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By Luke Y.L.

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22 If we multiply the generating function of the last example by z we find that z = z + 2z2 + 3z3 + 4z4 + · · · . (1 − z)2 We may conclude that the sequence {0, 1, 2, 3, . } = {r} has the generating function z . 1 Generating function library. We will build the library as the book progresses and it is summarized in Appendix A. 2 Recurrence to Generating Function Most enumerations start with a recurrence relation that has been derived, rather than the enumerative sequence itself. 3 Sequence to Generating Function 35 sequence.

Working with Generating Functions But we can expand the denominator by the Binomial Theorem and then ∑ Rr zr = r 0 1 − z + z2 (1 − z)−3 = 1 − z + z2 ∑ r 0 r+2 r z. 2 In effect this gives us an explicit form for the number of regions – a major breakthrough. All we need to do is to write out (as far as we please) the power series on the right: (1 − z + z2 ) 2 3 + z+ 2 2 2 = + 2 2 − 2 2 + 2 4 2 z +··· 2 3 4 2 z+ z + 2 2 4 3 2 z− z − 2 2 3 4 z3 + z2 + 2 2 5 3 z +··· 2 z3 + · · · z4 + · · · = 1 + (3 − 1)z + (6 − 3 + 1) z2 + (10 − 6 + 3) z3 + · · · .

Find the generating function of the sequence {ar }. 26. 15 Find a generating function for the number of strings of length r made up from the digits {0, 1, 2, 3} in which there is never a 3 anywhere to the right of 0. 16 We define the matrix M as the 2 × 2 array: M= 0 1 . 1 1 The powers of this matrix have a number of surprising connections with Fibonacci and Lucas numbers. 4 Miscellaneous Exercises 39 (i) Show that Mr = Fr−1 Fr Fr . Fr+1 (ii) show that the trace of M r is given by tr (M r ) = Lr ; (iii) by considering the determinant of M r prove Cassini’s identity Fr−1 Fr+1 − Fr2 = (−1)r .

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