# research in integer partitions: alive and well

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Research in Integer Partitions: Alive and Well. James Sellers Associate Professor and Director, Undergraduate Mathematics Penn State University. Opening Thoughts. Thanks for this opportunity to speak. It is a true privilege to do so. Who Is This Guy?. - PowerPoint PPT PresentationTRANSCRIPT

Research in Integer Partitions: Research in Integer Partitions: Alive and Well Alive and Well

James SellersJames SellersAssociate Professor and Director, Associate Professor and Director,

Undergraduate MathematicsUndergraduate Mathematics

Penn State UniversityPenn State University

Opening ThoughtsOpening Thoughts

Thanks for this opportunity to speak. Thanks for this opportunity to speak.

It is a true privilege to do so. It is a true privilege to do so.

Who Is This Guy?Who Is This Guy?

Professionally: Director of UG Studies and Professionally: Director of UG Studies and Associate Professor at PSU - administrator, Associate Professor at PSU - administrator, teacher, researcher, writer, …teacher, researcher, writer, …

Personally: Husband, father of five, assistant Personally: Husband, father of five, assistant football and baseball coach, webmaster for football and baseball coach, webmaster for four different organizations, Sunday School four different organizations, Sunday School teacher, …teacher, …

Who Is This Guy?Who Is This Guy?

Goals for the TalkGoals for the Talk

Share some important definitions and Share some important definitions and examples related to partitionsexamples related to partitions

Discuss the early study of integer Discuss the early study of integer partitions (Leonhard Euler)partitions (Leonhard Euler)

Share some 21Share some 21stst century research which century research which generalizes Euler’s early resultsgeneralizes Euler’s early results

Basic DefinitionsBasic Definitions

A A partitionpartition of an integer of an integer nn is a non- is a non-increasing sequence of positive integers increasing sequence of positive integers which sum to which sum to nn. .

Each integer used in a partition is Each integer used in a partition is known as a known as a partpart. .

ExampleExample

The partitions of The partitions of n n = 5= 5 are: are:

55

4+14+1

3+23+2

3+1+13+1+1

2+2+12+2+1

2+1+1+12+1+1+1

1+1+1+1+11+1+1+1+1

NotationNotation

Let Let pp((nn)) be the number of partitions of be the number of partitions of nn. .

pp(3) = 3(3) = 3

pp(5) = 7(5) = 7

pp(50) = 204,226(50) = 204,226

Historical MotivationHistorical Motivation

In 1740, Philippe Naude sent a letter to In 1740, Philippe Naude sent a letter to Leonhard Euler asking the following:Leonhard Euler asking the following:

““HHow many ways can the number 50 be written as a sum of seven different positive integers?”

What an impact Naude’s question had on What an impact Naude’s question had on mathematics!mathematics!

Leonhard EulerLeonhard Euler

Born April 15, 1707 in Born April 15, 1707 in Basel, Switzerland Basel, Switzerland

Died September 18, 1783 in Died September 18, 1783 in St. Petersburg, St. Petersburg, RussiaRussia

One of the most prolific One of the most prolific mathematicians of the 18mathematicians of the 18thth century century

Leonhard EulerLeonhard Euler

Devout Protestant (father Paul was a Devout Protestant (father Paul was a Protestant minister)Protestant minister)

Married Katharina Gsell in 1734Married Katharina Gsell in 1734

Had 13 children; only 5 survived infancyHad 13 children; only 5 survived infancy

Claimed that he made some of his greatest Claimed that he made some of his greatest mathematical discoveries while holding a mathematical discoveries while holding a baby in his arms with other children playing baby in his arms with other children playing round his feet round his feet

Leonhard EulerLeonhard Euler

Known for numerous mathematical Known for numerous mathematical discoveries!discoveries!

– Solution of the Basel ProblemSolution of the Basel Problem– Even perfect numbers (converse of Even perfect numbers (converse of

Euclid’s result)Euclid’s result)– Formula for polyhedra: Formula for polyhedra: vv – – e e + + f f = 2= 2

– Konigsberg Bridge Problem Konigsberg Bridge Problem (beginnings of graph theory)(beginnings of graph theory)

– Integer partitionsInteger partitions– Much, much more! Much, much more!

