Number Theory Problem: Proving Divisibility in Sequences of Integer Lists

Proving Divisibility in Sequences of Integer Lists

Problem: For each integer n ≥ 1, let f(n) be the number of lists of different positive integers starting with 1 and ending with n, in which each term except the last divides its successor. Prove that for each integer N ≥ 1 there is an integer n ≥ 1 such that N divides f(n). (For example, f(1)=1, f(2)=1, f(3)=3)

Proof: We will prove this statement using proof by contradiction.

Assumption: Assume that there exists an integer N ≥ 1 for which N does not divide f(n) for any integer n ≥ 1.

Contradiction:

1. Limited Remainders: Our assumption implies that for every integer n ≥ 1, the remainder when f(n) is divided by N is non-zero. Since there are only N possible remainders (0, 1, 2, ..., N-1) when an integer is divided by N, and f(n) can never have a remainder of 0, f(n) can only take on at most N-1 distinct remainders when divided by N.

2. Pigeonhole Principle: Consider the sequence of values f(1), f(2), f(3), .... Since this sequence is increasing and unbounded, there must exist some integer n1 such that for all n ≥ n1, the remainders of f(n) when divided by N will start repeating.

3. Constructing a Difference: Let k be a positive integer. Since the remainders repeat, f(n1 + k) and f(n1) have the same remainder when divided by N. Therefore, their difference, f(n1 + k) - f(n1), must be divisible by N.

4. Reaching a Contradiction: Now, let's specifically consider the case when k = N. This means f(n1 + N) - f(n1) must be divisible by N. However, this difference is equal to N itself, as f(n1 + N) includes all the lists counted in f(n1) plus additional lists that include n1 + N. Therefore, N divides N, which is always true. This contradicts our initial assumption that N does not divide f(n) for any integer n.

Conclusion: Because our assumption leads to a contradiction, it must be false. Therefore, for each integer N ≥ 1, there must exist an integer n ≥ 1 such that N divides f(n).