Thinking about it a bit further there are many more perfect cubes than that
because integers such as 8^8 would not be captured by the %3 rule, but are
still perfect cubes.
My new answer which is certainly brute force.
+/@:(=<.)@:(3&%:) n^n NB. take cube root of each n^n then set each
one that is an integer to 1 and then total up the 1's in the list
985
100 timespacex '+/@:(=<.)@:(3&%:) n^n'
5.526e_5 29056
Cheers, bob
> On Mar 20, 2021, at 20:14, 'robert therriault' via Programming
> <programm...@jsoftware.com> wrote:
>
> Same reasoning as Raul, but quicker and uses less space because we make use
> of the structure of n
>
> (1 + [: <. 3 %~ {:) n
> 334
> 100 timespacex '1++//0=3 |n'
> 2.69e_5 11200
> 100 timespacex '(1 +[: <. 3 %~ {:) n'
> 5.5e_7 1728
>
> Cheers, bob
>
>> On Mar 20, 2021, at 20:01, Raul Miller <rauldmil...@gmail.com> wrote:
>>
>> 1++/0=3 | n
>> 334
>>
>> p is a perfect cube if n is 1 or if n is a multiple of 3.
>>
>> (1 is the only whole power of 3 which is not a multiple of 3.)
>>
>> FYI,
>>
>> --
>> Raul
>>
>> On Sat, Mar 20, 2021 at 10:58 PM Skip Cave <s...@caveconsulting.com> wrote:
>>>
>>> How do you solve this problem using J (brute force)
>>> n =. >: i. 1000
>>> p =. n^n
>>> How many p are perfect cubes?
>>>
>>> Skip Cave
>>> Cave Consulting LLC
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