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Given $a_0=1$ and:$$a_n=a_{\frac{n}{2}}+a_{\frac{n}{3}}+a_{\frac{n}{6}}$$Find convergence or divergence of $\frac{a_n}{n}$.

Such a weird problem; I don't know how to attack it. I'm also fairly certain I typed it out correctly.

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    I am told this has to do with stochastic processes and probability theory. I do not actually know if this is the case.2012-03-26
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    How is $\frac{n}{6}$ defined when $n$ isn't a multiple of $6$?2012-03-26
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    The sequence needs more than just $a_0$ to be well-defined. You need to know beforehand the values of $a_n$ for all $n$ which are not multiples of $6$.2012-03-26
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    @AntonioVargas I think the same thing as well, but I'm writing down the problem exactly as it was given.2012-03-26
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    @user27572, then the problem is silly exactly as it was given.2012-03-26
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    @AntonioVargas Are you absolutely sure that this problem does not have a solution as given? That's what I think, but the context of how I got the question really strongly suggests otherwise.2012-03-26
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    @user27572, it would seem to me that a more sensible question would be "Is it possible to choose $a_n$ for all $n$ which are not multiples of $6$ such that the sequence $(a_n)$, as defined by the recurrence relation, is convergent?"2012-03-26
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    The last part of my comment should read "the sequence $(a_n/n)$, with $a_n$ defined by the recurrence relation, is convergent?"2012-03-26
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    @AntonioVargas I suspect that the fraction indexes are supposed to be rounded down. That is: $a_1 = a_0+a_0+a_0$, $a_5 = a_2+a_1+a_0$, $a_9 = a_4 + a_3 + a_1$ etc.2012-03-26
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    Alright, so I'm going to work under the assumption that the professor handed out a mistyped question, and everyone except me figured this out. :( In which case, thanks guys! It was still very helpful to understand why this question was demonstrably unsolvable. I believe my professor might've meant $a_n=a_{\lfloor\frac{n}2\rfloor}+a_{\lfloor\frac{n}3\rfloor}+a_{\lfloor\frac{n}6\rfloor}$, as specified by @Henry.2012-03-26

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