I have an interesting sequence whose properties I do not know yet(that is why I'm asking for your help). Define \(a_{1} = 1\) and \(a_{2} = 1\) for this sequence. From then on, \(a_{n}\) will be defined as \(a_{n-2} + a_{n-1}\) if \(n \equiv 3 \mod {4}\), \(a_{n-2} - a_{n-1}\) if \(n \equiv 0 \mod {4}\), \(a_{n-2} \times a_{n-1}\) if \(n \equiv 1 \mod {4}\), and \(a_{n-2} \div a_{n-1}\) if \(n \equiv 2 \mod {4}\). The first few terms of the sequence are \(1, 1, 2, -1, -2, \frac {1}{2}, -\frac {3}{2}, ...\). I haven't found much, but I've hypothesized that similar sequences(mixing up the order of adding, subtracting, multiplying, and dividing) will result in a finite sequence(stopping when a term is undefined) if and only if a 0 appears(the iff is important). Anyone want to find something about this?

Note by Tristan Shin
6 years, 10 months ago

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Well, the sequence starting from 1,0,1,0,\ldots defies your conjecture.

It is a recursive sequence with repeated segment [1,0,1,1,1][1,0,1,-1,-1]

Daniel Liu - 6 years, 10 months ago

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The sequence is defined with a2a_2 as 11, but your sequence shows a2a_2 as 00.

Nanayaranaraknas Vahdam - 6 years, 10 months ago

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@Daniel Liu Your sequence has A2 as 0 but it is supposed to be 1

Mardokay Mosazghi - 6 years, 9 months ago

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I don't see why a2a_2 can't be 00. The OP said a1=a2=1a_1=a_2=1, but I can surely change that, can I? Or can I only mix up the definitions for how to find later terms?

Daniel Liu - 6 years, 9 months ago

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@Daniel Liu I meant only definitions for later terms. The first two terms must stay at 1, otherwise there are several contradictions(including yours).

Tristan Shin - 6 years, 9 months ago

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