Vol. 10 No 5 - Pi Mu Epsilon

365 PI MU EPSILON JOURNALROUSSEAU, PARTITION PROBLEM 366Thus the resulting HT sequence satisfies the conditions of Problem C. It is clearthat the mapping just described is a bijection.Let a,, denote the common answer to Problems A through C. We shallproveif n = 2m,+m ( m )which agrees with the conjecture made in [I].2. The Method. Note that a function counted in Problem B satisfies f(n) = n -k + 1 if and only if the total number of T's in the corresponding HT sequence is(Rn) - 1)/2 = (n - k)/2. Equivalently, f(n) = n - k + 1 if and only if the player hasa fortune of (n - 1) - 2((n - k)/2) = k - 1 dollars after the last coin toss. Let c(n, k)denote the number cSHT sequences in Problem C so that the player's final fortuneis k - 1 dollars. Clearly, c(n, k) = 0 unless k and n have the same parity, andc(n, 2) +c(n,4) + ... + c(n,n) if n is even,c(n, 1) + c(n, 3) + .- + c(n, n) if n is odd.If the player's fortune after the last toss is k - 1 dollars, then the fortune just beforeobtaining the last head is k + 2(j - 1) for some j s 0, from which we have therecurrence relationc(n, k)=c(n- 1, k - 1) +c(n- 1, k+ 1) +- +c(n- 1, n- 1). (2)Comparing (1) and (2), we havec(n+1,1) ifn is even,c(n+1,2) if nis odd.Note that by two applications of (2),c(n, k)=c(n- 1, k- l)+c(n- 1, k+ I)+-+c(n- 1, n- 1)c(n,k+2)=c(n- l,k+ l)+c(n- l, k+3)+--+c(n-1,n- l),from which we havec(n, k) = c(n - 1, k - 1) + c(n, k + 2). (4)Thus our plan is mapped out. We want to determine the numbers c(n, k) using (4)and then find a,, using (3). Using (4) and the fact that c(n, n) = 1, we can easilycompute a table of values for c(n, k) for small values of n a la Pascal's triangle.(1)-Table 1 - Values for c(n, k)3. Solution by Generating Functions. We shall find c(n, k) using the methodof generaing functions. In part, the solution rests on the following result fromclassical analysis [3,97.32],[4, p. 1381.Lagrange's Expansion. Let f(z) and F(z) be analytic on and inside a contour Csurrounding the origin, and let w be such that lwf(z)l < lzl at all points z on C.Then the equation w = z/f(z) has one root z = z(w) in the interior of C, andF(z(w)) has the expansionAlso, the following fact is crucial.Lemma. Let Ck denote the class of all HI sequences for the game in Problem Csuch that the player is never in debt and has a fortune of k - 1 dollars after the lasttoss. Thai GI = {aHbl a e Cb b E Cl] . Here ah% denotes the sequence obtained