prog.sf

#!/usr/bin/ruby

# Smallest m such that the m-th Lucas number has exactly n divisors that are also Lucas numbers.
# https://oeis.org/A356666

# Known terms:
#   1, 0, 3, 6, 15, 30, 45, 90, 105, 210, 405

# New terms:
#   1, 0, 3, 6, 15, 30, 45, 90, 105, 210, 405, 810, 315, 630, 3645, 2025, 945, 1890, 1575, 3150, 2835, 5670
#   1, 0, 3, 6, 15, 30, 45, 90, 105, 210, 405, 810, 315, 630, 3645, 2025, 945, 1890, 1575, 3150, 2835, 5670, 36450, 25025, 3465, 6930

# Using the PARI/GP program, finding the first 26 terms, took 5 hours.

func count_lucas_divisors(n) {
    var count = 0
    for k in (0..Inf) {
        var t = k.lucas
        if (t.divides(n)) {
            ++count
        }
        break if (t >= n)
    }
    return count
}

func a(n) {
    return 1 if (n == 1)
    return 0 if (n == 2)
    0..Inf -> first_by {|k|
        count_lucas_divisors(k.lucas) == n
    }
}

for n in (1..100) {
    say "#{n} #{a(n)}"
}

__END__

# PARI/GP program:

countLd(n) = my(c=0,x=2,y=1); while(x <= n, if(n%x == 0, c++); [x,y]=[y,x+y]); c;
a(n) = if(n==1, return(1)); my(k=0,x=2,y=1); while(1, if(countLd(x) == n, return(k)); [x,y,k]=[y,x+y,k+1]); \\ ~~~~