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# Van Eck sequence

Van Eck sequence
You are encouraged to solve this task according to the task description, using any language you may know.

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`

...

1. Create a function/proceedure/method/subroutine/... to generate the Van Eck sequence of numbers.
1. The first ten terms of the sequence.
2. Terms 991 - to - 1000 of the sequence.
References

`with Ada.Text_IO; procedure Van_Eck_Sequence is    Sequence : array (Natural range 1 .. 1_000) of Natural;    procedure Calculate_Sequence is   begin      Sequence (Sequence'First) := 0;      for Index in Sequence'First .. Sequence'Last - 1 loop         Sequence (Index + 1) := 0;         for I in reverse Sequence'First .. Index - 1 loop            if Sequence (I) = Sequence (Index) then               Sequence (Index + 1) := Index - I;               exit;            end if;         end loop;      end loop;   end Calculate_Sequence;    procedure Show (First, Last : in Positive) is      use Ada.Text_IO;   begin      Put ("Element" & First'Image & " .." & Last'Image & " of Van Eck sequence: ");      for I in First .. Last loop         Put (Sequence (I)'Image);      end loop;      New_Line;   end Show; begin   Calculate_Sequence;   Show (First =>   1, Last =>    10);   Show (First => 991, Last => 1_000);end Van_Eck_Sequence;`
Output:
```Element 1 .. 10 of Van Eck sequence:  0 0 1 0 2 0 2 2 1 6
Element 991 .. 1000 of Van Eck sequence:  4 7 30 25 67 225 488 0 10 136```

## AppleScript

AppleScript is not the tool for the job, but here is a quick assembly from ready-made parts:

`use AppleScript version "2.4"use scripting additions  -- vanEck :: Int -> [Int]on vanEck(n)    script go        on |λ|(xxs)            maybe(0, elemIndex(item 1 of xxs, rest of xxs)) & xxs        end |λ|    end script    reverse of applyN(n - 1, go, {0})end vanEck  -- TEST ---------------------------------------------------on run    {vanEck(10), ¬        items 991 thru 1000 of vanEck(1000)}end run   -- GENERIC ------------------------------------------------ -- Just :: a -> Maybe aon Just(x)    -- Constructor for an inhabited Maybe (option type) value.    -- Wrapper containing the result of a computation.    {type:"Maybe", Nothing:false, Just:x}end Just  -- Nothing :: Maybe aon Nothing()    -- Constructor for an empty Maybe (option type) value.    -- Empty wrapper returned where a computation is not possible.    {type:"Maybe", Nothing:true}end Nothing  -- applyN :: Int -> (a -> a) -> a -> aon applyN(n, f, x)    script go        on |λ|(a, g)            |λ|(a) of mReturn(g)        end |λ|    end script    foldl(go, x, replicate(n, f))end applyN  -- elemIndex :: Eq a => a -> [a] -> Maybe Inton elemIndex(x, xs)    set lng to length of xs    repeat with i from 1 to lng        if x = (item i of xs) then return Just(i)    end repeat    return Nothing()end elemIndex  -- foldl :: (a -> b -> a) -> a -> [b] -> aon foldl(f, startValue, xs)    tell mReturn(f)        set v to startValue        set lng to length of xs        repeat with i from 1 to lng            set v to |λ|(v, item i of xs, i, xs)        end repeat        return v    end tellend foldl  -- maybe :: a -> Maybe a -> aon maybe(v, mb)    if Nothing of mb then        v    else        Just of mb    end ifend maybe  -- mReturn :: First-class m => (a -> b) -> m (a -> b)on mReturn(f)    -- 2nd class handler function lifted into 1st class script wrapper.     if script is class of f then        f    else        script            property |λ| : f        end script    end ifend mReturn  -- Egyptian multiplication - progressively doubling a list, appending-- stages of doubling to an accumulator where needed for binary -- assembly of a target length-- replicate :: Int -> a -> [a]on replicate(n, a)    set out to {}    if 1 > n then return out    set dbl to {a}     repeat while (1 < n)        if 0 < (n mod 2) then set out to out & dbl        set n to (n div 2)        set dbl to (dbl & dbl)    end repeat    return out & dblend replicate`
Output:
`{{0, 0, 1, 0, 2, 0, 2, 2, 1, 6}, {4, 7, 30, 25, 67, 225, 488, 0, 10, 136}}`

## AWK

` # syntax: GAWK -f VAN_ECK_SEQUENCE.AWK# converted from GoBEGIN {    limit = 1000    for (i=0; i<limit; i++) {      arr[i] = 0    }    for (n=0; n<limit-1; n++) {      for (m=n-1; m>=0; m--) {        if (arr[m] == arr[n]) {          arr[n+1] = n - m          break        }      }    }    printf("terms 1-10:")    for (i=0; i<10; i++) { printf(" %d",arr[i]) }    printf("\n")    printf("terms 991-1000:")    for (i=990; i<1000; i++) { printf(" %d",arr[i]) }    printf("\n")    exit(0)} `
Output:
```terms 1-10: 0 0 1 0 2 0 2 2 1 6
terms 991-1000: 4 7 30 25 67 225 488 0 10 136
```

## C

`#include <stdlib.h>#include <stdio.h> int main(int argc, const char *argv[]) {  const int max = 1000;  int *a = malloc(max * sizeof(int));  for (int n = 0; n < max - 1; n ++) {    for (int m = n - 1; m >= 0; m --) {      if (a[m] == a[n]) {        a[n+1] = n - m;        break;      }    }  }   printf("The first ten terms of the Van Eck sequence are:\n");  for (int i = 0; i < 10; i ++) printf("%d ", a[i]);  printf("\n\nTerms 991 to 1000 of the sequence are:\n");  for (int i = 990; i < 1000; i ++) printf("%d ", a[i]);  putchar('\n');   return 0;}`
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
```

