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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 previously.


Example

Using A:

0

Using B:

0 0

Using C:

0 0 1

Using B:

0 0 1 0

Using C: (zero last occurred two steps back - before the one)

0 0 1 0 2

Using B:

0 0 1 0 2 0

Using C: (two last occurred two steps back - before the zero)

0 0 1 0 2 0 2 2

Using C: (two last occurred 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/procedure/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



11l[edit]

Translation of: Python: List based
F van_eck(c)
[Int] r
V n = 0
V seen = [0]
V val = 0
L
r.append(val)
I r.len == c
R r
I val C seen[1..]
val = seen.index(val, 1)
E
val = 0
seen.insert(0, val)
n++
 
print(‘Van Eck: first 10 terms: ’van_eck(10))
print(‘Van Eck: terms 991 - 1000: ’van_eck(1000)[(len)-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]

8080 Assembly[edit]

	org	100h
lxi h,ecks ; Zero out 2000 bytes
lxi b,0
lxi d,2000
zero: mov m,b
inx h
dcx d
mov a,d
ora e
jnz zero
lxi b,-1 ; BC = Outer loop variable
outer: inx b
mvi a,3 ; Are we there yet? 1000 = 03E8h
cmp b ; Compare high byte
jnz go
mvi a,0E8h ; Compare low byte
cmp c
jz done
go: mov d,b ; DE = Inner loop variable
mov e,c
inner: dcx d
mov a,d ; <= 0?
ral
jc outer
push b ; Keep both pointers
push d
mov h,b ; Load BC = eck[BC]
mov l,c
call eck
mov c,m
inx h
mov b,m
xchg ; Load HL = -eck[DE]
call eck
xchg
ldax d
cma
mov l,a
inx d
ldax d
cma
mov h,a
inx h ; Two's complement
dad b ; -eck[DE] + eck[BC]
mov a,h ; Unfortunately this does not set flags
ora l ; Check zero
pop d ; Meanwhile, restore the pointers
pop b
jnz inner ; If no match, continue with inner loop
mov h,b ; If we _did_, then get &eck[BC + 1]
mov l,c
inx h
call eck
mov a,c ; Store BC - DE at that address
sub e
mov m,a
inx h
mov a,b
sbb d
mov m,a
jmp outer ; And continue the outer loop
done: lxi h,0 ; Print first 10 terms
call p10
lxi h,990 ; Print last 10 terms
p10: mvi b,10 ; Print 10 terms starting at term HL
call eck
ploop: mov e,m ; Load term into DE
inx h
mov d,m
inx h
push b ; Keep counter
push h ; Keep pointer
xchg ; Term in HL
call printn ; Print term
pop h ; Restore pointer and counter
pop b
dcr b
jnz ploop
lxi d,nl ; Print a newline afterwards
jmp prints
eck: push b ; Set HL = &eck[HL]
lxi b,ecks ; Base address
dad h ; Multiply by two
dad b ; Add base
pop b
ret
printn: lxi d,buf ; Print the number in HL
push d ; Buffer pointer on stack
lxi b,-10 ; Divisor
pdigit: lxi d,-1 ; Quotient
pdiv: inx d
dad b
jc pdiv
mvi a,'0'+10
add l ; Make ASCII digit
pop h
dcx h ; Store digit
mov m,a
push h
xchg
mov a,h ; Quotient nonzero?
ora l
jnz pdigit ; Then there are more digits
pop d ; Otherwise, print string using CP/M
prints: mvi c,9
jmp 5
nl: db 13,10,'$'
db '.....'
buf: db ' $'
ecks: equ $
Output:
0 0 1 0 2 0 2 2 1 6
4 7 30 25 67 225 488 0 10 136