An Advertising BreakAn Advertising Break

Read William Dunham’s Read William Dunham’s Euler: The Euler: The Master of Us All Master of Us All ! (MAA, 1999)! (MAA, 1999)

Generating FunctionsGenerating Functions

A A generating functiongenerating function for the sequence for the sequence ss((nn)) is is given by given by

SS((qq) = ) = ss(0) + (0) + ss(1)(1)qq + + ss(2)(2)qq22 + + ss(3)(3)qq33 + … + …

As a result of answering Naude’s question, As a result of answering Naude’s question, Euler found a very natural infinite product Euler found a very natural infinite product which served as the generating function for the which served as the generating function for the partition function partition function pp((nn))..

Generating FunctionsGenerating Functions

Consider the following: Let Consider the following: Let

PP((qq) = ) =

(1 + (1 + qq + + qq22 + + qq33 + + qq44 + + qq5 5 + …) + …) x x

(1 + (1 + qq22 + + qq44 + + qq66 + + qq88 + + qq1010 + …) + …) xx

(1 + (1 + qq33 + + qq66 + + qq99 + + qq1212 + + qq1515 + …) + …) xx

(1 + (1 + qq44 + + qq88 + + qq1212 + + qq1616 + + qq2020 + …) + …) xx

(1 + (1 + qq55 + + qq1010 + + qq1515 + + qq2020 + + qq2525 + …) + …) xx

……

Generating FunctionsGenerating Functions

This is, of course, the same asThis is, of course, the same as

PP((qq) = ) =

(1 + (1 + qq11 + + qq1+11+1 + + qq1+1+11+1+1 + + qq1+1+1+1 1+1+1+1 + …) + …) x x

(1 + (1 + qq22 + + qq2+22+2 + + qq2+2+22+2+2 + + qq2+2+2+22+2+2+2 + …) + …) xx

(1 + (1 + qq33 + + qq3+33+3 + + qq3+3+33+3+3 + + qq3+3+3+33+3+3+3 + …) + …) xx

(1 + (1 + qq44 + + qq4+44+4 + + qq4+4+44+4+4 + + qq4+4+4+44+4+4+4 + …) + …) xx

(1 + (1 + qq55 + + qq5+55+5 + + qq5+5+55+5+5 + + qq5+5+5+55+5+5+5 + …) + …) xx

……

Generating FunctionsGenerating Functions

In fact, Euler proved that In fact, Euler proved that PP((qq)) is the generating is the generating function for the partition function function for the partition function pp((nn).).

Thanks to the geometric series, note that Thanks to the geometric series, note that PP((qq)) can also be written ascan also be written as

2 3 4

1 1 1 1( ) ...

1 1 1 1P q

q q q q

A Historical SidenoteA Historical Sidenote

Euler considered the reciprocal of Euler considered the reciprocal of PP((qq) ) and proved that it equals a very nice and proved that it equals a very nice power series. This result became known power series. This result became known as Euler’s Pentagonal Number Theorem. as Euler’s Pentagonal Number Theorem.

2 3 4

1 1 1 1( ) ...

1 1 1 1P q

q q q q

A Historical SidenoteA Historical Sidenote

Euler’s Pentagonal Number Theorem Euler’s Pentagonal Number Theorem leads to a very nice recurrence for leads to a very nice recurrence for pp((nn)). .

It was this recurrence which led English It was this recurrence which led English mathematician Major Percy MacMahon mathematician Major Percy MacMahon (1854 – 1929) to write down a table of (1854 – 1929) to write down a table of values of values of pp((nn)) for for nn from from 11 to to 100100. .

A Historical SidenoteA Historical Sidenote

And it was MacMahon’s table, broken up And it was MacMahon’s table, broken up into groups of five values each, which led into groups of five values each, which led Srinivasa Ramanujan (1887 – 1920) to his Srinivasa Ramanujan (1887 – 1920) to his discoveries about congruences for discoveries about congruences for pp((nn).).

Back to Naude’s QuestionBack to Naude’s Question

Remember that Naude asked Euler Remember that Naude asked Euler about the number of partitions of about the number of partitions of nn where the parts are where the parts are distinctdistinct. How . How does that change the generating does that change the generating function in question?function in question?

