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Divisibility Chains Generalization

Based on this problem.


Define two positive integer sequences \(\{a_n\}\) and \(\{b_n\}\) be defined as \(a_1\ne b_1\), \(a_{n+1}=a_n+k\) and \(b_{n+1}=b_n+k\). These two sequences form an Order \(k\) Divisibility Chain of length \(n\) if \(a_i\mid b_i\) for \(i=1\to n\).

Prove that no matter what \(n\) and \(k\) you choose, there always exists an infinite number of sequences \(\{a_n\}\) and \(\{b_n\}\) that form an Order \(k\) Divisibility Chain of length \(n\).


Please only post hints, do not post the solution. If you do, it will give away the solution for the problem I based this on. Thanks.

Note by Daniel Liu
2 years, 8 months ago

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Take \(a_1=k, b_1=k+k\times n!\). This gives an example of a single such sequence for any n,k. Should not be hard to generalize this to produce infinitely many such sequences, Abhishek Sinha · 2 years, 8 months ago

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Hint: Consider the sequence in modulo \(lcm(a_1,a_2,...,a_n)\). From there rest will be pretty easy.

@Daniel Liu : Am I being too obvious? Siam Habib · 2 years, 8 months ago

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@Siam Habib No, you're not being too obvious as far as I can tell.

Now to think about it, I should have posted this problem for the Proofathon Sequences and Series competition. Dang! Daniel Liu · 2 years, 8 months ago

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Hint: \(a_1 = b_1\)

(Which, by the way, is why I reported the problem.) Ivan Koswara · 2 years, 8 months ago

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@Ivan Koswara I can't believe I didn't realize that any \(a_1=b_1\) works. Edited the problem and this note to show that \(a_1\ne b_1\). Daniel Liu · 2 years, 8 months ago

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