8086 Assembly[edit]

LIMIT:	equ	1000
cpu 8086
org 100h
section .text
mov di,eck ; Zero out the memory
xor ax,ax
mov cx,LIMIT
rep stosw
mov bx,eck ; Base address
mov cx,LIMIT ; Limit
xor ax,ax
mov si,-1 ; Outer loop index
outer: inc si
dec cx
jcxz done
mov di,si ; Inner loop index
inner: dec di
js outer
shl si,1 ; Shift the loop indices (each entry is 2 bytes)
shl di,1
mov ax,[si+bx] ; Find a match?
cmp ax,[di+bx]
je match
shr si,1 ; If not, shift SI and DI back and keep going
shr di,1
jmp inner
match: mov ax,si ; Calculate the new value
sub ax,di
shr ax,1 ; Compensate for shift
mov [si+bx+2],ax ; Store value
shr si,1 ; Shift SI back and calculate next value
jmp outer
done: xor si,si ; Print first 10 elements
call p10
mov si,LIMIT-10 ; Print last 10 elements⌈
p10: mov cx,10 ; Print 10 elements starting at SI
shl si,1 ; Items are 2 bytes wide
add si,eck
.item: lodsw ; Retrieve item
call printn ; Print it
loop .item
mov dx,nl ; Print a newline afterwards
jmp prints
printn: mov bx,buf ; Print AX
mov bp,10
.digit: xor dx,dx ; Extract digit
div bp
add dl,'0' ; ASCII digit
dec bx
mov [bx],dl ; Store in buffer
test ax,ax ; Any more digits?
jnz .digit
mov dx,bx
prints: mov ah,9 ; Print string in buffer
int 21h
ret
section .data
nl: db 13,10,'$'
db '.....'
buf: db ' $'
section .bss
eck: resw LIMIT
Output:
0 0 1 0 2 0 2 2 1 6
4 7 30 25 67 225 488 0 10 136

AArch64 Assembly[edit]

Works with: as version Raspberry Pi 3B version Buster 64 bits
 
/* ARM assembly AARCH64 Raspberry PI 3B */
/* program vanEckSerie64.s */
 
/*******************************************/
/* Constantes file */
/*******************************************/
/* for this file see task include a file in language AArch64 assembly */
.include "../includeConstantesARM64.inc"
 
.equ MAXI, 1000
 
/*********************************/
/* Initialized data */
/*********************************/
.data
sMessResultElement: .asciz " @ "
szCarriageReturn: .asciz "\n"
 
/*********************************/
/* UnInitialized data */
/*********************************/
.bss
sZoneConv: .skip 24
TableVanEck: .skip 8 * MAXI
/*********************************/
/* code section */
/*********************************/
.text
.global main
main: // entry of program
mov x2,#0 // begin first element
mov x3,#0 // current counter
ldr x4,qAdrTableVanEck // table address
str x2,[x4,x3,lsl 3] // store first zéro
1: // begin loop
mov x5,x3 // init current indice
2:
sub x5,x5,1 // decrement
cmp x5,0 // end table ?
blt 3f
ldr x6,[x4,x5,lsl 3] // load element
cmp x6,x2 // and compare with the last element
bne 2b // not equal
sub x2,x3,x5 // else compute gap
b 4f
3:
mov x2,#0 // first, move zero to next element
4:
add x3,x3,#1 // increment counter
str x2,[x4,x3,lsl 3] // and store new element
cmp x3,MAXI
blt 1b
 
mov x2,0
5: // loop display ten elements
ldr x0,[x4,x2,lsl 3]
ldr x1,qAdrsZoneConv
bl conversion10 // call décimal conversion
ldr x0,qAdrsMessResultElement
ldr x1,qAdrsZoneConv // insert conversion in message
bl strInsertAtCharInc
mov x1,0 // final zéro
strb w1,[x0,5] //
bl affichageMess // display message
add x2,x2,1 // increment indice
cmp x2,10 // end ?
blt 5b // no -> loop
ldr x0,qAdrszCarriageReturn
bl affichageMess
 
mov x2,MAXI - 10
6: // loop display ten elements 990-999
ldr x0,[x4,x2,lsl 3]
ldr x1,qAdrsZoneConv
bl conversion10 // call décimal conversion
ldr x0,qAdrsMessResultElement
ldr x1,qAdrsZoneConv // insert conversion in message
bl strInsertAtCharInc
mov x1,0 // final zéro
strb w1,[x0,5] //
bl affichageMess // display message
add x2,x2,1 // increment indice
cmp x2,MAXI // end ?
blt 6b // no -> loop
ldr x0,qAdrszCarriageReturn
bl affichageMess
 
