UMTYMP Advanced Topics – Introduction to Combinatorics April 20^th, 2005

Bona Chapter 8 #26. Find an explicit formula for the numbers a_n if if , and a₀ = 0.

First of all, there is a typo in the statement of the problem. should be , otherwise there is no way to compute any of the values!

Now, since a_{n + 1} depends on na_n , as well as on n!, we need to use an exponential generating function. Otherwise, the coefficients will grow too fast for the function to converge anywhere except at x = 0 and we will not be able to find a closed-form formula for the generating function. So let .

Next we multiply both sides of the recursion by and sum for all . (We choose so that the n + 1 will cancel with the factors of n + 1 appearing in the recursion.) This gives

Then, since we have

So .

Bona Chapter 8 #28. Find the exponential generating function D(x) for derangements, defined in Example 7.3. Look for several different ways to obtain D(x).

Method #1 Using the recursion D_{n + 1} = (n + 1)D_n + (– 1)^{n + 1}, .

Multiply both sides of the recursion by and sum for all . This gives

Method #2 Using the explicit formula .

Multiply both sides of the formula by and sum for all . This gives

Method #3 Using the combinatorial interpretation of the product of exponential generating functions.

The exponential generating function for the number of permutations is . Every permutation consists of a certain set of fixed points, and a derangement on the remaining points. By the Product Rule (for exponential generating functions), the exponential generating function for permutations is the exponential generating function for derangements, D(x), times the exponential generating function for fixed points. There is only one way to “permute” fixed points, so the function for the fixed points is just .

This means we have , or .

Method #4 Using the recursion D_{n + 2} = (n + 1)D_{n + 1} + (n + 1)D_n, .

Multiply both sides of the recursion by and sum for all . This gives

Now notice that

Similarly,

So we have

Setting x = 0 gives lnD₀ = 0 + ln(1) + C, i.e., 0 = C (recall that D₀ = 1).

Hence , and so .