fea-ahlgren.qxp 8/21/01 2:42 PM Page 978 Addition and Counting: The Arithmetic of Partitions Scott Ahlgren and Ken Ono At first glance the stuff of partitions seems like computed the values of p(n) for all n up to 200. child’s play: He found that 4=3+1=2+2=2+1+1=1+1+1+1. p(200) = 3, 972, 999, 029, 388, Therefore, there are 5 partitions of the number 4. and he did not count the partitions one-by-one: But (as happens in number theory) the seemingly 200 = 199 + 1 = 198 + 2 simple business of counting the ways to break a =198+1+1=197+3=............... number into parts leads quickly to some difficult and beautiful problems. Partitions play important Instead, MacMahon employed classical formal roles in such diverse areas of mathematics as power series identities due to Euler. combinatorics, Lie theory, representation theory, To develop Euler’s recurrence, we begin with the elementary fact that if |x| < 1, then mathematical physics, and the theory of special functions, but we shall concentrate here on 1 =1+x + x2 + x3 + x4 + ... their role in number theory (for which [A] is the 1 − x standard reference). Using this, Euler noticed that when we expand the In the Beginning, There Was Euler… infinite product ∞ 1 A partition of the natural number n is any nonin- =(1+x + x2 + x3 + ...) 1 − xn creasing sequence of natural numbers whose sum n=1 is n (by convention, we agree that p(0) = 1). The × (1 + x2 + x4 + ...) number of partitions of n is denoted by p(n). Eighty × (1 + x3 + x6 + ...) ..., years ago Percy Alexander MacMahon, a major in the British Royal Artillery and a master calculator, the coefficient of xn is equal to p(n) (think of the first factor as counting the number of 1’s in a par- Scott Ahlgren is assistant professor of mathematics at tition, the second as counting the number of 2’s, the University of Illinois, Urbana-Champaign. His e-mail and so on). In other words, we have the generat- address is [email protected]. ing function Ken Ono is professor of mathematics at the University of ∞ ∞ n 1 Wisconsin-Madison. His e-mail address is ono@math. p(n)x = 1 − xn wisc.edu. n=0 n=1 Both authors thank the National Science Foundation for =1+x +2x2 +3x3 +5x4 + ... its support. The second author thanks the Alfred P. Sloan Foundation, the David and Lucile Packard Foundation, and Moreover, Euler observed that the reciprocal of the Number Theory Foundation for their support. this infinite product satisfies a beautiful identity 978 NOTICES OF THE AMS VOLUME 48, NUMBER 9 fea-ahlgren.qxp 8/21/01 2:42 PM Page 979 (also known as Euler’s Pentagonal Number Theo- Rademacher’s series converges is remarkable; for rem): example, the first eight terms give the approxi- ∞ ∞ mation − n − k (3k2+k)/2 (1 x )= ( 1) x p(200) ≈ 3, 972, 999, 029, 388.004 n=1 k=−∞ =1− x − x2 + x5 + x7 − x12 − ... (compare with the exact value computed by MacMahon). These two identities show that To implement the circle method requires a ∞ detailed study of the analytic behavior of the p(n)xn generating function for p(n). Recall that we have n=0 ∞ n 1 × 1 − x − x2 + x5 + x7 − x12 − ... =1, F(x):= p(n)x = . (1 − x)(1 − x2)(1 − x3) ... n=0 which in turn implies, for positive integers n, that This is an analytic function on the domain |x| < 1. − − p(n)=p(n 1) + p(n 2) A natural starting point is Cauchy’s Theorem, − − − − − p(n 5) p(n 7) + p(n 12) + ... which gives 1 F(x) This recurrence enabled MacMahon to perform his p(n)= n+1 dx, massive calculation. 2πi C x Hardy-Ramanujan-Rademacher where C is any simple closed counterclockwise contour around the origin. One would hope to Asymptotic Formula for p(n) adjust the contour in relation to the singularities p(n) It is natural to ask about the size of . The an- of F(x) in order to obtain as much information as swer to this question is given by a remarkable as- possible about the integral. But consider for a mo- ymptotic formula, discovered by G. H. Hardy and ment these singularities; they occur at every root Ramanujan in 1917 and perfected by Hans of unity, forming an impenetrable barrier on the Rademacher two decades later. This formula is so unit circle. In our favor, however, it can be shown accurate that it can actually be used to compute that the size of F(x) near a primitive q-th root of individual values of p(n); Hardy called it “one of unity diminishes rapidly as q