Macaulay Expansion

Given natural numbers n and r , the “greedy” algorithm enables us to obtain an expansion of the integer n as a sum of binomial coefficients in the form (ar r ) + (ar−1 r−1 ) + · · · + (a1 1 ) . We give an alternate interpretation of this expansion, which also proves its uniqueness in an interesting manner. The 1996 Iranian mathematical olympiad competition contained the following problem. For natural numbers n and r, there is a unique expansion n = ( ar r ) + ( ar−1 r − 1 ) + · · · + ( a1 1 ) with each ai an integer and ar > ar−1 > · · · > a1 ≥ 0. The existence is fairly easy to prove using the “greedy” algorithm. This expansion is sometimes known as the Macaulay expansion. However, the following alternate interpretation does not seem to be well known; it gives uniqueness in an interesting manner. In what follows, the following well-known convention is used: the binomial coefficient (n r ) is equated to 0 if n < r . For each natural number r , denote by Sr the set of all r -digit numbers in some base b whose digits are in strictly decreasing order of size. Evidently, Sr is nonempty if and only if b ≥ r ; in this case, Sr has (b r ) elements. Let us now write the elements of Sr in increasing order. For instance, in base 10, the first few of the 120 members of S3 are: (2, 1, 0), (3, 1, 0), (3, 2, 0), (3, 2, 1), (4, 1, 0), (4, 2, 0), (4, 2, 1), (4, 3, 0), . . . . We will prove the following. Theorem. Given any positive integer n, and any base b such that (b r ) > n, the (n + 1)th member of Sr is (ar , . . . , a2, a1), where n = (ar r ) + (ar−1 r−1 ) + · · · + (a1 1 ) . In particular, for each n, the Diophantine equation (ar r ) + (ar−1 r−1 ) + · · · + (a1 1 ) = n has a unique solution in positive integers ar > ar−1 > · · · > a1 ≥ 0. Here are a couple of examples to illustrate the theorem. (i) Let r = 3 and n = 12. We may take any base b so that (b 3 ) > 12. For example, b = 6 is allowed because (6 3 ) = 20. Among the 20 members in S3, the 13th member is (5, 2, 1). Note that ( 5 3 ) + ( 2 2 ) + ( 1 1 ) = 12. http://dx.doi.org/10.4169/amer.math.monthly.121.04.359 MSC: Primary 05A10 April 2014] NOTES 359 (ii) Let r = 3, n = 74. We may take b = 10 as (10 3 ) = 120. The 75th member of S3 is (8, 6, 3). Note that ( 8 3 ) + ( 6 2 ) + ( 3 2 ) = 74. Proof of theorem. First of all, we notice that the number of members in Sr that have first digit < m equals (m r ) ; this is because we are choosing r numbers from {0, 1, . . . , m − 1} and arranging them in decreasing order. Now, suppose the (n + 1)th member of Sr is (ar , ar−1, . . . , a1). The number of members of Sr with first digit < ar is (ar r ) . The number of members of Sr , whose first digit is ar and which occur before the above member, is the number of members of Sr−1 occurring prior to (ar−1, . . . , a1). Inductively, it is clear that this equals ( ar−1 r − 1 ) + · · · + ( a2 2 ) + ( a1 1 ) . Therefore, the number of members of Sr occurring prior to the (n + 1)th member above (which must be n) is( ar r ) + ( ar−1 r − 1 ) + · · · + ( a1 1 ) . This proves our result. Remark. We may proceed in a slightly different direction, if we do not use the first observation in the proof. For any k, we can obtain by induction that the number of elements in Sk starting with some a is ( a k−1 ) . Indeed, to prove this by induction, we use the identity ( n r ) = n−1 ∑