Find an integer $k$ such that $a_{k} = 2^{261}$?

Question

Let $a_{1} := 2$ and $$a_{k} := \frac{2^{(k+1)(k+2)/2}}{\prod\limits_{j=1}^{k-1}a_{j}}$$ for all integers $k \geq 2.$ The problem is to find an integer $k$ such that $a_{k} = 2^{261}.$

The statement above is not in the original form of the problem, I have sieved the original one and made it the present more compact form.

But my problem is that, when I try to draw something useful from the fraction then I am stuck. Any ideas?

Answer 1

Let $a_n=2^{b_n}$. Then we can work out the recurrence

$$b_k=\frac{(k+1)(k+2)}2-\sum_1^{k-1}b_j$$

(note that $b_1=1$, as well as the original $a_1=2$, is actually redundant since this is a definition by complete recursion, and you can just plug in $k=1$ in the above formula). We can rearrange this into

$$\sum_1^kb_j=\frac{(k+1)(k+2)}2,$$

and by taking differences we get

$$b_k=\sum_1^kb_j-\sum_1^{k-1}b_j=\frac{(k+1)(k+2)}2-\frac{k(k+1)}2=k+1,$$

whence $b_{260}=261$ and $a_{260}=2^{261}$.

Answer 2

$$\frac{a_k}{a_{k-1}}=\frac{2^{(k+1)(k+2)/2-k(k+1)/2}}{\frac{a_1a_\cdots a_{k-2}a_{k-1}}{a_1a_2\cdots a_{k-2}}}=\frac{2^{k+1}}{a_{k-1}}.$$ Hence $$a_k=2^{k+1}.$$