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Output the Nth term of the Van Eck Sequence.

Van Eck Sequence is defined as:

  • Starts with 0.
  • If the last term is the first occurrence of that term the next term is 0.
  • If the last term has occurred previously the next term is how many steps back was the most recent occurrence.

https://oeis.org/A181391

https://www.youtube.com/watch?v=etMJxB-igrc

https://www.youtube.com/watch?v=8VrnqRU7BVU

Sequence: 0,0,1,0,2,0,2,2,1,6,0,5,0,2,...

Tests:

Input | Output

  • 1 | 0
  • 8 | 2
  • 19 | 5
  • 27 | 9
  • 52 | 42
  • 64 | 0

EDIT

1 indexed is preferred, 0 indexed is acceptable; that might change some of the already submitted solutions.

Just the Nth term please.

Same (except for the seeing it already posted part), it seems code golfers and numberphile watchers have a decent overlap.

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  • 10
    \$\begingroup\$ Watched the numpherphile video at work and was going to post this when I got home. Curse you for getting there first. :P \$\endgroup\$ Jun 10 '19 at 21:27
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    \$\begingroup\$ Does it have to be 1-indexed, or may we use 0-indexing? \$\endgroup\$ Jun 10 '19 at 22:02
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    \$\begingroup\$ May we return or output the infinite sequence instead? \$\endgroup\$
    – Jo King
    Jun 11 '19 at 2:10
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    \$\begingroup\$ ... or the first n terms? \$\endgroup\$
    – Shaggy
    Jun 11 '19 at 15:11
  • \$\begingroup\$ @Draco18s The same thing happened to me. :P \$\endgroup\$
    – VFDan
    Jul 5 '19 at 16:31

32 Answers 32

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DC, 94 91 90 bytes

Input is taken during the program. Save this to a file and then do "dc " to run. Definitely not the shortest, but I have fun with challenges like these in dc. Input is 1-based index, as preferred.

[st1si0swlbxltlwlu1-sulu0!=m]sm[dlt=qSsli1+siz0!=b0siLs]sb[0pq]sf[lisw2Q]sq?2-dsu1>f0dlmxp

Main control macro
[st                         ]sm   save top value as target
[  1si0sw                   ]sm   reset i to 1 and w to 0
[        lbx                ]sm   execute macro b to get next value in w
[           ltlw            ]sm   restore target to the stack and add w to the stack
[               lu1-su      ]sm   decrement the user inputted variable
[                     lu0!=m]sm   if the user inputted variable is not 0 recurse

Next value finder macro
[dlt=q                  ]sb     if the value on the stack is the target, quit
[     Ss                ]sb     save top value to s register
[       li1+si          ]sb     increment i register
[             z0!=b     ]sb     recurse if still more values            
[                  0si  ]sb     set i to 0 (will be saved to w if relevant)
[                     Ls]sb     move top value of s register to stack

[lisw2Q]sq   Load i, save it to w, and then quit this macro and the one that called it

[0pq]sf print 0 and quit the program
```
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C++ (clang), 241 235 234 219 197 189 bytes

197 -> 189 bytes, thanks to ceilingcat

#import<bits/stdc++.h>
int f(int n,int i=0,std::vector<int>l={0}){return~-n>i?l.push_back(find(begin(l),end(l)-1,l[i])-end(l)+1?find(rbegin(l)+1,rend(l),l[i])-rbegin(l):0),f(n,i+1,l):l[i];}

Try it online!

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