How to make an exponential progression

Question

I know aritmetic and geometric progression:

Aritmetic progression

$$2 \xrightarrow{+2}4\xrightarrow{+2}6\xrightarrow{+2}8\xrightarrow{+2}10\xrightarrow{+2}12\xrightarrow{+2}14\xrightarrow{+2}16\xrightarrow{+2}18\xrightarrow{+2}20\xrightarrow{+2}22\xrightarrow{+2}24\xrightarrow{+2}26\xrightarrow{+2}28$$

It's:
$$a_{n} = a_{1}+r(n-1)$$ Or: $$a_{n} = a_{m}+r(n-m)$$

Where:

• $a_{1}$ is the first term.
• $a_{n}$ is the nth term.
• $a_{m}$ is an nth term that it isn't $a_{1}$ nor $a_{n}$
• $r$ is the difference between terms.
• $n$ is the number of terms.
• $a_{m}$ is the number of terms of $a_{m}$.

Its difference or ratio is: $$r=a_{n}-a_{n-1}$$

And its sum is: $$S_{n} = \frac{n(a_{1}+a_{n})}{2}$$ Or if you don't want start from the first term: $$S_{n-m} = \frac{(n-m)(a_{m}+a_{n})}{2}$$

Geometric progression

$$2\xrightarrow{\times2}4\xrightarrow{\times2}8\xrightarrow{\times2}16\xrightarrow{\times2}32\xrightarrow{\times2}64\xrightarrow{\times2}128\xrightarrow{\times2}256\xrightarrow{\times2}512\xrightarrow{\times2}1024\xrightarrow{\times2}2048\xrightarrow{\times2}4096$$

It's $$a_{n} = a_{1}\times r^{n-1}$$ Or: $$a_{n} = a_{m}\times r^{n-m}$$

Where:

• $a_{1}$ is the first term.
• $a_{n}$ is the nth term.
• $a_{m}$ is an nth term that it isn't $a_{1}$ nor $a_{n}$
• $r$ is the ratio of multiplication between terms.
• $n$ is the number of terms.
• $m$ is the number of terms of $a_{m}$.

Its difference or ratio is:
$$r = \frac{a_{n}}{a_{n-1}}$$

And its sum is: $$S_{n} = {a_{1}\frac {(r^{n}-1)}{r-1}}$$ Or if you don't want start from the first term: $$S_{n-m} = {a_{1}\frac {(r^{n-m+1}-1)}{r-1}}$$

Now the progression I want: exponential

$$2\xrightarrow{x^2}4\xrightarrow{x^2}16\xrightarrow{x^2}256\xrightarrow{x^2}65,536\xrightarrow{x^2}4,294,967,296\xrightarrow{x^2}18,446,744,073,709,551,616$$ I have no idea of how to calculate anything related to it.

Answer 1

hint

For $n\ge 0$,

$$a_{n+1}=a_n^{2^1}=a_{n-1}^{2^2} $$ $$=a_1^{2^n}=(2)^{(2^n)} $$

Answer 2

These results can be synthetized and generalized by means of the logarithm:

$$a_n=a_{n-1}+r\leftrightarrow\log a_n=\log a_{n-1}+\log r\leftrightarrow\log\log a_n=\log\log a_{n-1}+\log\log r,$$ or equivalently

$$a_n=a_{n-1}+r\leftrightarrow a_n=a_{n-1}r\leftrightarrow a_n=a_{n-1}^{\log r}$$

and you can continue the nesting.

Following this logics, you get a "power progression" with power

$$r=e^{\log a_n/\log a_{n-1}}$$ and $$a_n=a_1^{\log^{n-1}r}.$$

As far as I know, there is no simple formula for the summation (but there is one for the product of the terms).