## C++

`#include <iostream>#include <map> class van_eck_generator {public:    int next() {        int result = last_term;        auto iter = last_pos.find(last_term);        int next_term = (iter != last_pos.end()) ? index - iter->second : 0;        last_pos[last_term] = index;        last_term = next_term;        ++index;        return result;    }private:    int index = 0;    int last_term = 0;    std::map<int, int> last_pos;}; int main() {    van_eck_generator gen;    int i = 0;    std::cout << "First 10 terms of the Van Eck sequence:\n";    for (; i < 10; ++i)        std::cout << gen.next() << ' ';    for (; i < 990; ++i)        gen.next();    std::cout << "\nTerms 991 to 1000 of the sequence:\n";    for (; i < 1000; ++i)        std::cout << gen.next() << ' ';    std::cout << '\n';    return 0;}`
Output:
```First 10 terms of the Van Eck sequence:
0 0 1 0 2 0 2 2 1 6
Terms 991 to 1000 of the sequence:
4 7 30 25 67 225 488 0 10 136
```

## 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))))`
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)```

## Common Lisp

` ;;Tested using CLISP (defun VanEck (x) (reverse (VanEckh x 0 0 '(0)))) (defun VanEckh (final index curr lst)	(if (eq index final) 		lst		(VanEckh final (+ index 1) (howfar curr lst) (cons curr lst)))) (defun howfar (x lst) (howfarh x lst 0)) (defun howfarh (x lst runningtotal) 	(cond		((null lst) 0)		((eq x (car lst)) (+ runningtotal 1))		(t (howfarh x (cdr lst) (+ runningtotal 1))))) (format t "The first 10 elements are ~a~%" (VanEck 9))(format t "The 990-1000th elements are ~a~%" (nthcdr 990 (VanEck 999))) `
Output:
```The first 10 elements are (0 0 1 0 2 0 2 2 1 6)
The 990-1000th elements are (4 7 30 25 67 225 488 0 10 136)
```

## D

Translation of: Java
`import std.stdio; void vanEck(int firstIndex, int lastIndex) {    int[int] vanEckMap;    int last = 0;    if (firstIndex == 1) {        writefln("VanEck[%d] = %d", 1, 0);    }    for (int n = 2; n <= lastIndex; n++) {        int vanEck = last in vanEckMap ? n - vanEckMap[last] : 0;        vanEckMap[last] = n;        last = vanEck;        if (n >= firstIndex) {            writefln("VanEck[%d] = %d", n, vanEck);        }    }} void main() {    writeln("First 10 terms of Van Eck's sequence:");    vanEck(1, 10);    writeln;    writeln("Terms 991 to 1000 of Van Eck's sequence:");    vanEck(991, 1000);}`
Output:
```First 10 terms of Van Eck's sequence:
VanEck = 0
VanEck = 0
VanEck = 1
VanEck = 0
VanEck = 2
VanEck = 0
VanEck = 2
VanEck = 2
VanEck = 1
VanEck = 6

Terms 991 to 1000 of Van Eck's sequence:
VanEck = 4
VanEck = 7
VanEck = 30
VanEck = 25
VanEck = 67
VanEck = 225
VanEck = 488
VanEck = 0
VanEck = 10
VanEck = 136```

## Dyalect

Translation of: Go
`const max = 1000var a = Array.empty(max, 0)for n in 0..(max-2) {    var m = n - 1    while m >= 0 {        if a[m] == a[n] {            a[n+1] = n - m            break        }        m -= 1    }}print("The first ten terms of the Van Eck sequence are: \(a[0..10])")print("Terms 991 to 1000 of the sequence are: \(a[991..1000])")`
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: {7, 30, 25, 67, 225, 488, 0, 10, 136}```

## F#

### The function

` // Generate Van Eck's Sequence. Nigel Galloway: June 19th., 2019let ecK()=let n=System.Collections.Generic.Dictionary<int,int>()          Seq.unfold(fun (g,e)->Some(g,((if n.ContainsKey g then let i=n.[g] in n.[g]<-e;e-i else n.[g]<-e;0),e+1)))(0,0) `

First 50
` ecK() |> Seq.take 50 |> Seq.iter(printf "%d "); printfn "";; `
Output:
```0 0 1 0 2 0 2 2 1 6 0 5 0 2 6 5 4 0 5 3 0 3 2 9 0 4 9 3 6 14 0 6 3 5 15 0 5 3 5 2 17 0 6 11 0 3 8 0 3 3
```
50 from 991
` ecK() |> Seq.skip 990 |> Seq.take 50|> Seq.iter(printf "%d "); printfn "";; `
Output:
```4 7 30 25 67 225 488 0 10 136 61 0 4 12 72 0 4 4 1 24 41 385 0 7 22 25 22 2 84 68 282 464 0 10 25 9 151 697 0 6 41 20 257 539 0 6 6 1 29 465
```
I thought the longest sequence of non zeroes in the first 100 million items might be interesting

It occurs between 32381749 and 32381774:

```9 47 47 1 10 33 27 548 548 1 6 33 6 2 154 15657 695734 270964 235721 238076 4896139 655158 7901804 146089 977945 21475977
```

### Alternative recursive procedure

` open System.Collections.Genericlet VanEck() =    let rec _vanEck (num:int) (pos:int) (lastOccurence:Dictionary<int, int>) =        match lastOccurence.TryGetValue num with        | (true, position) ->            set num pos (pos - position) lastOccurence        | _ ->             set num pos 0 lastOccurence    and set num pos next lastOccurenceByNumber = seq {            lastOccurenceByNumber.[num] <- pos            yield next            yield! _vanEck next (pos + 1) lastOccurenceByNumber        }     seq {         yield 0        yield! _vanEck 0 1 (new Dictionary<int, int>())    } VanEck() |> Seq.take 10 |> Seq.map (sprintf "%i") |> String.concat " " |> printfn "The first ten terms of the sequence : %s"VanEck() |> Seq.skip 990 |> Seq.take 10 |> Seq.map (sprintf "%i") |> String.concat " " |> printfn "Terms 991 - to - 1000 of the sequence : %s" `

## Factor

`USING: assocs fry kernel make math namespaces prettyprintsequences ; : van-eck ( n -- seq )    [        0 , 1 - H{ } clone '[            building get [ length 1 - ] [ last ] bi _ 3dup            2dup key? [ at - ] [ 3drop 0 ] if , set-at        ] times    ] { } make ; 1000 van-eck 10 [ head ] [ tail* ] 2bi [ . ] [email protected]`
Output:
```{ 0 0 1 0 2 0 2 2 1 6 }
{ 4 7 30 25 67 225 488 0 10 136 }
```

## FreeBASIC

` Const limite = 1000 Dim As Integer a(limite), n, m, i For n = 0 To limite-1    For m = n-1 To 0 Step -1        If a(m) = a(n) Then a(n+1) = n-m: Exit For    Next mNext n Print "Secuencia de Van Eck:" &Chr(10)Print "Primeros 10 terminos: ";For i = 0 To 9    Print a(i) &" ";Next iPrint Chr(10) & "Terminos 991 al 1000: ";For i = 990 To 999    Print a(i) &" ";Next i End `
Output:
```Secuencia de Van Eck:

