In the absence of a WRDSUM editorial, here are some thoughts.

The problem statement WRDSUM provides a definition of F(n) equivalent to smallest x for which n is a perfect power of x.

Define S(N) as F(2) + F(3) + ... + F(N).

WRDSUM asks for various values of S(N) for N \le 10^{2600} modulo a given prime.

For most n, we have F(n) = n, so a first step is to write

S(N) = \sum_{r=2}^N r - \it corrections

The first corrections come from the perfect squares. In the sum over r, the values 4, 9, 16, 25, 36, 49, 64, ... are replaced by 2,3,2,5,6,7,2,..., or better written as F(2), F(3), F(4), F(5), F(6), F(7), F(8), ....

Perfect cubes follow the same pattern: the values 8, 27, 64, ... are replaced by F(2), F(3), F(4), ...

So S(N) has become

S(N) = \sum_{r=2}^N r - \sum_ {p} \left( \sum_ {r=2}^{\sqrt[p] N} r^p - S(\sqrt[p]N) \right) + \it more\ corrections

where the sum over p is a sum over prime numbers.

Perfect sixth powers have been subtracted twice (because perfect sixth powers are both squares and cubes) so have to be added back. Continuing along these lines, and introducing the mobius function \mu(n) to handle the inclusion/exclusion arguments, we have

S(N) = \sum_{r=2}^N r + \sum_{n} \mu(n) \left( \sum_{r=2}^{\sqrt[n] N} r^n - S(\sqrt[n]N) \right)

where the sum over n is now over all integers >1. The \mu(n) is the number theory möbius function, defined as zero if n has a square divisor, -1 if n has an odd number of distinct prime divisors, and +1 in n has an even number of distinct prime divisors.

The largest value of n in which we are interested arises when 2^n = 10^{2600}, which is n=8638. Calculating primes and the möbius function to 8638 is fairly quick.

Calculating sums of n th powers can either be done by explicit calculation, or using Faulhaber’s formula, depending on how many nth powers are required. Formulae for the Bernoulli numbers from wiki or similar.

Integer n th roots can be calculated using a mix of logarithms, or iterative formulae.

Various numbers are needed repeatedly, so caching or memoizing of functions is useful.

I couldn’t answer the last group of test cases during the FEB16 competition. Since then I have

an untidy solution. Other answers seem to be smaller and faster, so must use better ideas.