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The closed form for $a_{n}$ is of the form $c_{1}(1)^{n}+c_{2}(-2)^{n}+c_{3}(-3)^{n}$ for some constants $c_{1}, c_{2},$ and $c_{3}.$ I won't go into much detail on how I got that, but it involves manipulating the recursion. For $n = 2014,$ we get $a_{2014} = 2014 = c_{1}(1)^{2014}+c_{2}(-2)^{2014}+c_{3}(-3)^{2014}.$ Now, notice the $c_{2}(-2)^{n}$ and $c_{3}(-3)^{n}$ terms in our closed form. If $c_{2}$ and/or $c_{3}$ are nonzero, then for some non-negative integer $n$ , $a_{n}$ will be negative. (For example, for $c_{2} = 1$ and $c_{3} = -100, n = 1000000...$ with a large number of $0$ s will guarantee that $a_{n}$ is negative.) Since all $a_{n}$ are positive, we must have $c_{2} = c_{3} = 0$ , so $c_{1}(1)^{n}+c_{2}(-2)^{n}+c_{3}(-3)^{n} = c_{1} = 2014 = a_{n}.$ Thus, $a_{10000}$ can only be $\boxed{2014}.$