By Peter J. Cameron
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Extra info for Notes on counting [Lecture notes]
1) where the sum is over all sequences a1 , a2 , . . of natural numbers which satisfy ∑ kak = n. Multiplying by xn and summing over n, we get 1 = 1 − qx = ∑ qn x n n≥0 ∑ ∏ a1 ,a2 ,... k≥1 = ∏∑ k≥1 a≥0 = mk + ak − 1 kak x ak mk + a − 1 (xk )a a ∏ (1 − xk )−mk . k≥1 Here the manipulations are similar to those for the sum of cycle indices in Chapter 2; we use the fact that the number of choices of a things from a set of m, with repetition allowed and order unimportant, is m+a−1 , and in the fourth line we a invoke the Binomial Theorem with negative exponent.
Thus, ek has nk terms, and ek (1, 1, . . , 1) = n . k For example, if n = 3, the elementary symmetric functions are e1 = x1 + x2 + x3 , e2 = x1 x2 + x2 x3 + x3 x1 , e3 = x1 x2 x3 . We adopt the convention that e0 = 1. 6 n n i=1 k=0 ∏(z − xi) = ∑ (−1)k ek (x1, . . , xn)zn−k . Consider the generating function for the ek : n E(z) = ∑ ek (x1, . . , xn)zk . k=0 A slight rewriting of Newton’s Theorem shows that n E(z) = ∏(1 + xi z). 7 (a) If x1 = . . = xn = 1, then E(z) = (1 + z)n = n ∑ k=0 so ek (1, 1, .
So fa (n) = ∑ ga (n + j), where the sum is over all j with 1 ≤ j ≤ k for which ca (k − j) = 1. This can be rewritten k fa (n) = ∑ ca(k − j)ga(n + j), j=1 or in terms of generating functions, Fa (x) = x−kCa (x)Ga (x). 3) gives the result. In the case where a = 11, we obtain F11 (x) = 1+x x2 + (1 − 2x)(1 + x) = 1+x , 1 − x − x2 so that f11 (n) = Fn + Fn−1 = Fn+1 , as previously noted. 2. OTHER RECURRENCE RELATIONS Unbounded recurrences We will give here just one example. Recall from the last chapter that the generating function for the number p(n) of partitions of the integer n is given by ∑ p(n)x n≥0 n = ∏ (1 − x ) k −1 .