Talk:Numbers which are the cube roots of the product of their proper divisors: Difference between revisions
Talk:Numbers which are the cube roots of the product of their proper divisors (view source)
Revision as of 21:58, 30 September 2022
, 1 year agoComments on comments, hints and OEIS...
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: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. --[[User:Tigerofdarkness|Tigerofdarkness]] ([[User talk: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 :) --[[User:PureFox|PureFox]] ([[User talk:PureFox|talk]]) 17:55, 30 September 2022 (UTC)
:::Thanks, both.
:::Purefox, I noticed your comment on the newer Wren solution and checked the OEIS page and it does indeed say that the sequence is 1 and the numbers with 8 divisors! So it really is simple and divisor products, cubes, cube roopts etc. are not needed.
:::BTW, I didn't mean "deceptively simple" as an insult... It started out looking simple but when the magnitude of the divisor products became apparent, it started to look hard but (as with a lot of things) it is actually simple if you use the right technique (i.e. someone shows you a better approach). --[[User:Tigerofdarkness|Tigerofdarkness]] ([[User talk:Tigerofdarkness|talk]]) 21:58, 30 September 2022 (UTC)
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