# Find the smallest value for n for which $a_n ≥ 1,000,000$

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

I got that $a_5$ is the smallest value before it is equal to or greater than $1,000,000$. If I'm correct $a_5$ is $65,536$.

I got that $a_6$ is the smallest value is 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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have you meant $a_n=(a_{n-1})^2,$ by any chance? –  lab bhattacharjee Nov 14 '13 at 18:10

I think you meant $a_n=(a_{n-1})^2$

As $a_1=2,a_2=2^2,a_3=(2^2)^2=2^4,a_4=(2^4)^2=2^8\implies a_n=2^{2^{n-1}}$ for $n\ge1$

$\implies a_5=2^{2^{4}}=2^{16}=65536$

We need $2^{2^{n-1}}>10^6$

One way to apply logarithm $$2^{n-1}\log_{10}2>6$$

Now, $\log_{10}2>0.3$ as $2^{10}>10^3$

$\implies 2^{n-1}\log_{10}2>2^{n-1}\cdot0.3$

which will be $>6 \iff 2^{n-2}>\frac6{0.3}=20\implies n\ge6$

Just validate with $n=5$

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