Question

(Recall that Cn is the number of positive lattice walks from (0,0) to (n,n) which do not go above the line y = x.) For all integers n > 0, n Cutz = C:Cn-i 2+1 = i=0 (Hint: Note that for every positive lattice walk from (0,0) to (n + 1, n + 1) which does not go above the line y = x, there exists some 0 < i <n such that (i + 1, i + 1) is the first time the walk touches the line y = x.)

Answer 1

Actually this is nth Catalan number. Here are several proofs.
Here are the formulas. It is a good idea to try plugging in the numbers you know to make certain that you haven't made a silly error. In this case, the formula for C3 indicates that it should be equal to C3 = 2.1+1:1 +1.2 = 5. C = Coco C2 = CiCo + CoC1 C3 = C2Co +C1C1 + C0C2 = C3Co + C2C + CC2 + CoC3

On = Cn-1Co + Cn-2C1 +...+C Cn-2 +CoCn-1 (5)

DEFINITION The Catalan numbers Co, C1,..., Cn,... are given by the formula Cn = n+1() The first few Catalan numbers are Co = 1 C1 = 1 C2 = 2 C3 = 5 C4 = 14 C5 = 42. It is not immediately clear from the definition that the Cn are integers,

Proof by reflection principle. Recall that C counts the number of Dyck paths that are above the x-axis from (0,0) to (2n, 0) First of all, if we remove the "above the x-axis" condition, we just want the number of paths that have n ups and n downs: this is just (2). We now want to remove all paths that were overcounted: how many non-Dyck paths are there? Each of these paths must intersect the line y = -1 at least once. Pick the first intersection, and take the portion of the path from that intersection to the end (2n, 0). Reflect it over y = -1, and now the endpoint of our path is (2n, -2). This new path has n - 1 up steps and n + 1 down steps, and such paths are in bijection with the non-Dyck paths! This is because we can find the first intersection with y = -1 and do the reflection again. Since there are (21) such paths, the number of Dyck paths is C. = (2,") - (,2-1) = n+1(2"). But as a combinatorialist, professor Postnikov doesn't like subtraction. Here's a proof that doesn't require subtraction! Proof by necklaces. Start with the binomial coefficient (2n+1), which is the number of sequences €1,..., E2n+1 with n +1s and n+1 -1s. For example, for n = 2, consider the sequence (+,-,-, +,-). Do a cyclic shift of the sequence: move the first term to the end to get (-:- + - +. Keep doing this: we get (-, +, -, +,-), (+,-.+,-,-), and (-, +, -, -, +). In general, there are 2n + 1 cyclic shifts. So we can arrange these on a necklace (as black and white beads). The necklace has n +s and n +1 -5, and each of the cyclic shifts corresponds to cutting the necklace at one spot.

First Proof by Andre's Reflection Method This method is based on counting the total number of monotonically increasing paths from bottom left corner to the top right corner in an nXn square. We need to count the number of paths possible without crossing the diagonal.Suppose we are given a monotonic path in an n n grid that does cross the diagonal. Find the first edge in the path that lies above the diagonal, and flip the portion of the path occurring after that edge, along a line parallel to the diagonal. Observe that now we have taken into account k+1 vertical edges and k horizontal edges for some k between 1 and n-1. This leaves l-1 vertical edges and 1 horizontal edges, where 1+ k = n. By reflecting the rest of the vertical and horizontal edges, we will have k +1+1= n +1 vertical edges and k +1-1= n - 1 horizontal edges. The resulting path is a monotonic path in an (n-1) (n + 1) grid. Figure 1 illustrates this procedure; the green portion of the path is the portion being flipped. This method can be mapped to the Dyke words problem where we start with a sequence of sequence of n X's and n Y's which is not a Dyck word, and exchanging all X's with Y's after the first Y that violates the Dyck condition. At that first Y, there are k +1 Y's and k X's for some k between 1 and n - 1. 1-1 Figure 1: The green portion of the path is flipped. The old grid is n x n. The new grid is (n - 1) x (n + 1). Since every monotonic path in the (n - 1) (n + 1) grid is of length 2n and every such path involves n + 1 vertical edges in any order, the number of these paths is equal to: | 2n In +1 (2) Therefore, to calculate the number of monotonic n n paths which do not cross the diagonal, we need to subtract this from the total number of monotonic n n paths, so we finally obtain (2) - (*') (3) which is the nth Catalan number Cn.

Counting Diagonal-Avoiding Paths Up to now we do not have an explicit formula for the Catalan numbers. We know that a large collection of problems all have the same answers, and we have a recursive formula for those numbers, but it would be nice to have an explicit form. s the easiest way to obtain an explicit formula for the Catalan numbers is to analyze the number of diagonal-avoiding paths discussed in Section 1.3. We will do so by counting the total number of paths through the grid and then subtract off the number of paths that hit the diagonal. Figure 8: Modifying a Bad Path Figure 8 illustrates a typical path that we do not want to count since it crosses the dotted diagonal line. Such a path may cross that line multiple times, but there is always a first time; in the figure, point P is the first grid point it touches on the wrong side of the diagonal. There will always be such a point P for every bad path. For every such path, reflect the path beginning at P-every time the original path goes to the right, go down instead, and when the original path goes down, go to the right. It is clear that by the time the path reaches the point P it will have traveled one more step down than across, so it will have moved k steps to the right and k + 1 steps down. The total path has n steps across and down, so there remain n – k steps to the right and n - k-1 steps down. But since we swap steps to the right and steps down, the modified path with have a total of (k) + (n - k-1) = n - 1 steps to the right and (k + 1) + (n - k) = n + 1 steps down. Thus every modified path ends at the same point, n - 1 steps to the right and n + 1 steps down. Every bad path can be modified this way, and every path from the original starting point to this point n - 1 to the right and n + 1 down corresponds to exactly one bad path. Thus the number of bad paths is the total number of routes in a grid that is (n – 1) by (n + 1). There are ("+") paths through an k x m grid'. Thus the total number of paths through the nxn grid is (2n) and the total number of bad paths is (,21,). Thus Cn, the nth Catalan number, or the total number of diagonal-avoiding paths through an n x n grid, is given by: C = (?") - (22+1) = (%) - -*1(?) - +1 (").