## Question

Consider all integer combinations of $a^b$ for $2 \leq a \leq 5$ and $2 \leq b \leq 5$:

\displaystyle \begin{aligned} & 2^2 = 4\text{, } 2^3 = 8\text{, } 2^4 = 16\text{, } 2^5 = 32 \\ & 3^2 = 9\text{, } 3^3 = 27\text{, } 3^4 = 81\text{, } 3^5 = 243 \\ & 4^2 = 16\text{, } 4^3 = 64\text{, } 4^4 = 256\text{, } 4^5 = 1024 \\ & 5^2 = 25\text{, } 5^3 = 125\text{, } 5^4 = 625\text{, } 5^5 = 3125 \end{aligned}

If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:

$\displaystyle 4\text{, } 8\text{, } 9\text{, } 16\text{, } 25\text{, } 27\text{, } 32\text{, } 64\text{, } 81\text{, } 125\text{, } 243\text{, } 256\text{, } 625\text{, } 1024\text{, } 3125$

How many distinct terms are in the sequence generated by $a^b$ for $2 \leq a \leq 100$ and $2 \leq b \leq 100$?

import Data.List (nub)

main ::  IO ()
main = print $length$ nub [a^n | a <- [2..100], n <- [2..100]]
$ghc -O2 -o exponentComb exponentComb.hs$ time ./exponentComb
real   0m0.523s
user   0m0.516s
sys    0m0.000s

## Python

#!/usr/bin/env python
import itertools
print(len(set(a**b for a,b in itertools.product(list(range(2,101)), repeat=2))))
$time python3 exponent-combinations.py real 0m0.041s user 0m0.036s sys 0m0.008s ## Ruby #!/usr/bin/env ruby puts (2..100).to_a.product((2..100).to_a).map { |a,n| a**n }.uniq.count $ time ruby exponent-combinations.rb
real   0m0.057s
user   0m0.048s
sys    0m0.008s