Van Eck sequence
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.
Clojure
<lang clojure>(defn van-eck
([] (van-eck 0 0 {})) ([val n seen] (lazy-seq (cons val (let [next (- n (get seen val n))] (van-eck next (inc n) (assoc seen val n)))))))
(println "First 10 terms:" (take 10 (van-eck))) (println "Terms 991 to 1000 terms:" (take 10 (drop 990 (van-eck))))</lang>
- Output:
First 10 terms: (0 0 1 0 2 0 2 2 1 6) Terms 991 to 1000 terms: (4 7 30 25 67 225 488 0 10 136)
Go
<lang go>package main
import "fmt"
func main() {
const max = 1000 a := make([]int, max) // all zero by default 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]
Alternatively, using a map to store the latest index of terms previously seen (output as before): <lang go>package main
import "fmt"
func main() {
const max = 1000 a := make([]int, max) // all zero by default seen := make(map[int]int) for n := 0; n < max-1; n++ { if m, ok := seen[a[n]]; ok { a[n+1] = n - m } seen[a[n]] = n } 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>
Perl 6
Implemented as a lazy, extendable list.
<lang perl6>sub next-van-eck ( $i, $k ) {
state %v; my $t = %v{$i}.defined ?? $k - %v{$i} !! 0; %v{$i} = $k; $t
}
my @Van-Ecks = 0, *.&next-van-eck($++) ... *;
- The task
put "Van-Eck sequence:
First 10 terms: {@Van-Ecks[^10]}
Terms 991 through 1000: {@Van-Ecks[990...999]}";
- And just for the heck of it.
for 1e3, 1e4, 1e5 {
my ($k, $v) = @Van-Ecks.first: * > $_, :kv; printf "%22s: %d at position %d\n", "First term > $_", $v, $k+1;
}</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 First term > 1000: 1024 at position 1150 First term > 10000: 10381 at position 11603 First term > 100000: 100881 at position 104511
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.
REXX
This REXX version allows the start and end of two Van Eck sequences (to be displayed). <lang rexx>/*REXX program generates/displays two 'start' to 'end' elements of the Van Eck sequence.*/ parse arg s.1 e.1 s.2 e.2 . /*obtain optional arguments from the CL*/ if s.1== | s.1=="," then s.1= 1 /*Not specified? Then use the default.*/ if e.1== | e.1=="," then e.1= 10 /* " " " " " " */ if s.2== | s.2=="," then s.2= 991 /* " " " " " " */ if e.2== | e.2=="," then e.2= 1000 /* " " " " " " */
mx= max(e.1, e.2) /*MX: is the largest (range) "end". */
$= 0 /*the initial (start) value of the seq.*/
do j=1 for mx; $= $ wordpos( reverse( word($, j)), reverse( subword($, 1, j-1)) ) end /*j*/
do r=1 for 2; L= length(mx) /*L: is the width of the largest "end"*/ say 'terms' right(s.r,L) " through " right(e.r,L)': ' subword($, s.r, e.r-s.r+1) end /*r*/ /*stick a fork in it, we're all done. */</lang>
- output when using the default inputs:
terms 1 through 10: 0 0 1 0 2 0 2 2 1 6 terms 991 through 1000: 4 7 30 25 67 225 488 0 10 136
Scala
<lang scala> object VanEck extends App {
def vanEck(n: Int): List[Int] = {
def vanEck(values: List[Int]): List[Int] = if (values.size < n) vanEck(math.max(0, values.indexOf(values.head, 1)) :: values) else values
vanEck(List(0)).reverse }
val vanEck1000 = vanEck(1000) println(s"The first 10 terms are ${vanEck1000.take(10)}.") println(s"Terms 991 to 1000 are ${vanEck1000.drop(990)}.")
} </lang>
- Output:
The first 10 terms are List(0, 0, 1, 0, 2, 0, 2, 2, 1, 6). Terms 991 to 1000 are List(4, 7, 30, 25, 67, 225, 488, 0, 10, 136).