Back to Naude’s QuestionBack to Naude’s Question

Well, instead of this…Well, instead of this…

PP((qq) = ) =

(1 + (1 + qq + + qq22 + + qq33 + + qq44 + + qq5 5 + …) + …) x x

(1 + (1 + qq22 + + qq44 + + qq66 + + qq88 + + qq1010 + …) + …) xx

(1 + (1 + qq33 + + qq66 + + qq99 + + qq1212 + + qq1515 + …) + …) xx

(1 + (1 + qq44 + + qq88 + + qq1212 + + qq1616 + + qq2020 + …) + …) xx

(1 + (1 + qq55 + + qq1010 + + qq1515 + + qq2020 + + qq2525 + …) + …) xx

……

we now want this…we now want this…

Back to Naude’s QuestionBack to Naude’s Question

DD((qq) = (1 + ) = (1 + qq )(1 + )(1 + qq22)(1 + )(1 + qq33)(1 + )(1 + qq44) …) …

since “distinct parts” means each part can since “distinct parts” means each part can appear at most one time.appear at most one time.

Back to Naude’s QuestionBack to Naude’s Question

To Euler, this would have been a very To Euler, this would have been a very satisfactory result given his excellent satisfactory result given his excellent calculation skills.calculation skills.

Even without Euler’s excellent calculation Even without Euler’s excellent calculation skills, this is indeed a very satisfactory skills, this is indeed a very satisfactory result today thanks to the advent of result today thanks to the advent of computer algebra systems.computer algebra systems.

Back to Naude’s QuestionBack to Naude’s Question

So, for example, we have:So, for example, we have:

dd(3) = 2(3) = 2

dd(5) = 3(5) = 3

dd(50) = 3658(50) = 3658

Back to Naude’s QuestionBack to Naude’s Question

One last comment on Naude’s question is One last comment on Naude’s question is in order. Naude asked: in order. Naude asked:

““HHow many ways can the number 50 be written as a sum of seven different positive integers?”

The answer given by Euler is 522. The answer given by Euler is 522.

Back to Naude’s QuestionBack to Naude’s Question

Here is a “snapshot” of the opening Here is a “snapshot” of the opening portion of one of Euler’s papers on portion of one of Euler’s papers on partitions.partitions.

This paper was originally published in This paper was originally published in 1753. 1753.

Euler’s Famous DiscoveryEuler’s Famous Discovery

Euler did not choose to be satisfied with Euler did not choose to be satisfied with solving Naude’s question. solving Naude’s question.

He considered how he could manipulate He considered how he could manipulate

DD((qq) = (1 + ) = (1 + qq )(1 + )(1 + qq22)(1 + )(1 + qq33)(1 + )(1 + qq44) …) …

and prove additional partitions results. and prove additional partitions results.

Euler’s Famous DiscoveryEuler’s Famous Discovery

Euler noted the following: Euler noted the following:

That meant that Euler could rewrite That meant that Euler could rewrite DD((qq)) as as

q

1

11

2

Euler’s Famous DiscoveryEuler’s Famous Discovery

DD((qq) = (1 + ) = (1 + qq )(1 + )(1 + qq22)(1 + )(1 + qq33)(1 + )(1 + qq44) …) …

Notice that all the numerators cancel with Notice that all the numerators cancel with some of the denominators, leaving only some of the denominators, leaving only those denominators with odd exponents! those denominators with odd exponents!

4

8

3

6

2

42

1

1

1

1

1

1

1

1

q

q

q

q

q

q

q

q

Euler’s Famous DiscoveryEuler’s Famous Discovery

That means we haveThat means we have

Euler realized that this way of writing Euler realized that this way of writing DD((qq)) had a never-before-seen partition-theoretic had a never-before-seen partition-theoretic interpretation! interpretation!

753 1

1

1

1

1

1

1

1)(

qqqqqD

Euler’s Famous DiscoveryEuler’s Famous Discovery

Theorem: For all positive integers Theorem: For all positive integers n,n, the the number of partitions of number of partitions of nn into distinct parts into distinct parts equals the number of partitions of equals the number of partitions of nn into into odd parts. odd parts.

Proof: The generating functions for the two Proof: The generating functions for the two partition functions are the same.partition functions are the same.

