Van Eck sequence: Difference between revisions

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Revision as of 12:51, 28 August 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

AppleScript

AppleScript is not the tool for the job, but here is a quick assembly from ready-made parts: <lang applescript>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 a on 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 a on 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 -> a on 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 Int on 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] -> a on 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 tell

end foldl

-- maybe :: a -> Maybe a -> a on maybe(v, mb)

   if Nothing of mb then
       v
   else
       Just of mb
   end if

end 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 if

end 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 & dbl

end replicate</lang>

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 Go

BEGIN {

   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)

} </lang>

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

<lang 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;

}</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

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)

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))) </lang>

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)

Dyalect

Translation of: Go

<lang dyalect>const max = 1000 var 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])")</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: {7, 30, 25, 67, 225, 488, 0, 10, 136}

F#

This example is incomplete. F# Please ensure that it meets all task requirements and remove this message.

The function

<lang fsharp> // Generate Van Eck's Sequence. Nigel Galloway: June 19th., 2019 let 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)

</lang>

The Task

First 50

<lang fsharp> ecK() |> Seq.take 50 |> Seq.iter(printf "%d "); printfn "";; </lang>

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

<lang fsharp> ecK() |> Seq.skip 990 |> Seq.take 50|> Seq.iter(printf "%d "); printfn "";; </lang>

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

Factor

<lang factor>USING: assocs fry kernel make math namespaces prettyprint sequences ;

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 [ . ] bi@</lang>

Output:

{ 0 0 1 0 2 0 2 2 1 6 }
{ 4 7 30 25 67 225 488 0 10 136 }

FreeBASIC

<lang 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 m

Next n

Print "Secuencia de Van Eck:" &Chr(10) Print "Primeros 10 terminos: "; For i = 0 To 9

   Print a(i) &" ";

Next i Print Chr(10) & "Terminos 991 al 1000: "; For i = 990 To 999

   Print a(i) &" ";

Next i End </lang>

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

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>

Haskell

<lang haskell>import Data.List (elemIndex) import Data.Maybe (maybe)

vanEck :: Int -> [Int] vanEck n = reverse $ iterate go [] !! n

 where
   go [] = [0]
   go xxs@(x:xs) = maybe 0 succ (elemIndex x xs) : xxs

main :: IO () main = do

 print $ vanEck 10
 print $ drop 990 (vanEck 1000)</lang>

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:

<lang haskell>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 =
     let v = maybe 0 (i -) (M.lookup x dct)
     in ((v, M.insert x i dct), v)

main :: IO () main =

 mapM_ print $
 (drop . subtract 10 <*> flip take vanEck) <$>
 [10, 1000, 10000, 100000, 1000000]</lang>

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)

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

The output

<lang j> VanEck 9 0 0 1 0 2 0 2 2 1 6

  990 }. VanEck 999

4 7 30 25 67 225 488 0 10 136</lang>

A structured derivation of the verb (function)

JavaScript

Either declaratively, without premature optimization:

Translation of: Python

<lang JavaScript>(() => {

   'use strict';

   // vanEck :: Int -> [Int]
   const vanEck = n =>
       reverse(
           churchNumeral(n)(
               xs => 0 < xs.length ? cons(
                   maybe(
                       0, succ,
                       elemIndex(xs[0], xs.slice(1))
                   ),
                   xs
               ) : [0]
           )([])
       );

   // 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();

})();</lang>

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

<lang javascript>(() => {

   'use strict';

   // vanEck :: Int -> [Int]
   const vanEck = n =>
       // First n terms of the vanEck series.
       [0].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)
       )[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[0], x, i);
           return [pair[0], a[1].concat(pair[1])];
       }, [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();

})();</lang>

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

<lang julia>function vanecksequence(N, startval=0)

   ret = zeros(Int, N)
   ret[1] = 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
   ret

end

println(vanecksequence(10)) println(vanecksequence(1000)[991:1000])

</lang>

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): <lang julia>function vanecksequence(N, startval=0)

   ret = zeros(Int, N)
   ret[1] = 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
   ret

end </lang>

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 <lang pascal>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[0];
 pSeen   := @LastSeen[0];
 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.</lang>

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 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 Perl 6 implementation will handle any integer starting value.