Primeros 10 terminos: 0 0 1 0 2 0 2 2 1 6
Terminos 991 al 1000: 4 7 30 25 67 225 488 0 10 136
```

## Fōrmulæ

Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text (more info). Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for transportation effects more than visualization and edition.

The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code.

## 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:])}`
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):

`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:])}`

`import Data.List (elemIndex)import Data.Maybe (maybe) vanEck :: Int -> [Int]vanEck n = reverse \$ iterate go [] !! n  where    go [] =     go xxs@(x:xs) = maybe 0 succ (elemIndex x xs) : xxs main :: IO ()main = do  print \$ vanEck 10  print \$ drop 990 (vanEck 1000)`
Output:
```[0,0,1,0,2,0,2,2,1,6]
[4,7,30,25,67,225,488,0,10,136]```

And if we wanted to look a little further than the 1000th term, we could accumulate a Map of most recently seen positions to improve performance:

`{-# LANGUAGE TupleSections #-} import qualified Data.Map.Strict as M hiding (drop)import Data.List (mapAccumL)import Data.Maybe (maybe) vanEck :: [Int]vanEck = 0 : snd (mapAccumL go (0, M.empty) [1 ..])  where    go (x, dct) i =      ((,) =<< (, M.insert x i dct)) (maybe 0 (i -) (M.lookup x dct)) main :: IO ()main =  mapM_ print \$  (drop . subtract 10 <*> flip take vanEck) <\$>  [10, 1000, 10000, 100000, 1000000]`
Output:
```[0,0,1,0,2,0,2,2,1,6]
[4,7,30,25,67,225,488,0,10,136]
[7,43,190,396,2576,3142,0,7,7,1]
[92,893,1125,47187,0,7,34,113,140,2984]
[8,86,172,8878,172447,0,6,30,874,34143]```

## J

### The tacit verb (function)

`VanEck=. (, (<:@:# - }: i: {:))^:(]`0:)`

### The output

`   VanEck 90 0 1 0 2 0 2 2 1 6    990 }. VanEck 9994 7 30 25 67 225 488 0 10 136`

### A structured derivation of the verb (function)

` next  =. <:@:# - }: i: {:    NB. Next term of the sequenceVanEck=. (, next)^:(]`0:) f. NB. Appending terms and fixing the verb`

## Java

Use map to remember last seen index. Computes each value of the sequence in O(1) time.

` import java.util.HashMap;import java.util.Map; public class VanEckSequence {     public static void main(String[] args) {        System.out.println("First 10 terms of Van Eck's sequence:");        vanEck(1, 10);        System.out.println("");        System.out.println("Terms 991 to 1000 of Van Eck's sequence:");        vanEck(991, 1000);    }     private static void vanEck(int firstIndex, int lastIndex) {        Map<Integer,Integer> vanEckMap = new HashMap<>();                int last = 0;        if ( firstIndex == 1 ) {            System.out.printf("VanEck[%d] = %d%n", 1, 0);        }        for ( int n = 2 ; n <= lastIndex ; n++ ) {            int vanEck = vanEckMap.containsKey(last) ? n - vanEckMap.get(last) : 0;            vanEckMap.put(last, n);            last = vanEck;            if ( n >= firstIndex ) {                System.out.printf("VanEck[%d] = %d%n", n, vanEck);            }        }     } } `
Output:
```First 10 terms of Van Eck's sequence:
VanEck = 0
VanEck = 0
VanEck = 1
VanEck = 0
VanEck = 2
VanEck = 0
VanEck = 2
VanEck = 2
VanEck = 1
VanEck = 6

Terms 991 to 1000 of Van Eck's sequence:
VanEck = 4
VanEck = 7
VanEck = 30
VanEck = 25
VanEck = 67
VanEck = 225
VanEck = 488
VanEck = 0
VanEck = 10
VanEck = 136
```

## JavaScript

Either declaratively, without premature optimization:

Translation of: Python
`(() => {    'use strict';     // vanEck :: Int -> [Int]    const vanEck = n =>        reverse(            churchNumeral(n)(                xs => 0 < xs.length ? cons(                    maybe(                        0, succ,                        elemIndex(xs, xs.slice(1))                    ),                    xs                ) :             )([])        );     // TEST -----------------------------------------------    const main = () => {        console.log('VanEck series:\n')        showLog('First 10 terms', vanEck(10))        showLog('Terms 991-1000', vanEck(1000).slice(990))    };     // GENERIC FUNCTIONS ----------------------------------     // Just :: a -> Maybe a    const Just = x => ({        type: 'Maybe',        Nothing: false,        Just: x    });     // Nothing :: Maybe a    const Nothing = () => ({        type: 'Maybe',        Nothing: true,    });     // churchNumeral :: Int -> (a -> a) -> a -> a    const churchNumeral = n => f => x =>        Array.from({            length: n        }, () => f)        .reduce((a, g) => g(a), x)     // cons :: a -> [a] -> [a]    const cons = (x, xs) => [x].concat(xs)     // elemIndex :: Eq a => a -> [a] -> Maybe Int    const elemIndex = (x, xs) => {        const i = xs.indexOf(x);        return -1 === i ? (            Nothing()        ) : Just(i);    };     // maybe :: b -> (a -> b) -> Maybe a -> b    const maybe = (v, f, m) =>        m.Nothing ? v : f(m.Just);     // reverse :: [a] -> [a]    const reverse = xs =>        'string' !== typeof xs ? (            xs.slice(0).reverse()        ) : xs.split('').reverse().join('');     // showLog :: a -> IO ()    const showLog = (...args) =>        console.log(            args            .map(JSON.stringify)            .join(' -> ')        );     // succ :: Int -> Int    const succ = x => 1 + x;     // MAIN ---    return main();})();`
Output:
```VanEck series:

"First 10 terms" -> [0,0,1,0,2,0,2,2,1,6]
"Terms 991-1000" -> [4,7,30,25,67,225,488,0,10,136]```

or as a map-accumulation, building a look-up table:

Translation of: Python
`(() => {    'use strict';     // vanEck :: Int -> [Int]    const vanEck = n =>        // First n terms of the vanEck series.        .concat(mapAccumL(            ([x, seen], i) => {                const                    prev = seen[x],                    v = 0 !== prev ? (                        i - prev                    ) : 0;                return [                    [v, (seen[x] = i, seen)], v                ];            }, [0, replicate(n - 1, 0)],             enumFromTo(1, n - 1)        ));     // TEST -----------------------------------------------    const main = () =>        console.log(fTable(            'Terms of the VanEck series:\n',            n => str(n - 10) + '-' + str(n),            xs => JSON.stringify(xs.slice(-10)),            vanEck,            [10, 1000, 10000]        ))      // GENERIC FUNCTIONS ----------------------------------     // enumFromTo :: Int -> Int -> [Int]    const enumFromTo = (m, n) =>        Array.from({            length: 1 + n - m        }, (_, i) => m + i);     // fTable :: String -> (a -> String) -> (b -> String) ->    //                     (a -> b) -> [a] -> String    const fTable = (s, xShow, fxShow, f, xs) => {        // Heading -> x display function ->        //           fx display function ->        //    f -> values -> tabular string        const            ys = xs.map(xShow),            w = Math.max(...ys.map(x => x.length));        return s + '\n' + zipWith(            (a, b) => a.padStart(w, ' ') + ' -> ' + b,            ys,            xs.map(x => fxShow(f(x)))        ).join('\n');    };     // Map-accumulation is a combination of map and a catamorphism;    // it applies a function to each element of a list, passing an accumulating    // parameter from left to right, and returning a final value of this    // accumulator together with the new list.     // mapAccumL :: (acc -> x -> (acc, y)) -> acc -> [x] -> (acc, [y])    const mapAccumL = (f, acc, xs) =>        xs.reduce((a, x, i) => {            const pair = f(a, x, i);            return [pair, a.concat(pair)];        }, [acc, []]);     // replicate :: Int -> a -> [a]    const replicate = (n, x) =>        Array.from({            length: n        }, () => x);     // str :: a -> String    const str = x => x.toString();     // zipWith :: (a -> b -> c) -> [a] -> [b] -> [c]    const zipWith = (f, xs, ys) => {        const            lng = Math.min(xs.length, ys.length),            as = xs.slice(0, lng),            bs = ys.slice(0, lng);        return Array.from({            length: lng        }, (_, i) => f(as[i], bs[i], i));    };     // MAIN ---    return main();})();`
Output:
```Terms of the VanEck series:

0-10 -> [0,0,1,0,2,0,2,2,1,6]
990-1000 -> [4,7,30,25,67,225,488,0,10,136]
9990-10000 -> [7,43,190,396,2576,3142,0,7,7,1]```

## Julia

`function vanecksequence(N, startval=0)    ret = zeros(Int, N)    ret = startval    for i in 1:N-1        lastseen = findlast(x -> x == ret[i], ret[1:i-1])        if lastseen != nothing            ret[i + 1] = i - lastseen        end    end    retend println(vanecksequence(10))println(vanecksequence(1000)[991:1000]) `
Output:
```[0, 0, 1, 0, 2, 0, 2, 2, 1, 6]
[4, 7, 30, 25, 67, 225, 488, 0, 10, 136]
```

Alternate version, with a Dict for memoization (output is the same):

`function vanecksequence(N, startval=0)    ret = zeros(Int, N)    ret = startval    lastseen = Dict{Int, Int}()    for i in 1:N-1        if haskey(lastseen, ret[i])            ret[i + 1] = i - lastseen[ret[i]]        end        lastseen[ret[i]] = i    end    retend `

## Kotlin

Translation of: Java
`fun main() {    println("First 10 terms of Van Eck's sequence:")    vanEck(1, 10)    println("")    println("Terms 991 to 1000 of Van Eck's sequence:")    vanEck(991, 1000)} private fun vanEck(firstIndex: Int, lastIndex: Int) {    val vanEckMap = mutableMapOf<Int, Int>()    var last = 0    if (firstIndex == 1) {        println("VanEck = 0")    }    for (n in 2..lastIndex) {        val vanEck = if (vanEckMap.containsKey(last)) n - vanEckMap[last]!! else 0        vanEckMap[last] = n        last = vanEck        if (n >= firstIndex) {            println("VanEck[\$n] = \$vanEck")        }    }}`
Output:
```First 10 terms of Van Eck's sequence:
VanEck = 0
VanEck = 0
VanEck = 1
VanEck = 0
VanEck = 2
VanEck = 0
VanEck = 2
VanEck = 2
VanEck = 1
VanEck = 6

Terms 991 to 1000 of Van Eck's sequence:
VanEck = 4
VanEck = 7
VanEck = 30
VanEck = 25
VanEck = 67
VanEck = 225
VanEck = 488
VanEck = 0
VanEck = 10
VanEck = 136```

## Lua

`-- Return a table of the first n values of the Van Eck sequencefunction vanEck (n)  local seq, foundAt = {0}  while #seq < n do    foundAt = nil    for pos = #seq - 1, 1, -1 do      if seq[pos] == seq[#seq] then        foundAt = pos        break      end    end    if foundAt then      table.insert(seq, #seq - foundAt)    else      table.insert(seq, 0)    end  end  return seqend -- Show the set of values in table t from key numbers lo to hifunction showValues (t, lo, hi)  for i = lo, hi do    io.write(t[i] .. " ")  end  print()end -- Main procedurelocal sequence = vanEck(1000)showValues(sequence, 1, 10)showValues(sequence, 991, 1000)`
Output:
```0 0 1 0 2 0 2 2 1 6
4 7 30 25 67 225 488 0 10 136```

## Nim

`const max = 1000var a: array[max, int]for n in countup(0, max - 2):  for m in countdown(n - 1, 0):    if a[m] == a[n]:      a[n + 1] = n - m      break echo "The first ten terms of the Van Eck sequence are:"echo \$a[..9]echo "\nTerms 991 to 1000 of the sequence are:"echo \$a[990..^1]`
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]
```

