generate.sf

#!/usr/bin/ruby

# Number of primitive abundant numbers (A071395) < 10^n.
# https://oeis.org/A306986

# Known terms:
#   0, 3, 14, 98, 441, 1734, 8667, 41653, 213087, 1123424

#var PERFECT = [6, 28, 496, 8128, 33550336, 8589869056, 137438691328, 2305843008139952128, 2658455991569831744654692615953842176].grep{ _ <= 1e10 }

func f(n, q, limit) {

    #PERFECT.none { n.is_div(_) } || return false

    var count = 0

    for(var p = q; true; p.next_prime!) {
        var t = n*p
        break if (t >= limit)
        # This includes terms with perfect divisors
        if (t.is_primitive_abundant) {
            count += 1
        }
        else {
            count += f(t, p, limit)
        }
    }

    return count
}

say f(1, 2, 1e4)

__END__

# PARI/GP programs

# Generate terms

prim_abundant(n, q, limit) = my(list=List()); forprime(p=q, oo, my(t = n*p); if(t >= limit, break); if(sigma(t) > 2*t, my(F=factor(t)[, 1], ok=1); for(i=1, #F, if(sigma(t\F[i], -1) > 2, ok=0; break)); if(ok, listput(list, t)), list = concat(list, prim_abundant(n*p, p, limit)))); list;

# Count only

prim_abundant(limit, n=1, q=2) = my(count=0); forprime(p=q, oo, my(t = n*p); if(t >= limit, break); if(sigma(t) > 2*t, my(F=factor(t)[, 1], ok=1); for(i=1, #F, if(sigma(t\F[i], -1) >= 2, ok=0; break)); if(ok, count += 1), count += prim_abundant(limit, n*p, p))); count;
a(n) = prim_abundant(10^n);