# Find the smallest value for n for which an ≥ 1,000,000. [duplicate]

This question already has an answer here:

The recursive formula is: $a_n = (a_n-1)^2$, with $a_1 = 2$.

I got that $a_6$ is the smallest value (equal to) or more than 1,000,000

If I am correct $a_6$ = 4294967296.

Can you tell me if I am correct? If not, why?

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–  lab bhattacharjee Nov 14 '13 at 18:23
@lab, I voted to close this one as a duplicate of the other one. FWIW –  The Chaz 2.0 Nov 14 '13 at 18:51
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## marked as duplicate by The Chaz 2.0, Lord_Farin, amWhy, Dennis Gulko, Henning MakholmNov 14 '13 at 19:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 1 Answer

I assume you mean $a_{n+1}=(a_n-1)^2$ with $a_1=2$. This sequence might not be right since $$a_1=2$$ $$a_2=(2-1)^2=1$$ $$a_3=(1-1)^2=0$$ $$a_4=(0-1)^2=1$$

$$\text{etc}...$$

If you mean $a_{n+1}=a_{n}^2$,then

$$a_1=2$$ $$a_2=4$$ $$a_3=16$$ $$a_4=256$$ $$a_5=65,536$$ $$a_6=4,294,967,296$$

In general,

$$a_n=2^{(2^{n-1})}$$

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I assume $a_n = a_{n-1}^2$ was what was meant. –  Daniel Fischer Nov 14 '13 at 18:40
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