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?

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TopNewestWell, the sequence starting from \(1,0,\ldots \) defies your conjecture.

It is a recursive sequence with repeated segment \([1,0,1,-1,-1]\) – Daniel Liu · 3 years, 1 month ago

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@Daniel Liu Your sequence has A2 as 0 but it is supposed to be 1 – Mardokay Mosazghi · 3 years ago

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– Daniel Liu · 3 years ago

I don't see why \(a_2\) can't be \(0\). The OP said \(a_1=a_2=1\), but I can surely change that, can I? Or can I only mix up the definitions for how to find later terms?Log in to reply

– Tristan Shin · 3 years ago

I meant only definitions for later terms. The first two terms must stay at 1, otherwise there are several contradictions(including yours).Log in to reply

– Nanayaranaraknas Vahdam · 3 years, 1 month ago

The sequence is defined with \(a_2\) as \(1\), but your sequence shows \(a_2\) as \(0\).Log in to reply