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Rev

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What's next, Achilles?

Powerful numbers are positive integers such that, when expressed as a prime factorisation:

$$a = p_1^{e_1} \times p_2^{e_2} \times p_3^{e_3} \cdots \times p_k^{e_k}$$

all exponents \$e_1, e_2, ...\$ are greater than or equal to \2ドル\$. Note that the exponents do not include zero exponents, as exampled by \200ドル = 2^3 \times 3^0\times 5^2 = 2^3 \times 5^2\$ being a powerful number. This includes all perfect powers, and a few "extra" numbers that are not perfect powers.

Achilles numbers are powerful numbers that are not perfect powers. The smallest Achilles number is \72ドル = 2^3 \times 3^2\$. The Achilles numbers less than or equal to \500ドル\$ are \72,ドル 108, 200, 288, 392, 432\$ and \500ドル\$.

You are to take an Achilles number as input and output the smallest Achilles number greater than the input.

You may input and output in any convenient method. This is code-golf so the shortest code in bytes wins

Test Cases

input output
 72 108
 108 200
 200 288
 800 864
 1152 1323
 4500 4563
 3456 3528
 4563 4608
43808 43904
90828 91592
28800 29403
64800 64827
29768 30375

The program I used to generate these test cases. Contains spoilers for anyone who can understand Jelly.

Answer*

# [Jelly], 12 [bytes]

 ‘ÆEg/ḟ$f1Ʋ1#

A full program which prints the next Achilles number.

**[Try it online!][TIO-klhe5lwk]** Or see the [test-suite].

### How?

 ‘ÆEg/ḟ$f1Ʋ1# - Main Link: Archiles number (or any integer), n
 ‘ - increment (n)
 1# - find the first k (starting at k=n+1) for which:
 Ʋ - last four links as a monad - i.e. f(k):
 ÆE - prime exponent vactor (e.g k=400 -> [3,0,2] since 2^3+3^0+5^2)
 $ - last two links as a monad:
 / - reduce by:
 g - Greatest Common Divisor -> x
 ḟ - filter discard (from [x] if in ÆE result)
 1 - one
 f - filter keep (i.e. [1] if g/ was 1 and 1 not in ÆE result)
 (empty list is falsey, so we keep incrementing k until we get [1])

[Jelly]: https://github.com/DennisMitchell/jelly
[bytes]: https://github.com/DennisMitchell/jelly/wiki/Code-page

[TIO-klhe5lwk]: https://tio.run/##y0rNyan8//9Rw4zDba7p@g93zFdJMzy2yVD5////pgYGBgA "Jelly – Try It Online"

[test-suite]: https://tio.run/##y0rNyan8//9Rw4zDba7p@g93zFdJMzy2yVD5/9E9h9sfNa1xU3H//9/cSMfQwELHyMBAxwKIDQ1NjXRMTIEsYxNTMyDLzFjHxNgCqMLSwMIIqM4CpMrMBEQaWZqbWQAA "Jelly – Try It Online"

If this is an answer to a challenge…

…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one. Explanations of your answer make it more interesting to read and are very much encouraged.
…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.

More generally…

…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).

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1

\$\begingroup\$ Very clever use of filtering--even if it had occurred to me that I could count on the GCD being 1 if it matters to check if the number is powerful, I probably still wouldn't have come up with that. \$\endgroup\$

Unrelated String
– Unrelated String

2021年02月23日 13:37:31 +00:00
Commented Feb 23, 2021 at 13:37
\$\begingroup\$ If you’d like to save a byte on this almost three-year-old answer, you can replace f1Ʋ with ỊƊ: tio.run/#%23y0rNyan8//… \$\endgroup\$

Nick Kennedy
– Nick Kennedy

2023年10月29日 13:26:39 +00:00
Commented Oct 29, 2023 at 13:26
\$\begingroup\$ @NickKennedy that won't work even though it passes all the test cases. A counter-example would be an input of \2ドル^3 \times 5^4 = 5000\$ which should find \2ドル^2 \times 3^3 \times 7^2 = 5292\$ but will find \2ドル^6 \times 3^4 = 5184\$ because [2]Ị is [0] which is truthy. \$\endgroup\$

Jonathan Allan
– Jonathan Allan

2023年10月29日 18:31:57 +00:00
Commented Oct 29, 2023 at 18:31
1

\$\begingroup\$ Quite right. I thought I’d tested that but must have missed something! Nice solution by the way. \$\endgroup\$

Nick Kennedy
– Nick Kennedy

2023年10月29日 23:16:45 +00:00
Commented Oct 29, 2023 at 23:16

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