## Pascal

I memorize the last position of each number that occured and use a circular buffer to remember last values. Running once through the list of last positions maybe faster Try it online! takes only 1.4 s for 32,381,775

`program VanEck;{* 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.}uses  sysutils;const  MAXNUM = 32381775;//1000*1000*1000;  MAXSEENIDX = (1 shl 7)-1;var  PosBefore : array of UInt32;  LastSeen  : array[0..MAXSEENIDX]of UInt32;// circular buffer  SeenIdx,HaveSeen : Uint32; procedure OutSeen(Cnt:NativeInt);var  I,S_Idx : NativeInt;Begin  IF Cnt > MAXSEENIDX then    Cnt := MAXSEENIDX;  If  Cnt > HaveSeen  then    Cnt := HaveSeen;  S_Idx := SeenIdx;  S_Idx := (S_Idx-Cnt);  IF S_Idx < 0 then    inc(S_Idx,MAXSEENIDX);  For i := 1 to Cnt do  Begin    write(' ',LastSeen[S_Idx]);    S_Idx:= (S_Idx+1) AND MAXSEENIDX;  end;  writeln;end; procedure Test(MaxTestCnt:Uint32);var  i,actnum,Posi,S_Idx: Uint32;  pPosBef,pSeen :pUint32;Begin  Fillchar(LastSeen,SizeOf(LastSeen),#0);  HaveSeen := 0;  IF MaxTestCnt> MAXNUM then    EXIT;  //setlength and clear  setlength(PosBefore,0);  setlength(PosBefore,MaxTestCnt);   pPosBef := @PosBefore;  pSeen   := @LastSeen;  S_Idx := 0;  i := 1;  actnum := 0;  repeat    // save value    pSeen[S_Idx] := actnum;    S_Idx:= (S_Idx+1) AND MAXSEENIDX;    //examine new value often out of cache    Posi := pPosBef[actnum];    pPosBef[actnum] := i;//  if Posi=0 ? actnum = 0:actnum = i-Posi    IF Posi = 0 then      actnum := 0    else      actnum := i-Posi;    inc(i);  until i > MaxTestCnt;  HaveSeen := i-1;  SeenIdx := S_Idx;end; Begin  Test(10)  ; OutSeen(10000);  Test(1000); OutSeen(10);  Test(MAXNUM); OutSeen(28);  setlength(PosBefore,0);end.`
Output:
``` 0 0 1 0 2 0 2 2 1 6
4 7 30 25 67 225 488 0 10 136
0 9 47 47 1 10 33 27 548 548 1 6 33 6 2 154 15657 695734 270964 235721 238076 4896139 655158 7901804 146089 977945 21475977 0
```

## Perl

Translation of: Raku
`use strict;use warnings;use feature 'say'; sub van_eck {    my(\$init,\$max) = @_;    my(%v,\$k);    my @V = my \$i = \$init;    for (1..\$max) {        \$k++;        my \$t  = \$v{\$i} ? \$k - \$v{\$i} : 0;        \$v{\$i} = \$k;        push @V, \$i = \$t;    }    @V;} for (    ['A181391', 0],    ['A171911', 1],    ['A171912', 2],    ['A171913', 3],    ['A171914', 4],    ['A171915', 5],    ['A171916', 6],    ['A171917', 7],    ['A171918', 8],) {    my(\$seq, \$start) = @\$_;    my @seq = van_eck(\$start,1000);    say <<~"END";    Van Eck sequence OEIS:\$seq; with the first term: \$start            First 10 terms: @{[@seq[0  ..  9]]}    Terms 991 through 1000: @{[@seq[990..999]]}    END}`
Output:
```Van Eck sequence OEIS:A181391; with the first term: 0
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

Van Eck sequence OEIS:A171911; with the first term: 1
First 10 terms: 1 0 0 1 3 0 3 2 0 3
Terms 991 through 1000: 0 6 53 114 302 0 5 9 22 71

Van Eck sequence OEIS:A171912; with the first term: 2
First 10 terms: 2 0 0 1 0 2 5 0 3 0
Terms 991 through 1000: 8 92 186 0 5 19 41 413 0 5

Van Eck sequence OEIS:A171913; with the first term: 3
First 10 terms: 3 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 5 5 1 17 192 0 6 34 38 179

Van Eck sequence OEIS:A171914; with the first term: 4
First 10 terms: 4 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 33 410 0 6 149 0 3 267 0 3

Van Eck sequence OEIS:A171915; with the first term: 5
First 10 terms: 5 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 60 459 0 7 13 243 0 4 10 211

Van Eck sequence OEIS:A171916; with the first term: 6
First 10 terms: 6 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 6 19 11 59 292 0 6 6 1 12

Van Eck sequence OEIS:A171917; with the first term: 7
First 10 terms: 7 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 11 7 2 7 2 2 1 34 24 238

Van Eck sequence OEIS:A171918; with the first term: 8
First 10 terms: 8 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 16 183 0 6 21 10 249 0 5 48```

## Phix

Just like the pascal entry, instead of searching/dictionaries use a fast direct/parallel lookup table, and likewise this can easily create a 32-million-long table in under 2s.
While dictionaries are pretty fast, there is a huge overhead adding/updating millions of entries compared to a flat list of int.

`constant lim = 1000sequence van_eck = repeat(0,lim),         pos_before = repeat(0,lim)for n=1 to lim-1 do    integer vn = van_eck[n]+1,            prev = pos_before[vn]    if prev!=0 then        van_eck[n+1] = n - prev    end if    pos_before[vn] = nend forprintf(1,"The first ten terms of the Van Eck sequence are:%v\n",{van_eck[1..10]})printf(1,"Terms 991 to 1000 of the sequence are:%v\n",{van_eck[991..1000]})`
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}
```

