By Rudenskaya O. G.

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**Sample text**

Consider the permutation 2 S4 given in two-line form by 12 24 31 43 . This means that 1 D 2; 2 D 4, etc. 1 2 4 3/. ) Recall that every permutation can be decomposed into a product of disjoint cycles. For example, consider 2 S8 given by 13 22 35 48 56 67 71 84 . Under , 1 goes to 3, 3 goes to 5, 5 goes to 6, 6 goes to 7, and 7 goes back to 1, closing a cycle. Now 2 goes to 2, which makes a cycle unto itself, and finally, 4 goes to 8 and 8 goes back to 4. 4 8/; in the latter form, the convention is that every number that does not appear at all forms a cycle unto itself.

For each k 2 N, let BnIk denote the event that 2 the first integer chosen is divisible by pk and let BnIk denote the event that the second integer chosen is divisible by pk . i; j / 2 n n W pk jj g: 1 2 Note of course that the above sets are empty if pk > n. i; j / 2 n n W pk ji and pk jj g is the event that both selected integers have pk as a factor. There are Œ pnk integers in n that are divisible by pk , namely, ; Œ pnk pk . BnIk \ BnIk / is the event that the two selected integers have at least one common prime factor.

A) For fixed x 2 Z, use the fact that with probability one the random walk visits 0 infinitely often to show that with probability one the random walk visits x infinitely often. (Hint: Every time the process returns to 0, it has probability . ) (b) Show that with probability one the random walk visits every x 2 Z infinitely often. 2. In this exercise, you will prove that ET0 D 1, where T0 is the first return time to 0. We can consider the random walk starting from any j 2 Z, rather than just from 0.