Kiriti's sum

First of all, note that, by simple computation, $a_2 =2^{2013}$ .

Next, note that for $k \geq 3$ , we can make the following manipulation. Given:

$\frac{1}{a_k}=\frac{1}{a_1}+\ldots+\frac{1}{a_{k-2}}+\frac{1}{a_{k-1}}$ $\frac{1}{a_{k-1}}=\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_{k-2}}$

We can group together the first few terms in the first recursion to find:

$\frac{1}{a_k}=(\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_{k-2}})+\frac{1}{a_{k-1}} =\frac{1}{a_{k-1}}+\frac{1}{a_{k-1}}=\frac{2}{a_{k-1}}$

Note the penultimate step used the recursion for $\frac{1}{a_{k-1}}$ to collect together the terms of a sum.

From this manipulation, we unearth the recursion $a_k =\frac{a_{k-1}}{2}$ valid for $k>2$ .

From here, we can easily find an explicit formula, valid for $k \geq 2$ :

$a_k=2^{2-k} a_2=2^{2-k} \times 2^{2013}= 2^{2015-k}$

Plugging in $k=2013$ gives our desired answer of $\mathbf{4}$

Notice,

$\frac{1}{a_{2}} = \frac{1}{a_{1}}$

$\frac{1}{a_{3}} = \frac{1}{a_{1}} + \frac{1}{a_{2}} = \frac{2}{a_{1}}$

$\frac{1}{a_{4}} = \frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}} = \frac{2}{a_{3}} = \frac{4}{a_{1}}$

Through deductive reasoning,

$\frac{1}{a_{n}} = \frac{2^{n-2}}{a_{1}}$

$\Rightarrow a_{2013} = 2^{2013 - (2013 - 2)} = 4$

With complete induction $(2 \le n)$ it can be shown that $a_{n} = \frac{2^{2013}}{2^{n-2}}= 2^{2015-n}$ .

So: $a_{2013} = 2^{2015-2013} = 4.$

Let $2^{2013} = a$ , then we find that:

$a_1 = a$ $a_2 = a$ $a_3 = \frac{a}{2}$ $a_4 = \frac{a}{4}$

In general, it can be stated that:

$a_n = \frac{a}{2^{n-2}}, n > 1$

where $n$ and $a$ are the term number and starting value respectivley. Therefore, to calculate the $2013^{th}$ term, we find that:

$a_{2013} = \frac{2^{2013}}{2^{2011}} = 2^2 = \fbox4$

It is not obvious how to proceed immediately, so let's evaluate the first few terms of the sequence. Using the recursion, we get $1/a_2 = 1/2^{2013}, 1/a_3 = 2/2^{2013}, 1/a_4 = 4/2^{2013}$ and so on. One can easily show that $\frac{1}{a_k} = \frac{1}{2^{2013}} \cdot \sum_{k=0}^{2^{k-3}}$ for every integer $k$ greater than $1$ by induction. The reciprocal of the $2013$ th term is therefore $\frac{1}{2^{2013}} \cdot \sum_{k = 0}^{k=2^{2010}} = 1/4$ , so the answer is $\frac{1}{1/4} = \boxed{004}$ .

$\frac{1}{a_{k}} = \frac{1}{a_{1}} + \frac{1}{a_{2}} + \dots + \frac{1}{a_{k - 1}}$

$\Rightarrow \frac{2}{a_{k}} = \frac{1}{a_{1}} + \frac{1}{a_{2}} + \dots + \frac{1}{a_{k - 1}} +\frac{1}{a_{k}} = \frac{1}{a_{k + 1}}$

$a_{k + 1} = \frac{a_{k}}{2}$ for $k > 1$ , and $a_{2} = a_{1}$

Hence, $a_{2013} = \frac{a_{2}}{2^{2011}} = \frac{a_{1}}{2^{2011}} = \fbox{4}$

we've :

$a_1 = 2^{2013}$ and $\frac{1}{a_1} = \frac{1}{2^{2013}}$

so:

$\frac{1}{a_2} = \frac{1}{a_1} = \frac{1}{2^{2013}}$

$\frac{1}{a_3} = \frac{1}{a_2} + \frac{1}{a_1} = \frac{1}{2^{2013}} + \frac{1}{2^{2013}} = \frac{2}{2^{2013}}$

$\frac{1}{a_4} = \frac{1}{a_3} + \frac{1}{a_2} + \frac{1}{a_1} = \frac{2}{2^{2013}} + \frac{1}{2^{2013}} + \frac{1}{2^{2013}} = \frac{4}{2^{2013}} = \frac{2^2}{2^{2013}}$

$\frac{1}{a_5} = \frac{1}{a_4} + \frac{1}{a_3} + \frac{1}{a_2} + \frac{1}{a_1} = \frac{4}{2^{2013}} + \frac{2}{2^{2013}} + \frac{1}{2^{2013}} + \frac{1}{2^{2013}} = \frac{8}{2^{2013}} = \frac{2^3}{2^{2013}}$

so the pattern is $\frac{1}{a_k} = \frac{2^{k-2}}{a_1}$ for $k > 1$ so :

$\frac{1}{a_{2013}} = \frac{2^{2013-2}}{a_1} = \frac{2^{2013-2}}{2^{2013}} = \frac{1}{4}$ , $a_{2013} = 4$