increases; moreover the rare formulae which are both asymptotic and the behavior of F(x) near each root of unity can be exact.” It stands out further in importance since described with precision. Indeed, with an appro- it marks the birth of the circle method, which has priate choice of C, the contribution to the integral grown into one of the most powerful tools in an- from all of the primitive q-th roots of unity can be alytic number theory. calculated quite precisely. The main contribution Here we introduce Rademacher’s result. He de- is the function T (n); a detailed analysis of the fined explicit functions T (n) such that for all n we q q errors involved yields the complete formula. have ∞ The circle method has been of extraordinary im- p(n)= Tq(n). portance over the last eighty years. It has played q=1 a fundamental role in additive number theory (in Waring type problems, for instance), analysis, and The functions T (n) are too complicated to write q even the computation of black hole entropies. down here, but we mention that T1(n) alone yields the asymptotic formula Ramanujan’s Congruences √ 1 After a moment’s reflection on the combinatorial p(n) ∼ √ eπ 2n/3. 4n 3 definition of the partition function, we have no particular reason to believe that it possesses any (In their original work, Hardy and Ramanujan used interesting arithmetic properties (the analytic slightly different functions in place of the Tq(n). ∞ formula of the last section certainly does nothing As a result, their analogue of the series q=1 Tq(n) to change this opinion). There is nothing, for was divergent, although still useful.) Moreover, example, which would lead us to think that Rademacher computed precisely the error incurred p(n) should exhibit a preference to be even by truncating this series after Q terms. In partic- rather than odd. A natural suspicion, therefore, ular, there exist explicit constants A and B such might be that the values of p(n) are distributed that √ evenly modulo 2. A quick computation of the An B first 10,000 values confirms this suspicion: of p(n) − Tq(n) < . n1/4 q=1 these 10,000 values, exactly 4,996 are even and 5,004 are odd. This pattern continues with Since p(n) is an integer, this determines the exact 2 replaced by 3: of the first 10,000 values, 3,313; value of p(n) for large n. The rate at which 3,325; and 3,362 (in each case almost exactly OCTOBER 2001 NOTICES OF THE AMS 979 fea-ahlgren.qxp 8/21/01 2:42 PM Page 980 Euler one-third) are congruent respectively to 0, 1, and modulo 5 and 7 are quite ingenious but are not 2 modulo 3. When we replace 3 by 5, however, terribly difficult, while the proof of the congruence something quite different happens: we discover modulo 11 is much harder). In these same that 3,611 (many more than the expected papers he sketched proofs of extensions of one-fifth) of the first 10,000 values of p(n) are these congruences. For example, we have divisible by 5. What is the explanation for this p(25n +24)≡ 0 (mod 25), aberration? ≡ The answer must have been clear to Ramanu- p(49n + 47) 0 (mod 49). jan when he saw MacMahon’s table of values of Ramanujan noticed the beginnings of other p(n). So Ramanujan would have seen something patterns in these first 200 values: like the following. p(116) ≡ 0 (mod 121),p(99) ≡ 0 (mod 125). 1123 5 From such scant evidence he made the following conjecture: 7 111522 30 If δ =5a7b11c and 24λ ≡ 1 (mod δ), 4256 77 101 135 then p(δn + λ) ≡ 0 (mod δ). 176 231 297 385 490 When δ =125, for example, we have λ =99. So 627 792 1002 1255 1575 Ramanujan’s conjecture is that 1958 2436 3010 3718 4565 p(125n + 99) ≡ 0 (mod 125). What is striking, of course, is that every entry in We note that the general conjecture follows easily the last column is a multiple of 5. This phenome- from the cases when the moduli are powers of 5, non, which persists, explains the apparent aber- 7, or 11. ration above and was the first of Ramanujan’s It is remarkable that Ramanujan was able to ground-breaking discoveries on the arithmetic of formulate a general conjecture based on such p(n). Here is his own account. little evidence and therefore unsurprising that the conjecture was not quite correct (in the 1930s I have proved a number of arithmetic Chowla and Gupta discovered the counterexample properties of p(n)…in particular that p(243) ≡ 0 (mod 73) ).