Vol. 10 No 5 - Pi Mu Epsilon

367 PI MU EPSILON JOURNALobtained by concatenation of a e Cb a single H, and b e C,.Proof. Given a sequence in Cw locate the last point at which the player's fortuneis k - 1 dollars, and let a denote the HT sequence up to that point. Then a e Cband the next term of the sequence must be H; otherwise the player's fortune wouldbe k - 1 dollars at some later point. Letting b denote the remainder of the givensequence, we see that, had the player started with no funds at the beginning of b,he or she would never go in debt and have a final fortune of 1 - 1 dollars. Henceb e CI. Thus every member of GI can be written aHb where a e C, and b E CI.We claim that this representation is unique. To see this, suppose for the moment1hat a'AFb' = aflb where a, a' e Ck and b, b' e Ci. We may assume that the lengthsof a' and a are p and a, respectively, with p < q. Then toss p + 1 yields H, and theplayer begins at toss p + 2 with a fortune of k dollars, The ensuing sequence, b',leads to a fortune of k - 1 dollars after q - p - 1 tosses, so had the player startedthis sequence with no funds, he or she would then be in debt. This contradicts b'e CI and proves that the representation aHb with a e Ck and b e is unique.Note that if there are r - 1 H's in a and s - 1 H's in b then aHb hasr + s - 1 H's. In view of the uniqueness of the aHb decomposition, we have thefollowing important result.Corollary. For all n, k, 1 s 1,ROUSSEAU, PARTITION PROBLEM 368Using (9, we obtainGk+kx) = Gk(x)G,(x).For simplicity, let Gl(x) = G(x). Simple induction using (6) yields. aG&) = [G(x)lk (k = 1, 2, ...). (7)Multiplying both sides of (4) by x011"10'2 and summing on n, we findGi;(x) =xGi;.,(x) + xGw(x). (8)In view of (7) and (8), G(x) satisfies the cubic equationG(x) = x + xG 3 (x), (9)sox=G/f(G)wheref(z)=~+ 1. With~(z)=z"andf(z)=z'+l,Lagrange's expansion yieldsLet [9']Gk(x) denote the cofficient of ??" in the expansion of Q;(x). To simplifythe notation, let r = (n - k)/2. Thenccn, k) = tx^lGk(x)(6)Let us introduce the following generating function for column k of thetable of c(n, k) values:Thus, for example,G,(x) = x + x 4 + 3x 7 + 1 2x1Â + 55xI3 + 2andThus we have / \n-k40) = - 2Note: Having found a fonnula for c(n, k) using generating functions andLagrange's expansion, it is easy enough to verify that this is indeed the solutionof (4) satisfying c(n, n) = 1. It is clear that (1 0) gives c(n, n) = 1. Also, assumingthat the formula holds for c(n - 1, k - 1) and c(n, k + 2) and setting r = (n - k)/2,