Van Eck sequence

From Rosetta Code
Task
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

...

Task
  1. Create a function/proceedure/method/subroutine/... to generate the Van Eck sequence of numbers.
  2. Use it to display here, on this page:
  1. The first ten terms of the sequence.
  2. Terms 991 - to - 1000 of the sequence.
References


AppleScript[edit]

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

AWK[edit]

 
# 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)
}
 
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

Clojure[edit]

(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[edit]

 
;;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)

F#[edit]

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

The function[edit]

 
// 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)
 

The Task[edit]

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

Factor[edit]

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 [ . ] [email protected]
Output:
{ 0 0 1 0 2 0 2 2 1 6 }
{ 4 7 30 25 67 225 488 0 10 136 }

FreeBASIC[edit]

 
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
 
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[edit]

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:])
}

Haskell[edit]

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)
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:

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]
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[edit]

The tacit verb (function)[edit]

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

The output[edit]

   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

A structured derivation of the verb (function)[edit]

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

JavaScript[edit]

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[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();
})();
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.
[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();
})();
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[edit]

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])
 
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[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
 

Pascal[edit]

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[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.
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[edit]

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.

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

Phix[edit]

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 = 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]})
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[edit]

Python: Using a dict[edit]

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[edit]

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 = [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
Output:

As before.

Python: Composition of pure functions[edit]

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 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()
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 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()
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]

REXX[edit]

using a list[edit]

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[edit]

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).

/*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)
output   is identical to the 1st REXX version.


Ruby[edit]

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

Scala[edit]

 
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[edit]

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

zkl[edit]

Translation of: Perl6
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