Greatest prime dividing the n-th cubefree number

Greatest prime dividing the n-th cubefree number is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

Definitions

A cubefree number is a positive integer whose prime factorization does not contain any third (or higher) power factors. If follows that all primes are trivially cubefree and the first cubefree number is 1 because it has no prime factors.

Let a[n] be the greatest prime dividing the n-th cubefree number for n >= 2. By convention, let a[1] = 1 even though the first cubefree number, 1, has no prime factors.

Examples

a[2] is clearly 2 because it is the second cubefree number and also prime. The fourth cubefree number is 4 and it's highest prime factor is 2, hence a[4] = 2.

Task

Compute and show on this page the first 100 terms of a[n]. Also compute and show the 1,000th, 10,000th and 100,000th members of the sequence.

Stretch

Compute and show the 1 millionth and 10 millionth terms of the sequence.

This may take a while for interpreted languages.

Reference

• OEIS sequence: A370833: a(n) is the greatest prime dividing the n-th cubefree number, for n >= 2; a(1)=1.

RPL

I confirm it does take a while for an interpreted language like RPL. Getting the 100,000th term is likely to be a question of hours, even on an emulator.

Works with: HP version 49

« 1 { } DUP2 + → n res2 res1
  « 2
    WHILE 'n' INCR 10000 ≤ REPEAT
      WHILE DUP FACTORS DUP 1 « 3 < NSUB 2 MOD NOT OR » DOSUBS ΠLIST NOT 
      REPEAT DROP 1 + END
      HEAD
      CASE 
         n 100 ≤      THEN 'res1' OVER STO+ END
         n LOG FP NOT THEN 'res2' OVER STO+ END
      END
      DROP 1 +
    END
    DROP res1 res2
» » 'TASK' STO 

Output:

2: { 1 2 3 2 5 3 7 3 5 11 3 13 7 5 17 3 19 5 7 11 23 5 13 7 29 5 31 11 17 7 3 37 19 13 41 7 43 11 5 23 47 7 5 17 13 53 11 19 29 59 5 61 31 7 13 11 67 17 23 7 71 73 37 5 19 11 13 79 41 83 7 17 43 29 89 5 13 23 31 47 19 97 7 11 5 101 17 103 7 53 107 109 11 37 113 19 23 29 13 59 }
1: { 109 101 }

Wren

Library: Wren-math

Library: Wren-fmt

This takes just under 4 minutes to run on my system (Core i7).

Although it looks odd that the 1 millionth and 10 millionth terms are the same (and I have no independent source to check against), it would appear that the 1 millionth cubefree number is 1,202,057 which is prime and the 10 millionth such number is coincidentally 10 times this.

import "./math" for Int
import "./fmt" for Fmt

var res = [1]
var count = 1
var i = 2
var lim1 = 100
var lim2 = 1000
var max = 1e7
while (count < max) {
    var cubeFree = false
    var factors = Int.primeFactors(i)
    if (factors.count < 3) {
        cubeFree = true
    } else {
        cubeFree = true
        for (i in 2...factors.count) {
            if (factors[i-2] == factors[i-1] && factors[i-1] == factors[i]) {
                cubeFree = false
                break
            }
        }
    }
    if (cubeFree) {
        if (count < lim1) res.add(factors[-1])
        count = count + 1
        if (count == lim1) {
            System.print("First %(lim1) terms of a[n]:")
            Fmt.tprint("$3d", res, 10)
        } else if (count == lim2) {
            Fmt.print("\nThe $,r term of a[n] is $,d.", count, factors[-1])
            lim2 = lim2 * 10
        }
    }
    i = i + 1
}

Output:

First 100 terms of a[n]:
  1   2   3   2   5   3   7   3   5  11
  3  13   7   5  17   3  19   5   7  11
 23   5  13   7  29   5  31  11  17   7
  3  37  19  13  41   7  43  11   5  23
 47   7   5  17  13  53  11  19  29  59
  5  61  31   7  13  11  67  17  23   7
 71  73  37   5  19  11  13  79  41  83
  7  17  43  29  89   5  13  23  31  47
 19  97   7  11   5 101  17 103   7  53
107 109  11  37 113  19  23  29  13  59

The 1,000th term of a[n] is 109.

The 10,000th term of a[n] is 101.

The 100,000th term of a[n] is 1,693.

The 1,000,000th term of a[n] is 1,202,057.

The 10,000,000th term of a[n] is 1,202,057.