By Marshall, Jr. Hall
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10r + 4nr . This is the required recurrence. Once we know one initial value we can use the recurrence to creep forwards, finding successive terms of the sequence, one at a time. 102 + 4n2 = 532; we may continue this as far as we please. Here is another example, this time with a geometric flavour – and with a sequence as answer that turns out to be very important. 26 2. 11 A car park consists of a row of r spaces; a motorbike (m) takes one space, a car (c) two. 5 Parking cars and motorbikes. We seek the number of ways there are of filling the spaces.
The first comes in this way (1 − z) ∑ Rr zr = r 0 1 − z + z2 = (1 − z + z2 )(1 − z)−2 . (1 − z)2 Expanding the right-hand side and then comparing coefficients of zr we find that, Rr − Rr−1 = r+1 r r−1 − + = r. 1 1 1 That, of course, was the recurrence we started with. The second comes from (1 − z)2 ∑ Rr zr = r 0 1 − z + z2 = (1 − z + z2 )(1 − z)−1 (1 − z) ⇒ Rr − 2Rr−1 + Rr−2 = 1 − 1 + 1 = 1 which is the recurrence Rr − 2Rr−1 + Rr−2 = 1. Finally, there is the recurrence already derived from the whole denominator (1 − z)3 ∑ Rr zr = 1 − z − z2 r 0 ⇒ Rr − 3Rr−1 + 3Rr−2 − Rr−3 = 0.
3). We have, ∑ 5ur−1 zr = 5z ∑ ur−1 zr−1 = 5z ∑ ur−1 zr−1 − u0 r 2 r 2 = 5zU(z) − 25z. r 1 36 2. Generating Functions Count The final term is now easy ∑ 6ur−2 zr = 6z2 ∑ ur−2 zr−2 = 6z2U(z). 3): U(z) − 5 − 12z = 5zU(z) − 25z − 6z2U(z) and then solve this for U(z). We find that U(z) = 5 − 13z . 1 − 5z + 6z2 We have quickly passed over a very important idea which we now make explicit. This important result, whose proof is immediate, should become second nature. 25 (Re-indexing a sum) We have U(z) = ∑ ur zr = ∑ ur−1 zr−1 = ∑ ur−2 zr−2 .
Combinatorial Theory MAc by Marshall, Jr. Hall