The following algorithm can be applied to a list of fractions and an integer input. You go down the list and multiply the input by the first fraction that will result in an integer. Taking this product as the new input, you repeat, using the same list of fractions. The algorithm ends when none of the fractions will result in an integer.

For example, if the list of fractions is {5/6, 5/2, 5/3}, then inputting an integer 2^a * 3^b will result in 5^max(a,b). A more complicated example: inputting 2^a * 3^b into {7/11, 11/(3*7), 1/7, (5*7)/2, 3/5} will result in 3^a (if a>0).

Find a list of fractions such that inputting an integer 2^a * 3^b will result in 5^ab.

How about a list that, when the input is 2^a * 3^b, results in 5^{a^b} (with b>0)?

Here is the a*b list:
{11/13,  (7*13)/(11*2), 1/11, (5*2)/7, (7*11)/(2*3), 1/2, 1/3}

and the a^b list :
{(11*11)/(13*13), (7*13*13)/(11*11*2), 13/(11*11),  11/13, (17*2*13)/(11*7), (7*13*13)/(2*5*11), 1/11, 5/17, (7*11*11)/(2*5*3), (7*11*11)/(2*3), 17/(2*5), 5/2, 1/3}

Note that this solution actually handles b=0 (it only took one extra fraction) treating 0^0 as 0

Posted by Joel on 2006-11-21 22:55:04