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I have a sequence $a_n$: $1,2, 2^2, 2^{2^2}, 2^{2^{2^2}}, ...$

I would like to know how to factor $b_i = a_i-a_{i-1}$ where $a_0=1$

All I've been able to figure out so far is that 1 + $\sum\limits_{i=1}^nb_i = a_n$, which doesn't seem to go anywhere...

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  • $\begingroup$ Do you mean that $a_n: 1, 2, 2^2, (2^2)^2, ((2^2)^2)^2$ or $a_n: 1, 2, 2^2, 2^{(2^2)}, 2^{(2^{(2^2)})}$? $\endgroup$ – AlkaKadri Jul 3 '18 at 22:48
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Let $$a_i = 2^{x_i}, x_i = p_1^{p_2^{\cdots ^{p_i}}}$$

Then $b_i = a_i - a_{i-1} = a_{i-1} \cdot (2^{(x_i - x_{i-1})} - 1)$

Remove the trivial factor as:

$$\frac{b_i}{a_{i-1}} = 2^{(x_i - x_{i-1})}-1$$

Also it's divisible by $3$ since difference between exponents is even.

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  • $\begingroup$ How would I then factor $\frac{b_i}{a_{i-1}}$ ? You've mentioned that it is divisible by 3, but how would I figure out other divisors? $\endgroup$ – AAC Aug 29 '18 at 18:18

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