100: // standard end of the program
mov x0, 0 // return code
mov x8, EXIT // request to exit program
svc 0 // perform the system call
qAdrszCarriageReturn: .quad szCarriageReturn
qAdrsMessResultElement: .quad sMessResultElement
qAdrsZoneConv: .quad sZoneConv
qAdrTableVanEck: .quad TableVanEck
/********************************************************/
/* File Include fonctions */
/********************************************************/
/* for this file see task include a file in language AArch64 assembly */
.include "../includeARM64.inc"
 
 0  0  1  0  2  0  2  2  1  6
 4  7  30  25  67  225  488  0  10  136

Ada[edit]

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

ALGOL 68[edit]

BEGIN # find elements of the Van Eck Sequence - first term is 0, following  #
# terms are 0 if the previous was the first appearance of the element #
# or how far back in the sequence the last element appeared #
# returns the first n elements of the Van Eck sequence #
OP VANECK = ( INT n )[]INT:
BEGIN
[ 1 : IF n < 0 THEN 0 ELSE n FI ]INT result; FOR i TO n DO result[ i ] := 0 OD;
[ 0 : UPB result ]INT pos; FOR i FROM 0 TO n DO pos[ i ] := 0 OD;
FOR i FROM 2 TO n DO
INT j = i - 1;
INT prev = result[ j ];
IF pos[ prev ] /= 0 THEN
# not a new element #
result[ i ] := j - pos[ prev ]
FI;
pos[ prev ] := j
OD;
result
END # VANECK # ;
# construct the first 1000 terms of the sequence #
[]INT seq = VANECK 1000;
# show the first and last 10 elements #
FOR i TO 10 DO print( ( " ", whole( seq[ i ], 0 ) ) ) OD;
print( ( newline ) );
FOR i FROM UPB seq - 9 TO UPB seq DO print( ( " ", whole( seq[ i ], 0 ) ) ) OD;
print( ( newline ) )
END
Output:
 0 0 1 0 2 0 2 2 1 6
 4 7 30 25 67 225 488 0 10 136

ALGOL-M[edit]

begin
integer array eck[1:1000];
integer i, j;
 
for i := 1 step 1 until 1000 do
eck[i] := 0;
 
for i := 1 step 1 until 999 do
begin
j := i - 1;
while j > 0 and eck[i] <> eck[j] do
j := j - 1;
if j <> 0 then
eck[i+1] := i - j;
end;
 
for i := 1 step 1 until 10 do
writeon(eck[i]);
write("");
for i := 991 step 1 until 1000 do
writeon(eck[i]);
 
end
Output:
     0     0     1     0     2     0     2     2     1     6
     4     7    30    25    67   225   488     0    10   136

ALGOL W[edit]

Translation of: ALGOL 68
begin % find elements of the Van Eck Sequence - first term is 0, following  %
 % terms are 0 if the previous was the first appearance of the element %
 % or how far back in the sequence the last element appeared  %
 % sets s to the first n elements of the Van Eck sequence  %
procedure VanEck ( integer array s ( * ) ; integer value n ) ;
begin
integer array pos ( 0 :: n );
for i := 1 until n do s( i ) := 0;
for i := 0 until n do pos( i ) := 0;
for i := 2 until n do begin
integer j, prev;
j  := i - 1;
prev := s( j );
if pos( prev ) not = 0 then begin
 % not a new element  %
s( i ) := j - pos( prev )
end if_pos_prev_ne_0 ;
pos( prev ) := j
end for_j;
end VanEck ;
 % construct the first 1000 terms of the sequence  %
integer MAX_VAN_ECK;
MAX_VAN_ECK := 1000;
begin
integer array seq ( 1 :: MAX_VAN_ECK );
VanEck( seq, MAX_VAN_ECK );
 % show the first and last 10 elements  %
for i := 1 until 10 do writeon( i_w := 1, s_w := 0, " ", seq( i ) );
write();
for i := MAX_VAN_ECK - 9 until MAX_VAN_ECK do writeon( i_w := 1, s_w := 0, " ", seq( i ) );
write()
end
end.
Output:
 0 0 1 0 2 0 2 2 1 6
 4 7 30 25 67 225 488 0 10 136

APL[edit]

Works with: Dyalog APL
(10∘↑,[.5]¯10∘↑)(⊢,(⊃∘⌽∊¯1∘↓)∧(1↓⌽)⍳⊃∘⌽)⍣999⊢,0
Output:
0 0  1  0  2   0   2 2  1   6
4 7 30 25 67 225 488 0 10 136