Specifically handles:

among others.

Implemented as lazy, extendable lists.

<lang perl6>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 <

> -> $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

}</lang>

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. <lang Phix>constant lim = 1000 sequence 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] = n

end for printf(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]})</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}

Python

Python: Using a dict

<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: 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. <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.

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

<lang python>Van Eck sequence

from functools import reduce from 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[0])(xs[1:])
           )
       )(xs) if xs else [0]
   )([])[::-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 a

def 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 a

def 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 -> a

def 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 Int

def 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 lambda xs: go(xs)

maybe :: b -> (a -> b) -> Maybe a -> b

def 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 -> a

def succ(x):

   The successor of a value.
      For numeric types, (1 +).
   
   return 1 + x if isinstance(x, int) else (
       chr(1 + ord(x))
   )

MAIN ---

if __name__ == '__main__':

   main()</lang>

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.

<lang python>Van Eck series

from functools import reduce from 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 [0] + mapAccumL(go)((0, list(repeat(0, n))))(
       range(1, n)
   )[1]

TEST ----------------------------------------------------
main :: IO ()

def main():

   The last 10 of the first N vanEck terms

   print(
       fTable(main.__doc__ + ':\n')(
           lambda n: 'N=' + str(n)
       )(repr)(
           lambda n: vanEck(n)[-10:]
       )([10, 1000, 10000])
   )

FORMATTING ----------------------------------------------
fTable :: String -> (a -> String) ->
(b -> String) -> (a -> b) -> [a] -> String

def 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 lambda xShow: lambda fxShow: lambda f: lambda xs: go(
       xShow, fxShow, f, xs
   )

GENERIC -------------------------------------------------

insert :: Array Int -> Int -> Int -> Array Int

def 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[0], x)
       return (tpl[0], a[1] + [tpl[1]])
   return lambda acc: lambda xs: (
       reduce(go, xs, (acc, []))
   )

MAIN ---

if __name__ == '__main__':

   main()</lang>

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 </lang> Output: <lang Racket> "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) </lang>

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). <lang rexx>/*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)</lang>

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 times faster than using a list (in finding the previous
location of an "old" number (term). <lang rexx>/*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 /* " " " " " " */

                            x=$;       @.=.     /*$:  the Van Eck sequence as a list.  */
    do #=1  for HI                              /*X:  is the last term being examined. */
    if @.x==.  then do;   @.x= #;      $= $ 0;           x= 0;   end     /* a new term.*/
               else do;   z= # - @.x;  $= $ z;  @.x= #;  x= z;   end     /*an old term.*/
    end   /*#*/                                 /*Z:  the new term being added to list.*/
                                                /*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)</lang>

output is identical to the 1^st REXX version.

Ruby

<lang ruby>van_eck = Enumerator.new do |y|

 ar = [0]
 loop do
   y << (term = ar.last)  # yield
   ar << (ar.count(term)==1 ? 0 : ar.size - 1 - ar[0..-2].rindex(term))
 end

end

ve = van_eck.take(1000) p ve.first(10), ve.last(10) </lang>

Output:

[0, 0, 1, 0, 2, 0, 2, 2, 1, 6]
[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).

Sidef

<lang ruby>func van_eck(n) {

   var seen = Hash()
   var seq  = [0]
   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)</lang>

Output:

[0, 0, 1, 0, 2, 0, 2, 2, 1, 6]
[4, 7, 30, 25, 67, 225, 488, 0, 10, 136]

zkl

Translation of: Perl6

<lang zkl>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)

}</lang> <lang zkl>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(","));

}</lang>

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