## Prolog

Works with: SWI Prolog
`van_eck_init(v(0, 0, _assoc)):-    empty_assoc(_assoc). van_eck_next(v(Index, Last_term, Last_pos), v(Index1, Next_term, Last_pos1)):-    (get_assoc(Last_term, Last_pos, V) ->        Next_term is Index - V        ;        Next_term = 0    ),    Index1 is Index + 1,    put_assoc(Last_term, Last_pos, Index, Last_pos1). van_eck_sequence(N, Seq):-    van_eck_init(V),    van_eck_sequence(N, V, Seq). van_eck_sequence(0, _, []):-!.van_eck_sequence(N, V, [Term|Rest]):-    V = v(_, Term, _),    van_eck_next(V, V1),    N1 is N - 1,    van_eck_sequence(N1, V1, Rest). write_list(From, To, _, _):-    To < From,    !.write_list(_, _, _, []):-!.write_list(From, To, N, [_|Rest]):-    From > N,    !,    N1 is N + 1,    write_list(From, To, N1, Rest).write_list(From, To, N, [E|Rest]):-    writef('%t ', [E]),    F1 is From + 1,    N1 is N + 1,    write_list(F1, To, N1, Rest). write_list(From, To, List):-    write_list(From, To, 1, List),    nl. main:-    van_eck_sequence(1000, Seq),    writeln('First 10 terms of the Van Eck sequence:'),    write_list(1, 10, Seq),    writeln('Terms 991 to 1000 of the Van Eck sequence:'),    write_list(991, 1000, Seq).`
Output:
```First 10 terms of the Van Eck sequence:
0 0 1 0 2 0 2 2 1 6
Terms 991 to 1000 of the Van Eck sequence:
4 7 30 25 67 225 488 0 10 136
```

## Python

### Python: Using a dict

`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:])`
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: List based

The following alternative stores the sequence so far in a list `seen` rather than the first example that just stores last occurrences in a dict.

`def van_eck():    n = 0    seen =     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`
Output:

As before.

### Python: Composition of pure functions

As an alternative to the use of generators, a declarative definition in terms of a Church numeral function:

Works with: Python version 3.7
`'''Van Eck sequence''' from functools import reducefrom itertools import repeat  # vanEck :: Int -> [Int]def vanEck(n):    '''First n terms of the van Eck sequence.'''     return churchNumeral(n)(        lambda xs: cons(            maybe(0)(succ)(                elemIndex(xs)(xs[1:])            )        )(xs) if xs else     )([])[::-1]  # TEST ----------------------------------------------------def main():    '''Terms of the Van Eck sequence'''    print(        main.__doc__ + ':\n\n' +        'First 10: '.rjust(18, ' ') + repr(vanEck(10)) + '\n' +        '991 - 1000: '.rjust(18, ' ') + repr(vanEck(1000)[990:])    )  # GENERIC ------------------------------------------------- # Just :: a -> Maybe adef Just(x):    '''Constructor for an inhabited Maybe (option type) value.       Wrapper containing the result of a computation.    '''    return {'type': 'Maybe', 'Nothing': False, 'Just': x}  # Nothing :: Maybe adef Nothing():    '''Constructor for an empty Maybe (option type) value.       Empty wrapper returned where a computation is not possible.    '''    return {'type': 'Maybe', 'Nothing': True}  # churchNumeral :: Int -> (a -> a) -> a -> adef churchNumeral(n):    '''n applications of a function    '''    return lambda f: lambda x: reduce(        lambda a, g: g(a), repeat(f, n), x    )  # cons :: a -> [a] -> [a]def cons(x):    '''Construction of a list from a head and a tail.    '''    return lambda xs: [x] + xs  # elemIndex :: Eq a => a -> [a] -> Maybe Intdef elemIndex(x):    '''Just the index of the first element in xs       which is equal to x,       or Nothing if there is no such element.    '''    def go(xs):        try:            return Just(xs.index(x))        except ValueError:            return Nothing()    return go  # maybe :: b -> (a -> b) -> Maybe a -> bdef maybe(v):    '''Either the default value v, if m is Nothing,       or the application of f to x,       where m is Just(x).    '''    return lambda f: lambda m: v if None is m or m.get('Nothing') else (        f(m.get('Just'))    )  # succ :: Enum a => a -> adef succ(x):    '''The successor of a value.       For numeric types, (1 +).    '''    return 1 + x  # MAIN ---if __name__ == '__main__':    main()`
Output:
```Terms of the Van Eck sequence:

First 10: [0, 0, 1, 0, 2, 0, 2, 2, 1, 6]
991 - 1000: [4, 7, 30, 25, 67, 225, 488, 0, 10, 136]```

Or if we lose sight, for a moment, of the good advice of Donald Knuth, and fall into optimising more than is needed for the first 1000 terms, then we can define the vanEck series as a map accumulation over a range, with an array of positions as the accumulator.

`'''Van Eck series''' from functools import reducefrom itertools import repeat  # vanEck :: Int -> [Int]def vanEck(n):    '''First n terms of the vanEck sequence.'''    def go(xns, i):        (x, ns) = xns         prev = ns[x]        v = i - prev if 0 is not prev else 0        return (            (v, insert(ns, x, i)),            v        )     return  + mapAccumL(go)((0, list(repeat(0, n))))(        range(1, n)    )  # TEST ----------------------------------------------------# main :: IO ()def main():    '''The last 10 of the first N vanEck terms'''    print(        fTable(main.__doc__ + ':\n')(            lambda m: 'N=' + str(m), repr,            lambda n: vanEck(n)[-10:], [10, 1000, 10000]        )    )  # FORMATTING ----------------------------------------------# fTable :: String -> (a -> String) ->#                     (b -> String) -> (a -> b) -> [a] -> Stringdef fTable(s):    '''Heading -> x display function -> fx display function ->                     f -> xs -> tabular string.    '''    def go(xShow, fxShow, f, xs):        ys = [xShow(x) for x in xs]        w = max(map(len, ys))        return s + '\n' + '\n'.join(map(            lambda x, y: y.rjust(w, ' ') + ' -> ' + fxShow(f(x)),            xs, ys        ))    return go  # GENERIC ------------------------------------------------- # insert :: Array Int -> Int -> Int -> Array Intdef insert(xs, i, v):    '''An array updated at position i with value v.'''    xs[i] = v    return xs  # mapAccumL :: (acc -> x -> (acc, y)) -> acc -> [x] -> (acc, [y])def mapAccumL(f):    '''A tuple of an accumulation and a list derived by a       combined map and fold,       with accumulation from left to right.    '''    def go(a, x):        tpl = f(a, x)        return (tpl, a + [tpl])    return lambda acc: lambda xs: (        reduce(go, xs, (acc, []))    )  # MAIN ---if __name__ == '__main__':    main()`
Output:
```The last 10 of the first N vanEck terms:

