partition of n=. n n into vector with at leat one component not diviible by r i equal to the number of partition of n in which there are le than r copie of each part. 3

33 Proof. By the generating function given in (2), we have p(n le than r copie of each vector)x n x n x n = k i 0( + x k x k x k + x 2k x 2k x 2k + + x (r )k x (r )k x (r )k ) = k i 0( + x k x k x k + (x k x k x k ) (x k x k x k ) r ) = k i 0 = k i 0 (x k x k x k ) r x k x k x k x k x k x k, where at leat one k i i not diviible by r. Thi prove the reult, ince by the generating function given in (3) we have p(n vector with at leat one component not diviible by r)x n x n x n = k i 0 x k x k x k where at leat one k i i not diviible by r., Another partition identity give u that every vector ha the ame amount of partition into vector where every component i even a partition in which each vector i repeated an even number of time. Thi can alo be generalized to partition into vector where every component i a multiple of r and partition in which each vector i repeated a multiple of r time. Thi identity parallel Theorem 2.5. Table 7 give an example of thi identity for the even cae. The Sage code in Figure 7 and Figure 8 can be ued to verify thi identity for other vector and r value. 32

34 def componentmultipleofr(x,y,r): E=[] L=VectorPartition([x,y]) for p in L: if all(v[0]%r==0 and v[]%r==0 for v in p): E.append(p) return(e) Figure 7: Sage code for the number of partition in which every component i a multiple of r. def repeatedmultipleofr(x,y,r): L=VectorPartition([x,y]) i=0 for p in L: l=lit(p) if all(l.count(part)%r==0 for part in l): i+= return i Figure 8: Sage code for the number of partition in which each vector i repeated a multiple of r time. 33

35 Table 7: Example illutrating Theorem 3.4 for r=2. n Even Component Vector Repeated Even Number of Time [ ] 4 [ ] 4 [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] 2 [ ] 0 + [ ] + 0 [ ] 0 [ ] 0 [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] [ ] Theorem 3.4. The number of partition of n=. n n into vector with all component a multiple of r i equal to the number of partition of n in which each part i repeated a multiple of r time. Proof. By the generating function given in (2) we have p(n vector repeated a multiple of r time )x n x n x n = k i 0( + x rk x rk x rk + x 2rk x 2rk x 2rk +... ) = k i 0( + x rk x rk x rk + (x rk x rk x rk ) ) = k i 0 x rk x rk x rk = p(n vector with all component a multiple of r)x n x n x n, 34

36 where the lat line follow from the generating function given in (3). Thi prove the reult. The partition in the tatement of the next theorem parallel the partition from the econd half of the proof of Theorem 2.6. The firt half of the proof ha not yet been proven for vector. In order to complete the proof, one would need to determine how to meaure the ize of vector and a well defined notion of conjugation for vector. Theorem 3.5. Let A r (n) denote the number of partition of n=. n n in which any vector appearing an odd number of time appear at leat 2r+ time. Let B r (n) be the number of partition of n into vector of one of the following form: all component are even, or at leat one component i congruent to 2r + (mod 4r + 2) with all other component congruent to 0 (mod 4r + 2). Then A r (n) = B r (n). Proof. By the generating function given in (2), we have A r (n)x n x n x n = k i 0[ + (x k... x k ) (x k... x k ) 2r + (x k... x k ) 2r+ + (x k... x k ) 2r ] = k i 0[ + x 2k... x 2k + + (x 2k... x 2k ) r + (x k... x k ) 2r+ + (x k... x k ) 2r ] = k i 0 = k i 0 = k i 0 (x 2k... x 2k ) r+ x 2k... x 2k (x k... x k ) 2r+2 x 2k... x 2k + (x k... x k x 2k... x 2k ) 2r+. + (xk... x k ) 2r+ x k... x k + (xk... x k ) 2r+ + (x k... x k ) 2r+2 x 2k... x 2k 35

37 Note by Cheema Theorem (Theorem 3.2), we have k i 0 ( + (x k... x k ) 2r+ ) = k i 0 (x k... x k ), 2r+ where at leat one k i i odd. That i, k i 0 ( + (xk... x k ) 2r+ ) repreent vector with at leat one component of the form (2j + )(2r + ) = (4r + 2)j + (2r + ), and other component are of the form 2j(2r + ) = (4r + 2)j. Thu, k i 0 Thi prove the reult. + (x k... x k ) 2r+ x 2k... x 2k = B r (n)x n x n x n The Sage code in Figure 9 and Figure 0 can be ued to verify the partition [ ] n identity in Theorem 3.5 for vector of the form n=. def a(x,y,r): L=VectorPartition([x,y]) i=0 for p in L: l=lit(p) if all((l.count(part)%2==0) or (l.count(part)>=(2r+)) for part in l): i+= return i n 2 Figure 9: Sage code for partition enumerated by A r (n). 36

38 def b(x,y,r): E=[] L=VectorPartition([x,y]) for p in L: if all((v[0]%2==0 and v[]%2==0) or (v[0]%(4r+2)==(2r+) and v[]%(4r+2)==(2r+)) or (v[0]%(4r+2)==(2r+) and v[]%(4r+2)==0) or (v[0]%(4r+2)==0 and v[]%(4r+2)==(2r+)) for v in p): E.append(p) return(e) Figure 0: Sage code for partition enumerated by B r (n). 37

39 4 FUTURE WORK There i much more reearch that can be done in the area of vector partition. One poibility for future work would be to prove identitie imilar to the one in the previou chapter. Many integer partition identitie are known. Thee may be able to be expanded to vector uing generating erie argument like the one in Chapter 3. Another poibility for future work would be to attempt to come up with a welldefined notion of conjugation of vector partition in order to complete Theorem 3.5 to parallel Andrew theorem given in Theorem 2.6. Conjugation of vector partition could alo lead to the dicovery of other vector partition identitie. Conjugation of integer partition give u a way to prove identitie involving the ize of the part or the ize of the partition. If one could come up with a way to conjugate vector partition, thi could lead to identitie parallel to thoe of integer partition. 38

40 BIBLIOGRAPHY [] George E. Andrew. A Generalization of a Partition Theorem of MacMahon. Journal of Combinatorial Theory 3:00-0, 967. [2] George E. Andrew and Kimmo Erikkon. Integer Partition. Cambridge Univerity Pre, Cambridge, New York, [3] M.S. Cheema. Vector Partition and Combinatorial Identitie. Mathematic of Computation 8(87):44-420, 963. [4] P.A. MacMahon, Combinatory Analyi, Vol. 2. Cambridge Univerity Pre, Cambridge, New York, 96. [5] Ken Ono. The Web of Modularity: Arithmetic of the Coefficient of Modular Form and q-erie. American Mathematical Society, [6] D. Singh, A.M. Ibrahim, J.N. Singh, S.M. Ladan. Integer Partition: An Overview. Journal of Mathematical Science and Mathematic Education 7(2):9-3, 202. [7] William A. Stein et al. Sage Mathematic Software (Verion 8.), The Sage Development Team, 208, [8] Herbert S. Wilf. generatingfunctionology. AK Peter/CRC Pre,

41 VITA JENNIFER FRENCH Education: Profeional Experience: B.S. Mathematic, Univerity of Virginia College at Wie, Wie, Virginia, 2006 M.S. Mathematical Science, Eat Tenneee State Univerity, Johnon City, Tenneee, May 208 Teacher, Gate City High School and Middle School, Gate City, Virginia, Graduate Aitant, Eat Tenneee State Univerity Johnon City, Tenneee,