Van Eck sequence: Difference between revisions
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[4 7 30 25 67 225 488 0 10 136] |
[4 7 30 25 67 225 488 0 10 136] |
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</pre> |
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=={{header|Perl 6}}== |
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<lang perl6>sub next-van-eck ( $i ) { |
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state @v = 0; |
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my @key = @v[^*-1].grep( $i, :k ); |
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@v.push: +@key ?? +@v - @key.tail !! 0; |
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@v.tail |
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} |
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my @Van-Ecks = 0, *.&next-van-eck ... *; |
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put "Van-Eck sequence:\n First 10 terms: {@Van-Ecks[^10]}" ~ |
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"\nTerms 991 through 1000: {@Van-Ecks[990...999]}";</lang> |
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{{out}} |
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<pre>Van-Eck sequence: |
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First 10 terms: 0 0 1 0 2 0 2 2 1 6 |
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Terms 991 through 1000: 4 7 30 25 67 225 488 0 10 136</pre> |
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=={{header|Python}}== |
=={{header|Python}}== |
Revision as of 18:42, 17 June 2019
The sequence is generated by following this pseudo-code:
A: The first term is zero. Repeatedly apply: If the last term is *new* to the sequence so far then: B: The next term is zero. Otherwise: C: The next term is how far back this last term occured previousely.
- Example
Using A:
0
Using B:
0 0
Using C:
0 0 1
Using B:
0 0 1 0
Using C: (zero last occured two steps back - before the one)
0 0 1 0 2
Using B:
0 0 1 0 2 0
Using C: (two last occured two steps back - before the zero)
0 0 1 0 2 0 2 2
Using C: (two last occured one step back)
0 0 1 0 2 0 2 2 1
Using C: (one last appeared six steps back)
0 0 1 0 2 0 2 2 1 6
...
- Task
- Create a function/proceedure/method/subroutine/... to generate the Van Eck sequence of numbers.
- Use it to display here, on this page:
- The first ten terms of the sequence.
- Terms 991 - to - 1000 of the sequence.
- References
- Don't Know (the Van Eck Sequence) - Numberphile video.
- Wikipedia Article: Van Eck's Sequence.
- OEIS sequence: A181391.
Go
<lang go>package main
import "fmt"
func main() {
const max = 1000 a := make([]int, max) // all zero by default a[0] = 0 for n := 0; n < max-1; n++ { for m := n - 1; m >= 0; m-- { if a[m] == a[n] { a[n+1] = n - m break } } } fmt.Println("The first ten terms of the Van Eck sequence are:") fmt.Println(a[:10]) fmt.Println("\nTerms 991 to 1000 of the sequence are:") fmt.Println(a[990:])
}</lang>
- Output:
The first ten terms of the Van Eck sequence are: [0 0 1 0 2 0 2 2 1 6] Terms 991 to 1000 of the sequence are: [4 7 30 25 67 225 488 0 10 136]
Perl 6
<lang perl6>sub next-van-eck ( $i ) {
state @v = 0; my @key = @v[^*-1].grep( $i, :k ); @v.push: +@key ?? +@v - @key.tail !! 0; @v.tail
}
my @Van-Ecks = 0, *.&next-van-eck ... *;
put "Van-Eck sequence:\n First 10 terms: {@Van-Ecks[^10]}" ~
"\nTerms 991 through 1000: {@Van-Ecks[990...999]}";</lang>
- Output:
Van-Eck sequence: First 10 terms: 0 0 1 0 2 0 2 2 1 6 Terms 991 through 1000: 4 7 30 25 67 225 488 0 10 136
Python
<lang python>def van_eck():
n, seen, val = 0, {}, 0 while True: yield val last = {val: n} val = n - seen.get(val, n) seen.update(last) n += 1
- %%
if __name__ == '__main__':
print("Van Eck: first 10 terms: ", list(islice(van_eck(), 10))) print("Van Eck: terms 991 - 1000:", list(islice(van_eck(), 1000))[-10:])</lang>
- Output:
Van Eck: first 10 terms: [0, 0, 1, 0, 2, 0, 2, 2, 1, 6] Van Eck: terms 991 - 1000: [4, 7, 30, 25, 67, 225, 488, 0, 10, 136]
Python: Alternate
The following stores the sequence so far in a list seen
rather than the first example that just stores last occurrences in a dict.
<lang python>def van_eck():
n = 0 seen = [0] val = 0 while True: yield val if val in seen[1:]: val = seen.index(val, 1) else: val = 0 seen.insert(0, val) n += 1</lang>
- Output:
As before.