N=10 -> [0, 0, 1, 0, 2, 0, 2, 2, 1, 6]
N=1000 -> [4, 7, 30, 25, 67, 225, 488, 0, 10, 136]
N=10000 -> [7, 43, 190, 396, 2576, 3142, 0, 7, 7, 1]```

## Racket

`#lang racket(require racket/stream) (define (van-eck)  (define (next val n seen)    (define val1 (- n (hash-ref seen val n)))    (stream-cons val (next val1 (+ n 1) (hash-set seen val n))))  (next 0 0 (hash))) (define (get m n s)  (stream->list   (stream-take (stream-tail s m)                (- n m)))) "First 10 terms:"          (get 0     10 (van-eck))"Terms 991 to 1000 terms:" (get 990 1000 (van-eck)) ; counting from 0`
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)
```

## Raku

(formerly Perl 6) There is not a Van Eck sequence, rather a series of related sequences that differ in their starting value. This task is nominally for the sequence starting with the value 0. This Raku implementation will handle any integer starting value.

Specifically handles:

among others.

Implemented as lazy, extendable lists.

`sub n-van-ecks (\$init) {    \$init, -> \$i, {        state %v;        state \$k;        \$k++;        my \$t  = %v{\$i}.defined ?? \$k - %v{\$i} !! 0;        %v{\$i} = \$k;        \$t    } ... *} for <    A181391 0    A171911 1    A171912 2    A171913 3    A171914 4    A171915 5    A171916 6    A171917 7    A171918 8> -> \$seq, \$start {     my @seq = n-van-ecks(\$start);     # The task    put qq:to/END/     Van Eck sequence OEIS:\$seq; with the first term: \$start            First 10 terms: {@seq[^10]}    Terms 991 through 1000: {@seq[990..999]}    END}`
Output:
```Van Eck sequence OEIS:A181391; with the first term: 0
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

Van Eck sequence OEIS:A171911; with the first term: 1
First 10 terms: 1 0 0 1 3 0 3 2 0 3
Terms 991 through 1000: 0 6 53 114 302 0 5 9 22 71

Van Eck sequence OEIS:A171912; with the first term: 2
First 10 terms: 2 0 0 1 0 2 5 0 3 0
Terms 991 through 1000: 8 92 186 0 5 19 41 413 0 5

Van Eck sequence OEIS:A171913; with the first term: 3
First 10 terms: 3 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 5 5 1 17 192 0 6 34 38 179

Van Eck sequence OEIS:A171914; with the first term: 4
First 10 terms: 4 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 33 410 0 6 149 0 3 267 0 3

Van Eck sequence OEIS:A171915; with the first term: 5
First 10 terms: 5 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 60 459 0 7 13 243 0 4 10 211

Van Eck sequence OEIS:A171916; with the first term: 6
First 10 terms: 6 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 6 19 11 59 292 0 6 6 1 12

Van Eck sequence OEIS:A171917; with the first term: 7
First 10 terms: 7 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 11 7 2 7 2 2 1 34 24 238

Van Eck sequence OEIS:A171918; with the first term: 8
First 10 terms: 8 0 0 1 0 2 0 2 2 1
Terms 991 through 1000: 16 183 0 6 21 10 249 0 5 48```

## REXX

### using a list

This REXX version allows the specification of the   start   and   end   of the   Van Eck   sequence   (to be displayed)   as
well as the initial starting element   (the default is zero).

`/*REXX pgm generates/displays the   'start ──► end'    elements of the Van Eck sequence.*/parse arg LO HI \$ .                              /*obtain optional arguments from the CL*/if LO=='' | LO==","  then LO=   1                /*Not specified?  Then use the default.*/if HI=='' | HI==","  then HI=  10                /* "      "         "   "   "     "    */if  \$=='' |  \$==","  then  \$=   0                /* "      "         "   "   "     "    */\$\$=;               z= \$                          /*\$\$: old seq:  \$: initial value of seq*/     do HI-1;      z= wordpos( reverse(z), reverse(\$\$) );          \$\$= \$;          \$= \$ z     end   /*HI-1*/                              /*REVERSE allows backwards search in \$.*/                                                 /*stick a fork in it,  we're all done. */say 'terms '  LO  " through "  HI  ' of the Van Eck sequence are: '  subword(\$,LO,HI-LO+1)`
output   when using the default inputs:
```terms  1  through  10  of the Van Eck sequence are:  0 0 1 0 2 0 2 2 1 6
```
output   when using the inputs of:     991   1000
```terms  991  through  1000  of the Van Eck sequence are:  4 7 30 25 67 225 488 0 10 136
```
output   when using the inputs of:     1   20   6
```terms  1  through  20  of the Van Eck sequence are:  6 0 0 1 0 2 0 2 2 1 6 10 0 6 3 0 3 2 9 0
```

### using a dictionary

This REXX version   (which uses a dictionary)   is about   20,000   times faster   (when using larger numbers)   than
using a list   (in finding the previous location of an "old" number (term).

`/*REXX pgm generates/displays the   'start ──► end'    elements of the Van Eck sequence.*/parse arg LO HI sta .                            /*obtain optional arguments from the CL*/if  LO=='' |  LO==","  then  LO=  1              /*Not specified?  Then use the default.*/if  HI=='' |  HI==","  then  HI= 10              /* "      "         "   "   "     "    */if sta=='' | sta==","  then sta=  0              /* "      "         "   "   "     "    */\$.0= sta;                    x= sta;      @.=.   /*\$.: the  Van Eck  sequence as a list.*/     do #=1 for HI-1                             /*X:  is the last term being examined. */     if @.x==.  then do;   @.x= #;        \$.#= 0;             x= 0;   end    /*new term.*/                else do;     z= # - @.x;  \$.#= z;   @.x= #;   x= z;   end    /*old term.*/     end   /*#*/                                 /*Z:  the new term being added to list.*/          LOw= LO - 1;     out= \$.LOw            /*initialize the output value.         */     do j=LO  to HI-1;     out= out \$.j          /*build a list for the output display. */     end   /*j*/                                 /*stick a fork in it,  we're all done. */say 'terms '     LO     " through "     HI    ' of the Van Eck sequence are: '     out`
output   is identical to the 1st REXX version.

