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? Professionally: Director of UG Studies and Associate Professor at PSU - administrator, teacher, researcher, writer, … Personally: Husband, father of five, assistant football and baseball coach, webmaster for four different organizations, Sunday School teacher, …

Who Is This Guy?

Goals for the Talk Share some important definitions and examples related to partitions Discuss the early study of integer partitions (Leonhard Euler) Share some 21 st century research which generalizes Euler’s early results

Basic Definitions A partition of an integer n is a non- increasing sequence of positive integers which sum to n. Each integer used in a partition is known as a part.

Example The partitions of n = 5 are:

Notation Let p(n) be the number of partitions of n. p(3) = 3 p(5) = 7 p(50) = 204,226

Historical Motivation In 1740, Philippe Naude sent a letter to Leonhard Euler asking the following: “H “How many ways can the number 50 be written as a sum of seven different positive integers?” What an impact Naude’s question had on mathematics!

Leonhard Euler Born April 15, 1707 in Basel, Switzerland Died September 18, 1783 in St. Petersburg, Russia One of the most prolific mathematicians of the 18 th century

Leonhard Euler Devout Protestant (father Paul was a Protestant minister) Married Katharina Gsell in 1734 Had 13 children; only 5 survived infancy Claimed that he made some of his greatest mathematical discoveries while holding a baby in his arms with other children playing round his feet

Leonhard Euler Known for numerous mathematical discoveries! –Solution of the Basel Problem –Even perfect numbers (converse of Euclid’s result) –Formula for polyhedra: v – e + f = 2 –Konigsberg Bridge Problem (beginnings of graph theory) –Integer partitions –Much, much more!

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

Generating Functions A generating function for the sequence s(n) is given by S(q) = s(0) + s(1)q + s(2)q 2 + s(3)q 3 + … As a result of answering Naude’s question, Euler found a very natural infinite product which served as the generating function for the partition function p(n).

Generating Functions Consider the following: Let P(q) = (1 + q + q 2 + q 3 + q 4 + q 5 + …) x (1 + q 2 + q 4 + q 6 + q 8 + q 10 + …) x (1 + q 3 + q 6 + q 9 + q 12 + q 15 + …) x (1 + q 4 + q 8 + q 12 + q 16 + q 20 + …) x (1 + q 5 + q 10 + q 15 + q 20 + q 25 + …) x …

Generating Functions This is, of course, the same as P(q) = (1 + q 1 + q q q …) x (1 + q 2 + q q q …) x (1 + q 3 + q q q …) x (1 + q 4 + q q q …) x (1 + q 5 + q q q …) x …

Generating Functions In fact, Euler proved that P(q) is the generating function for the partition function p(n). Thanks to the geometric series, note that P(q) can also be written as

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

A Historical Sidenote Euler’s Pentagonal Number Theorem leads to a very nice recurrence for p(n). It was this recurrence which led English mathematician Major Percy MacMahon (1854 – 1929) to write down a table of values of p(n) for n from 1 to 100.

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

Back to Naude’s Question Remember that Naude asked Euler about the number of partitions of n where the parts are distinct. How does that change the generating function in question?

Back to Naude’s Question Well, instead of this… P(q) = (1 + q + q 2 + q 3 + q 4 + q 5 + …) x (1 + q 2 + q 4 + q 6 + q 8 + q 10 + …) x (1 + q 3 + q 6 + q 9 + q 12 + q 15 + …) x (1 + q 4 + q 8 + q 12 + q 16 + q 20 + …) x (1 + q 5 + q 10 + q 15 + q 20 + q 25 + …) x … we now want this…

Back to Naude’s Question D(q) = (1 + q )(1 + q 2 )(1 + q 3 )(1 + q 4 ) … since “distinct parts” means each part can appear at most one time.

Back to Naude’s Question To Euler, this would have been a very satisfactory result given his excellent calculation skills. Even without Euler’s excellent calculation skills, this is indeed a very satisfactory result today thanks to the advent of computer algebra systems.

Back to Naude’s Question So, for example, we have: d(3) = 2 d(5) = 3 d(50) = 3658

Back to Naude’s Question One last comment on Naude’s question is in order. Naude asked: “H “How many ways can the number 50 be written as a sum of seven different positive integers?” The answer given by Euler is 522.

Back to Naude’s Question Here is a “snapshot” of the opening portion of one of Euler’s papers on partitions. This paper was originally published in 1753.

Euler’s Famous Discovery Euler did not choose to be satisfied with solving Naude’s question. He considered how he could manipulate D(q) = (1 + q )(1 + q 2 )(1 + q 3 )(1 + q 4 ) … D(q) = (1 + q )(1 + q 2 )(1 + q 3 )(1 + q 4 ) … and prove additional partitions results.

Euler’s Famous Discovery Euler noted the following: That meant that Euler could rewrite D(q) as

Euler’s Famous Discovery D(q) = (1 + q )(1 + q 2 )(1 + q 3 )(1 + q 4 ) … Notice that all the numerators cancel with some of the denominators, leaving only those denominators with odd exponents!

Euler’s Famous Discovery That means we have Euler realized that this way of writing D(q) had a never-before-seen partition-theoretic interpretation!

Euler’s Famous Discovery Theorem: For all positive integers n, the number of partitions of n into distinct parts equals the number of partitions of n into odd parts. Proof: The generating functions for the two partition functions are the same.

Euler’s Famous Discovery Example: n = 7 Distinct parts: Odd parts:

Extension of Euler’s Theorem To close this talk, I want to show you two different sets of results related to Euler’s famous theorem. These were both recently published (within the last few years).

Extension of Euler’s Theorem Theorem: (Guy, 1958) The number of partitions of n into distinct parts with no powers of 2 as parts equals the number of partitions of n into odd parts with no 1s as parts.

Extension of Euler’s Theorem Theorem: (S, Sills, Mullen – EJC, 2004) Let J be a set of non-multiples of m. Let p 1 (n; m, J) be the number of partitions of n with no part of the form m k j where j is an element of J and where no part is allowed to appear more than m - 1 times in any partition. Let p 2 (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, p 1 (n; m, J) = p 2 (n; m, J).

Extension of Euler’s Theorem How does this fit with Euler and Guy? Guy’s result is Guy’s result is p 1 (n; 2, {1}) = p 2 (n; 2, {1}). Euler’s result is p 1 (n; 2, {}) = p 2 (n; 2, {}).

Extension of Euler’s Theorem 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 p 1 (n; 2, J) = p 2 (n; 2, J) where J is the set of odd primes.

Extension of Euler’s Theorem Theorem: (Plinio-Santos, 2001) The number of partitions of n of the form p 1 + p 2 + p 3 + p 4 + … such that equals the number of partitions of n into odd parts.

Extension of Euler’s Theorem Theorem: (S – Integers, 2003) The number of partitions of n of the form p 1 + p 2 + p 3 + p 4 + … such that equals the number of partitions of n into 1s or numbers of the form (k 2 +1)+ (k 3 +1) +…+ (k m +1) for some m.

Extension of Euler’s Theorem Plinio-Santos’ result is the special case k 2 = 2, k 3 = k 4 = … = 1. The proof technique in my paper involves MacMahon’s Partition Analysis, made popular in recent years by Andrews, Paule, and Riese.

One More Advertising Break! Conference on Undergraduate Research in Mathematics (CURM) Penn State University November 9-10,

Closing Thoughts Thanks again for the opportunity to share! I would be happy to answer any questions you might have.