AppleScript[edit]

Functional[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)
-- First n terms of the vanEck sequence.
 
script go
on |λ|(xns, i)
set {x, ns} to xns
set prev to item (1 + x) of ns
 
if 0 ≠ prev then
set v to i - prev
else
set v to 0
end if
 
{{v, insert(ns, x, i)}, v}
end |λ|
end script
 
 
{0} & item 2 of mapAccumL(go, ¬
{0, replicate(n, 0)}, enumFromTo(1, n - 1))
end vanEck
 
 
--------------------------- TEST ---------------------------
on run
unlines({¬
"First 10 terms:", ¬
showList(vanEck(10)), ¬
"", ¬
"Terms 990 to 1000:", ¬
showList(items -10 thru -1 of vanEck(1000))})
end run
 
 
 
------------------------- GENERIC --------------------------
 
-- enumFromTo :: Int -> Int -> [Int]
on enumFromTo(m, n)
if m ≤ n then
set lst to {}
repeat with i from m to n
set end of lst to i
end repeat
lst
else
{}
end if
end enumFromTo
 
 
-- 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
 
 
-- insert :: [Int] -> Int -> Int -> [Int]
on insert(xs, i, v)
-- A list updated at position i with value v.
set item (1 + i) of xs to v
xs
end insert
 
 
-- intercalate :: String -> [String] -> String
on intercalate(delim, xs)
set {dlm, my text item delimiters} to ¬
{my text item delimiters, delim}
set s to xs as text
set my text item delimiters to dlm
s
end intercalate
 
 
-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
-- The list obtained by applying f
-- to each element of xs.
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map
 
 
-- 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
 
 
-- 'The mapAccumL function behaves like a combination of map and foldl;
-- 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.' (see Hoogle)
-- mapAccumL :: (acc -> x -> (acc, y)) -> acc -> [x] -> (acc, [y])
on mapAccumL(f, acc, xs)
script
on |λ|(a, x, i)
tell mReturn(f) to set pair to |λ|(item 1 of a, x, i)
{item 1 of pair, (item 2 of a) & {item 2 of pair}}
end |λ|
end script
 
foldl(result, {acc, []}, xs)
end mapAccumL
 
 
-- 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
 
 
-- showList :: [a] -> String
on showList(xs)
"[" & intercalate(", ", map(my str, xs)) & "]"
end showList
 
 
-- str :: a -> String
on str(x)
x as string
end str
 
 
-- unlines :: [String] -> String
on unlines(xs)
-- A single string formed by the intercalation
-- of a list of strings with the newline character.
set {dlm, my text item delimiters} to ¬
{my text item delimiters, linefeed}
set s to xs as text
set my text item delimiters to dlm
s
end unlines
Output:
First 10 terms:
[0, 0, 1, 0, 2, 0, 2, 2, 1, 6]

Terms 999 to 1000:
[4, 7, 30, 25, 67, 225, 488, 0, 10, 136]

Idiomatic[edit]

On the contrary, it's right up AppleScript's street.

on vanEckSequence(limit)
script o
property sequence : {}
property lookup : {}
end script
 
set term to 0
repeat with i from 1 to (limit - 1) -- 1-based indices.
set end of o's sequence to term
set t to term + 1 -- 1-based index.
repeat (t - (count o's lookup)) times
set end of o's lookup to missing value
end repeat
set previous_i to item t of o's lookup
set item t of o's lookup to i
if (previous_i is missing value) then
set term to 0
else
set term to i - previous_i
end if
end repeat
set end of o's sequence to term
 
return o's sequence
end vanEckSequence
 
-- Task code:
tell vanEckSequence(1000) to return {items 1 thru 10, items 991 thru 1000}
Output:
{{0, 0, 1, 0, 2, 0, 2, 2, 1, 6}, {4, 7, 30, 25, 67, 225, 488, 0, 10, 136}}

ARM Assembly[edit]

Works with: as version Raspberry Pi
 
/* ARM assembly Raspberry PI */
/* program vanEckSerie.s */
 
/* REMARK 1 : this program use routines in a include file
see task Include a file language arm assembly
for the routine affichageMess conversion10
see at end of this program the instruction include */
/* for constantes see task include a file in arm assembly */
/************************************/
/* Constantes */
/************************************/
.include "../constantes.inc"
 
