nth number with n distinct prime factors

Question

A073329 is a sequence where the \$n\$th term is the \$n\$th number with exactly \$n\$ distinct prime factors. For example, \$a(3) = 60\$ as the first \$3\$ integers with \$3\$ distinct prime factors are \$30 = 2 \times 3 \times 5\$, \$42 = 2\times3\times7\$ and \$60 = 2^2 \times 3 \times 5\$.

You are to take a positive integer \$n\$ and output the first \$n\$ terms of the sequence. You may assume the input will never exceed the bounds of your language's integers. This is code-golf so the shortest code in bytes wins.

The first 10 terms of the sequence are

2, 10, 60, 420, 4290, 53130, 903210, 17687670, 406816410, 11125544430

Answer 1

Python, 144 chars

R=range
P=[x for x in R(2,99)if all(x%i for i in R(2,x))]
for a in R(input()):
 x=n=0
 while n<=a:n+=sum(x%p==0for p in P)==a+1;x+=1
 print x-1

It takes about 2 minutes to run to completion for x=8.

Answer 2

Java, 170 characters in one line

int a(int n) {
    int a = 2, t = a, m = 1, i = 1;
    Set s = new HashSet();
    while (i++ > 0) {
        if (t % i == 0) {
            s.add(i);
            t /= i;
            if (t == 1) {
                if (s.size() == n) {
                    if (n == m) {
                        break;
                    }
                    m++;
                }
                t = ++a;
                s.clear();
            }
            i = 1;
        }
    }
    return a;
}

Update, +77 characters IOL

int[] f(int n) {
    int[] f = new int[n];
    for (int i = 0; i < n; i++) {
        f[i] = a(i+1);
    }
    return f;
}

Answer 3

Java (Ungolfed)

public class M {

    public static void main(String[] a) {
        final int N = 100000000;
        int[] p = new int[N];
        for(int f = 2; f * f < N; f++) {
            if(p[f] == 0)
                for(int k = f; k < N; k += f)
                    p[k]++;
        }
        for(int i = 1; i <= 8; i++) {
            int c = 0, j;
            for(j = 1; j < N && c < i; j++)
                if(p[j] == i)
                    c++;
            if(c == i)
                System.out.println(i + " = " + --j);
        }
    }
}

Uses a sieve algorithm. It's pretty quick. (6 Seconds) Will work accurately for upto 8, will probably fail for anything higher.

Answer 4

JavaScript, 149 chars

function(n){function f(x){for(r=[],o=x,z=2;z<=o;x%z?++z:(x/=z,r.indexOf(z)+1?0:r.push(z)));return r}for(c=0,i=1;c<n;)f(++i).length==n?c++:0;return i}

Feels unresponsive for n >= 6 so I haven't tested how long it takes (my browser pops up a hung script notification every 10 seconds or so therefore I can't time it accurately and I don't want to completely hang if I check "don't show this again"...)

Edit: To return array is 200 characters (+51):

function(n){function F(){function f(x){for(r=[],o=x,z=2;z<=o;x%z?++z:(x/=z,r.indexOf(z)+1?0:r.push(z)));return r}for(c=0,i=1;c<n;)F(++i).length==n?c++:0;return i}for(a=[];n>0;n--)a.push(f());return a}

Answer 5

Jelly, 9 bytes

Æv=¥ḷ#)Ṫ€

Try it online!

How it works

Æv=¥ḷ#)Ṫ€ - Main link. Takes n on the left
      )   - For each integer 1 ≤ k ≤ n, so the following:
   ¥ḷ#    -   Find the first k integers for which the following is true:
Æv        -     The number of distinct prime factors
  =       -     equals k
       Ṫ€ - Return the last element from each list

Answer 6

J, 32 bytes

({"0 1~i.@#)(]/.~#@~.@q:)

But since I'm answering my own question so late, we'll just leave this answer as a curiosity.