## Ruby

`van_eck = Enumerator.new do |y|  ar =   loop do    y << (term = ar.last)  # yield    ar << (ar.count(term)==1 ? 0 : ar.size - 1 - ar[0..-2].rindex(term))  endend ve = van_eck.take(1000)p ve.first(10), ve.last(10) `
Output:
```[0, 0, 1, 0, 2, 0, 2, 2, 1, 6]
[4, 7, 30, 25, 67, 225, 488, 0, 10, 136]
```

## Rust

`fn van_eck_sequence() -> impl std::iter::Iterator<Item = i32> {    let mut index = 0;    let mut last_term = 0;    let mut last_pos = std::collections::HashMap::new();    std::iter::from_fn(move || {        let result = last_term;        let mut next_term = 0;        if let Some(v) = last_pos.get_mut(&last_term) {            next_term = index - *v;            *v = index;        } else {            last_pos.insert(last_term, index);        }        last_term = next_term;        index += 1;        Some(result)    })} fn main() {    let mut v = van_eck_sequence().take(1000);    println!("First 10 terms of the Van Eck sequence:");    for n in v.by_ref().take(10) {        print!("{} ", n);    }    println!("\nTerms 991 to 1000 of the Van Eck sequence:");    for n in v.skip(980) {        print!("{} ", n);    }    println!();}`
Output:
```First 10 terms of the Van Eck sequence:
0 0 1 0 2 0 2 2 1 6
Terms 991 to 1000 of the Van Eck sequence:
4 7 30 25 67 225 488 0 10 136
```

## 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)}.")} `
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).
```

## Sidef

`func van_eck(n) {     var seen = Hash()    var seq  =     var prev = seq[-1]     for k in (1 ..^ n) {        seq << (seen.has(prev) ? (k - seen{prev}) : 0)        seen{prev} = k        prev = seq[-1]    }     seq} say van_eck(10)say van_eck(1000).slice(991-1, 1000-1)`
Output:
```[0, 0, 1, 0, 2, 0, 2, 2, 1, 6]
[4, 7, 30, 25, 67, 225, 488, 0, 10, 136]
```

## Tcl

`## Mathematically, the first term has index "0", not "1".  We do that, also. set ::vE 0 proc vanEck {n} {    global vE vEocc    while {\$n >= [set k [expr {[llength \$vE] - 1}]]} {        set kv [lindex \$vE \$k]        ## value \$kv @ \$k is not yet stuffed into vEocc()        lappend vE [expr {[info exists vEocc(\$kv)] ? \$k - \$vEocc(\$kv) : 0}]        set vEocc(\$kv) \$k    }    return [lindex \$vE \$n]} proc show {func from to} {    for {set n \$from} {\$n <= \$to} {incr n} {        append r " " [\$func \$n]    }    puts "\${func}(\$from..\$to) =\$r"} show vanEck 0 9show vanEck 990 999`
Output:
```vanEck(0..9) = 0 0 1 0 2 0 2 2 1 6
vanEck(990..999) = 4 7 30 25 67 225 488 0 10 136
```

## Wren

Translation of: Go
`var max = 1000var a = List.filled(max, 0)var seen = {}for (n in 0...max-1) {    var m = seen[a[n]]    if (m != null) a[n+1] = n - m    seen[a[n]] = n}System.print("The first ten terms of the Van Eck sequence are:")System.print(a[0...10])System.print("\nTerms 991 to 1000 of the sequence are:")System.print(a[990..-1])`
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]
```

## zkl

Translation of: Raku
`fcn vanEck(startAt=0){	// --> iterator   (startAt).walker(*).tweak(fcn(n,seen,rprev){      prev,t := rprev.value, n - seen.find(prev,n);      seen[prev] = n;      rprev.set(t);      t   }.fp1(Dictionary(),Ref(startAt))).push(startAt)}`
`foreach n in (9){   ve:=vanEck(n);   println("The first ten terms of the Van Eck (%d) sequence are:".fmt(n));   println("\t",ve.walk(10).concat(","));   println("   Terms 991 to 1000 of the sequence are:");   println("\t",ve.drop(990-10).walk(10).concat(","));}`
Output:
```The first ten terms of the Van Eck (0) 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
The first ten terms of the Van Eck (1) sequence are:
1,0,0,1,3,0,3,2,0,3
Terms 991 to 1000 of the sequence are:
0,6,53,114,302,0,5,9,22,71
The first ten terms of the Van Eck (2) sequence are:
2,0,0,1,0,2,5,0,3,0
Terms 991 to 1000 of the sequence are:
8,92,186,0,5,19,41,413,0,5
The first ten terms of the Van Eck (3) sequence are:
3,0,0,1,0,2,0,2,2,1
Terms 991 to 1000 of the sequence are:
5,5,1,17,192,0,6,34,38,179
The first ten terms of the Van Eck (4) sequence are:
4,0,0,1,0,2,0,2,2,1
Terms 991 to 1000 of the sequence are:
33,410,0,6,149,0,3,267,0,3
The first ten terms of the Van Eck (5) sequence are:
5,0,0,1,0,2,0,2,2,1
Terms 991 to 1000 of the sequence are:
60,459,0,7,13,243,0,4,10,211
The first ten terms of the Van Eck (6) sequence are:
6,0,0,1,0,2,0,2,2,1
Terms 991 to 1000 of the sequence are:
6,19,11,59,292,0,6,6,1,12
The first ten terms of the Van Eck (7) sequence are:
7,0,0,1,0,2,0,2,2,1
Terms 991 to 1000 of the sequence are:
11,7,2,7,2,2,1,34,24,238
The first ten terms of the Van Eck (8) sequence are:
8,0,0,1,0,2,0,2,2,1
Terms 991 to 1000 of the sequence are:
16,183,0,6,21,10,249,0,5,48
```