By Louis Comtet

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**Sample text**

1. , for n = 9, (2, 1, 1, 2, 2, 1) is acceptable, but not (1, 2, 3, 1, 2). 6 Exponential generating function 15 2. , for n = 9, (3, 1, 5) is acceptable, but not (1, 2, 1, 5). 3. , for n = 9, (3, 4, 2) and (3, 3, 2, 1) are acceptable, but not (3, 3, 1, 2). 3. Show that the Fibonacci numbers satisfy the following identity: fn = k≥0 n−k . 2. 4. For n ≥ 1, let φn = fn /fn−1 , where fn is the nth Fibonacci number. Using the Fibonacci recurrence, ﬁnd a recurrence for φn and use it to compute the limit: φ = lim φn .

C(t, z) := n≥0 We have: Cn (t)z n , C(t, z) = n≥0 n−2 =1+ Ci (t)Cn−1−i (t) z n , Cn−1 (t) + t i=0 n≥1 n−2 Cn−1 z n−1 + tz =1+z n≥1 Ci (t)z i Cn−1−i (t)z n−1−i , n≥1 i=0 = 1 + zC(t, z) + tzC(t, z)(C(t, z) − 1). From this we can conclude that C(t, z) satisﬁes: tzC(t, z)2 − (1 + z(t − 1))C(t, z) + 1 = 0. Solving for C(t, z) gives: C(t, z) = 1 + z(t − 1) − 1 − 2z(t + 1) + z 2 (t − 1)2 . 6) 26 2 Narayana numbers The 231-avoiding permutations are one combinatorial interpretation for the Catalan numbers, but there are many, many others.

8 (Noncrossing matchings, balanced parenthesizations). Show that Cn counts the number of noncrossing matchings on [2n]. A noncrossing matching is a noncrossing partition with all the blocks having size two. For example, here are the ﬁve noncrossing matchings on {1, 2, 3, 4, 5, 6}: 1 2 3 4 5 6 ; 1 2 3 4 5 6 1 2 3 4 5 6 ; ; 1 2 3 4 5 6 1 2 3 4 5 6 : ; 42 2 Narayana numbers The noncrossing matchings can also be thought of as n pairs of parentheses, by mapping the beginning of an arc to a left parenthesis, “(”, and mapping the end of an arc to a right parenthesis, “)”.

### Analyse Combinatoire by Louis Comtet

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