Talk:Numbers which are the cube roots of the product of their proper divisors

Hint

Currently one of the task implementations has the text "Whilst deceptively simple, the task reuires large integers as some of the divisor prioducts needed are large" -- but this is not actually the case, as the divisor products are not needed for this task. (Or at least the requirement for large integers is unnecessary -- I am agnostic about whether the task is deceptively simple.) --Rdm (talk) 15:20, 30 September 2022 (UTC)

Hi Rdm, Thanks for the hint. Before I posted the Algol 68 sample, I saw your comment that we just need to inspect the powers of the prime factors but at the time I didn't realise what you meant and not being au-fait with J, couldn't read your code to find out. However after googling stuff about cubes I now realise what you meant. I shall reword that comment.

Considering that 41^3 is greater than 65535, I wasn't going to attempt the task in a 16 bit language but I think I will now... thanks. --Tigerofdarkness (talk) 16:05, 30 September 2022 (UTC)

Hmm, it's just dawned on me that, despite obtaining the correct answer for the 50,000th term, the Wren solution is unsafe for n > 208063 as n^3 will then exceed 2^53 which is the maximum integer the language can safely deal with. I've changed it now so that it is safe. BTW I did intend the task to be simple but not deceptively so :) --PureFox (talk) 17:55, 30 September 2022 (UTC)