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“mcs” — 2017/3/3 — 11:21 — page 696 — #704<br />

696<br />

Chapter 16<br />

Generating Functions<br />

(c) 1, 0, 1, 0, 1, 0, 1,. . .<br />

(d) 1, 4, 6, 4, 1, 0, 0, 0,. . .<br />

(e) 1, 2, 3, 4, 5,. . .<br />

(f) 1, 4, 9, 16, 25,. . .<br />

(g) 1, 1, 1/2, 1/6, 1/24, 1/120,. . .<br />

Class Problems<br />

Problem 16.5.<br />

Let A.x/ D P 1<br />

nD0 a nx n . Then it’s easy to check that<br />

a n D A.n/ .0/<br />

;<br />

nŠ<br />

where A .n/ is the nth derivative of A. Use this fact (which you may assume) instead<br />

of the Convolution Counting Principle 16.2.3, to prove that<br />

1<br />

1 !<br />

.1 x/ D X n C k 1<br />

x n :<br />

k k 1<br />

nD0<br />

So if we didn’t already know the Bookkeeper Rule from Section 15.6, we could<br />

have proved it from this calculation and the Convolution Rule <strong>for</strong> generating functions.<br />

Problem 16.6. (a) Let<br />

S.x/ WWD x2 C x<br />

.1 x/ 3 :<br />

What is the coefficient of x n in the generating function series <strong>for</strong> S.x/?<br />

(b) Explain why S.x/=.1 x/ is the generating function <strong>for</strong> the sums of squares.<br />

That is, the coefficient of x n in the series <strong>for</strong> S.x/=.1 x/ is P n<br />

kD1 k2 .<br />

(c) Use the previous parts to prove that<br />

nX<br />

k 2 D<br />

kD1<br />

n.n C 1/.2n C 1/<br />

:<br />

6

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