Problem Statement

Zsigmondy numbers n to a, b, are the greatest divisor of an - bn that is coprime to am - bm for all positive integers m < n.

Suppose we set a = 2 and b = 1. (Zs(n,2,1)) For each n, find the divisors of an - bn and return the largest that is coprime to all of am - bm, where m is each of the positive integers 1 to n - 1. When n = 4, 24 - 14 = 15. The divisors of 15 are 1, 3, 5, and 15. For m = 1, 2, 3 we get 2-1, 22-12, 23-13, or 1, 3, 7. The divisors of 15 that are coprime to each are 5 and 1, (1 is always included). The largest coprime divisor is 5, so Zs(4,2,1) = 5.

When n = 6, 26 - 16 = 63; its divisors are 1, 3, 7, 9, 21, 63. The largest divisor coprime to all of 1, 3, 7, 15, 31 is 1, (1 is always included), so Zs(6,2,1) = 1.

If a particular an - bn is prime, then Zs(n,a,b) will be equal to that prime. 25 - 15 = 31 so Zs(5,2,1) = 31.

Let's start with the solution:

func zsigmondy (a, b, c) {

    var aexp = (0..c -> map {|k| a**k })
    var bexp = (0..c -> map {|k| b**k })

    var z = []
    for n in (1 .. c) {
        divisors(aexp[n] - bexp[n]).flip.each {|d|
            1 ..^ n -> all {|k|
                aexp[k] - bexp[k] -> is_coprime(d)
            } && (z << d; break)
        }
    }
    return z
}

for name,a,b in ([
    'A064078: Zsigmondy(n,2,1)', 2,1,
    'A064079: Zsigmondy(n,3,1)', 3,1,
    'A064080: Zsigmondy(n,4,1)', 4,1,
    'A064081: Zsigmondy(n,5,1)', 5,1,
    'A064082: Zsigmondy(n,6,1)', 6,1,
    'A064083: Zsigmondy(n,7,1)', 7,1,
    'A109325: Zsigmondy(n,3,2)', 3,2,
    'A109347: Zsigmondy(n,5,3)', 5,3,
    'A109348: Zsigmondy(n,7,3)', 7,3,
    'A109349: Zsigmondy(n,7,5)', 7,5,
].slices(3)) {
    say ("\n#{name}:\n", zsigmondy(a, b, 20))
}