Euler’s Famous DiscoveryEuler’s Famous Discovery

Example: Example: n n = 7= 7

Distinct parts:Distinct parts:

77

6+16+1

5+25+2

4+34+3

4+2+14+2+1

Odd parts:Odd parts:

77

5+1+15+1+1

3+3+13+3+1

3+1+1+1+13+1+1+1+1

1+1+1+1+1+1+11+1+1+1+1+1+1

Extension of Euler’s TheoremExtension of Euler’s Theorem

To close this talk, I want to show you To close this talk, I want to show you two different sets of results related to two different sets of results related to Euler’s famous theorem. Euler’s famous theorem.

These were both recently published These were both recently published (within the last few years).(within the last few years).

Extension of Euler’s TheoremExtension of Euler’s Theorem

Theorem: (Guy, 1958)Theorem: (Guy, 1958)

The number of partitions of The number of partitions of nn into distinct parts with no into distinct parts with no powers of 2 as parts equals the powers of 2 as parts equals the number of partitions of number of partitions of nn into into odd parts with no 1s as parts. odd parts with no 1s as parts.

Extension of Euler’s TheoremExtension of Euler’s Theorem

Theorem: (S, Sills, Mullen – EJC, 2004)Theorem: (S, Sills, Mullen – EJC, 2004)

Let J be a set of non-multiples of m.

Let p1(n; m, J) be the number of partitions of n with no part of the form mkj where j is an element of J and where no part is allowed to appear more than m - 1 times in any partition.

Let p2(n; m, J) be the number of partitions of n with no part divisible by m and no part equal to j for each element j of J.

Then, for all n, p1(n; m, J) = p2(n; m, J).

Extension of Euler’s TheoremExtension of Euler’s Theorem

How does this fit with Euler and Guy? How does this fit with Euler and Guy?

Guy’s result is Guy’s result is p1(n; 2, {1}) = p2(n; 2, {1}).

Euler’s result is p1(n; 2, {}) = p2(n; 2, {}).

Extension of Euler’s TheoremExtension of Euler’s Theorem

Nifty corollary:Nifty corollary:

The number of partitions of n into distinct parts where no part is the product of an odd prime and a power of 2 equals the number of partitions of n using only 1s and odd composites as parts.

This is just This is just p1(n; 2, J) = p2(n; 2, J) where J is the set of odd primes.

Extension of Euler’s TheoremExtension of Euler’s Theorem

Theorem: (Plinio-Santos, 2001)Theorem: (Plinio-Santos, 2001)

The number of partitions of The number of partitions of nn of the of the form form pp1 1 + p+ p2 2 + p+ p3 3 + p+ p4 4 + …+ … such that such that

equals the number of partitions of equals the number of partitions of nn into odd parts.into odd parts.

4321 2 pppp

Extension of Euler’s TheoremExtension of Euler’s Theorem

Theorem: (S Theorem: (S –– Integers, 2003) Integers, 2003)

The number of partitions of The number of partitions of nn of the form of the form pp1 1 + p+ p2 2 + + pp3 3 + p+ p4 4 + …+ … such that such that

equals the number of partitions of equals the number of partitions of nn into into 11s or s or numbers of the form numbers of the form (k(k22+1)+ (k+1)+ (k33+1) +…+ (k+1) +…+ (kmm+1)+1) for for some some mm. .

4433221 pkpkpkp

Extension of Euler’s TheoremExtension of Euler’s Theorem

Plinio-Santos’ result is the special Plinio-Santos’ result is the special case case kk22 = 2, = 2, kk33 = = kk44 = … = 1 = … = 1. .

The proof technique in my paper The proof technique in my paper involves MacMahon’s Partition involves MacMahon’s Partition Analysis, made popular in recent years Analysis, made popular in recent years by Andrews, Paule, and Riese. by Andrews, Paule, and Riese.

One More Advertising Break!One More Advertising Break!

Conference on Undergraduate Research in Conference on Undergraduate Research in Mathematics (CURM)Mathematics (CURM)

Penn State UniversityPenn State University

November 9-10, 2007November 9-10, 2007

www.math.psu.edu/ug/curm/www.math.psu.edu/ug/curm/

Closing ThoughtsClosing Thoughts

Thanks again for the opportunity to Thanks again for the opportunity to share!share!

I would be happy to answer any I would be happy to answer any questions you might have. questions you might have.