July 28, 2016 17:25 ws-book9x6 A Walk Through Combinatorics book page 180

180 AWalkThroughCombinatorics

x/(1 x), and B(x)=1/(1 2x). So − − 1 1 x 1 x G(x)=B(A(x)) = 2x = − = , 1 1 x 1 3x 1 3x − 1 3x − − − − −

n n n 1 n n 1 n G(x)= 3 x 3 − x =1+ 2 3 − x . − · n 0 n 1 n 1 !≥ !≥ !≥ n 1 Consequently, if n 1, the officer in charge has 2 3 − options. ≥ ·

Quick Check

(1) Let a = 1, and let a =3a 1forn 0. Find an explicit formula 0 n+1 n − ≥ for a . n (2) Let bn be the number of partitions of the integer n into even parts that are at most 6, and at most one odd part (of any size). Find an explicit n formula for the ordinary generating function B(x)= n 0 bnx . (3) Let us revisit Example 8.16 with the additional restriction≥ that the " non-empty units that the officer forms cannot consist of more than three people each. Let gn be the total number of the ways in which the officer can proceed. Find an explicit formula for the ordinary generating n function Gn(x)= n 0 gnx . ≥ Combinatoricsclass " 8.2 Exponential Generating Functions 35nA's textbook 8.2.1 Recurrence Relations and Exponential Generating Functions

Not all recurrence relations can be turned into a closed formula by using an ordinary generating function. Sometimes, a closed formula may not exist. Some other times, it could be that we have to use a different kind of generating function.

Example 8.17. Let a = 1, and let a =(n +1)(a n + 1), if n 0. 0 n+1 n − ≥ Find a closed formula for an. If we try to solve this recurrence relation by ordinary generating func- tions, we run into trouble. The reason for this is that this sequence grows too fast, and its ordinary generating function will therefore not have a closed form. Let us instead make the following definition.

July 28, 2016 17:25 ws-book9x6 A Walk Through Combinatorics book page 181

AFunctionIsWorthManyNumbers.GeneratingFunctions 181

Definition 8.18. Let fn n 0 be a sequence of real numbers. Then the { } ≥ xn formal power series F (x)= n 0 fn n! is called the exponential generating ≥ function of the sequence fn n 0. { "} ≥ The word “exponential” is due to the fact that the exponential gener- x ating function of the constant sequence fn = 1 is e .Letususethisnew kind of generating function to solve the example at hand.

xn Solution. (of Example 8.17.) Let A(x)= ∞ a be the exponential I n=0 n n! generating function of the sequence an n 0. From this point on, we pro- { } ≥" ceed in a way that is very similar to the method of the previous section. Let us multiply both sides of our recursive formula by xn+1/(n + 1)!, and sum over all n 0toget ≥ ∞ xn+1 ∞ xn+1 ∞ xn+1 a = a (n 1) . (8.15) n+1 (n +1)! n n! − − n! n=0 n=0 n=0 ! ! ! Note that the left-hand side is A(x) 1, while the first term of the right-hand − side is xA(x). This leads to A(x) 1=xA(x) x2ex + xex, − − 1 xn+1 A(x)= + xex = xn + . 1 x n! n 0 n 0 − !≥ !≥ n n n The coefficient of x /n! in n 0 x is n!, while the coefficient of x /n! n+1 ≥ in x is n.Indeed,thissecondtermhassummandxn/(n 1)!. n 0 n! " − Therefore,≥ the coefficient of xn/n! in A(n) is a = n!+n. " n

Example 8.19. Let f = 0, and let f =2(n +1)f +(n + 1)! if n 0. 0 n+1 n ≥ Find an explicit formula for fn.

n Solution. Let F (x)= f x be the exponential generating function n 0 n n! of the sequence f .Letusmultiplybothsidesofourrecursiveformulaby≥ n " xn+1/(n + 1)!, then sum over all n 0. We get ≥ xn+1 xn f =2x f + xn+1. (8.16) n+1 (n +1)!Madge n n! n 0 n 0 n 0 !≥ !≥ !≥ As f0 = 0, the left-hand side of (8.16) is equal to F (x), while the first term of the right-hand side is 2xF (x), and the second term of the right-hand side is x/(1 x). Therefore, we get − x F (x)=2xF (x)+ , 1 x −

Ex do 1 Anti htt an nti n l for n o i Find the exponential generating function EGF for an Ii Find a closed formula for an Solution C Let A Ino an do a az Recursive Anti htt an na n i formula ntl ntl ntl X htt an I nti n DI ultiplyby an nth htt Htt a a E an i cnn.inii nEa iyi.iicsm E an i E intranet a'III simplify E an i Ioane Ea ne ft t the LHS is a t 92 So the LHS is ACx a A Note y Note term the is an the first of RHS o Zan Note the second term oftheRHS is a x nx t x

because xnE o.x o Eii I i x.E.IT x x Ii x E e nee ex Ei 1 A we xe So eq is equivalent to ACN A x XAK I X ex x ex A X 1 x l X ex x ex A x I x2 ex xe x I t xe f x 11 l X This is the exponential generating EGF an 1 function for t e the answer to part i Tx

Ti To find a closed formula for an use part i A x x ex

since and x Eno E FEox't E E.oxnt z.oxn.tt

Ex Ideas ns.Eaotxn n o E FEI D

n 1 Rewrite in such a way that on we see the 1 factor

Enon t holds because Ion n theqq.uaty E moth n Since the coefficient of nX in A x is th we have an htn7 End of Ex 8.17J as I have done for EX 8 17

Note Use Partial Fraction Decomposition to get t 4 NG 2 5 2X Il l X