New answers tagged atomic-code-golf
1
Here are a few solutions not written by me! If one of these is yours, please feel free to post it separately and I'll remove it from this answer. These have all been in this comment for a while, but I'm converting it to an answer for greater visibility.
Collected Answers
48 fracs by Chris Lomont –– SE profile, I think.
84 fracs by Hai-Anh Trinh –– SE ...
18
30 fractions
37^1/41^1
43^1/47^1
41^10*31^1*13^1/37^10*29^1
1/37^10
47^9*5^1/43^9*7^1
47^9/43^9*17^1
47^9/43^9*13^1
47^9/43^9*23^1
41^1*47^8/37^1*43^8*29^1
41^6*47^3/37^1*43^8*53^2
41^2*47^7*23^1*7^1/37^2*43^7*13^1*5^1
47^9*53^2/37^2*43^7*13^1
41^3*47^6*13^1/37^3*43^6*23^1
41^2*47^7*23^1*7^1*29^1/37^3*43^6*13^1*5^1
41^3*47^6/37^3*43^6*13^1
41^4*47^5*11^1/37^...
9
16 15 characters
6704196%(n*8+3)
One-indexed. A simple application of a generalisation of the Chinese Remainder Theorem, with the formula n*8+3 chosen to make it work (that is, where moduli have a common factor, the corresponding values' difference is also divisible by that factor), as well as to get a shorter number on the left.
Previously: 485559068%(n+41),...
8
15 bytes
n*4+703%n^3%7*3
Try it online!
Based off @Arnauld's answer.
The idea was to look for a magic formula A%n^B%C that maps (1, 2, 3, 4, 5, 6) to (0, 0, 1, 0, 1, 6) to combine the last two terms in Arnauld's formula into one:
n | 1 | 2 | 3 | 4 | 5 | 6 | (x4)
703%n^3%7 | 0 | 0 | 1 | 0 | 1 | 6 | (x3)
-------------+----+----+----+...
9
20 characters
Assuming that the precedence of the operators is similar to JS. The input is expected to be 1-indexed.
n*4+16%n*3+6*0^(6-n)
Try it online! (interpreted in JS ES7)
Breakdown
n | 1 | 2 | 3 | 4 | 5 | 6 | (x4)
16 % n | 0 | 0 | 1 | 0 | 1 | 4 | (x3)
0 ^ (6 - n) | 0 | 0 | 0 | 0 | 0 | 1 | (x6)
-------------+----+----...
Top 50 recent answers are included