.equ MAXI, 1000
 
/*********************************/
/* Initialized data */
/*********************************/
.data
sMessResultElement: .asciz " @ "
szCarriageReturn: .asciz "\n"
 
/*********************************/
/* UnInitialized data */
/*********************************/
.bss
sZoneConv: .skip 24
TableVanEck: .skip 4 * MAXI
/*********************************/
/* code section */
/*********************************/
.text
.global main
main: @ entry of program
mov r2,#0 @ begin first element
mov r3,#0 @ current counter
ldr r4,iAdrTableVanEck @ table address
str r2,[r4,r3,lsl #2] @ store first zéro
1: @ begin loop
mov r5,r3 @ init current indice
2:
sub r5,#1 @ decrement
cmp r5,#0 @ end table ?
movlt r2,#0 @ yes, move zero to next element
blt 3f
ldr r6,[r4,r5,lsl #2] @ load element
cmp r6,r2 @ and compare with the last element
bne 2b @ not equal
sub r2,r3,r5 @ else compute gap
3:
add r3,r3,#1 @ increment counter
str r2,[r4,r3,lsl #2] @ and store new element
cmp r3,#MAXI
blt 1b
 
mov r2,#0
4: @ loop display ten elements
ldr r0,[r4,r2,lsl #2]
ldr r1,iAdrsZoneConv
bl conversion10 @ call décimal conversion
ldr r0,iAdrsMessResultElement
ldr r1,iAdrsZoneConv @ insert conversion in message
bl strInsertAtCharInc
mov r1,#0 @ final zéro
strb r1,[r0,#5] @
bl affichageMess @ display message
add r2,#1 @ increment indice
cmp r2,#10 @ end ?
blt 4b @ no -> loop
ldr r0,iAdrszCarriageReturn
bl affichageMess
 
mov r2,#MAXI - 10
5: @ loop display ten elements 990-999
ldr r0,[r4,r2,lsl #2]
ldr r1,iAdrsZoneConv
bl conversion10 @ call décimal conversion
ldr r0,iAdrsMessResultElement
ldr r1,iAdrsZoneConv @ insert conversion in message
bl strInsertAtCharInc
mov r1,#0 @ final zéro
strb r1,[r0,#5] @
bl affichageMess @ display message
add r2,#1 @ increment indice
cmp r2,#MAXI @ end ?
blt 5b @ no -> loop
ldr r0,iAdrszCarriageReturn
bl affichageMess
 
100: @ standard end of the program
mov r0, #0 @ return code
mov r7, #EXIT @ request to exit program
svc #0 @ perform the system call
iAdrszCarriageReturn: .int szCarriageReturn
iAdrsMessResultElement: .int sMessResultElement
iAdrsZoneConv: .int sZoneConv
iAdrTableVanEck: .int TableVanEck
 
/***************************************************/
/* ROUTINES INCLUDE */
/***************************************************/
.include "../affichage.inc"
 
 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

BASIC[edit]

10 DEFINT A-Z
20 DIM E(1000)
30 FOR I=0 TO 999
40 FOR J=I-1 TO 0 STEP -1
50 IF E(J)=E(I) THEN E(I+1)=I-J: GOTO 80
60 NEXT J
70 E(I+1)=0
80 NEXT I
90 FOR I=0 TO 9: PRINT E(I);: NEXT
95 PRINT
100 FOR I=990 TO 999: PRINT E(I);: NEXT
Output:
 0  0  1  0  2  0  2  2  1  6
 4  7  30  25  67  225  488  0  10  136

BCPL[edit]

get "libhdr"
 
let start() be
$( let eck = vec 999
 
for i = 0 to 999 do eck!i := 0
for i = 0 to 998 do
for j = i-1 to 0 by -1 do
if eck!i = eck!j then
$( eck!(i+1) := i-j
break
$)
 
for i = 0 to 9 do writed(eck!i, 4)
wrch('*N')
for i = 990 to 999 do writed(eck!i, 4)
wrch('*N')
$)
Output:
   0   0   1   0   2   0   2   2   1   6
   4   7  30  25  67 225 488   0  10 136

C[edit]

#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#[edit]

Translation of: C
using System.Linq; class Program { static void Main() {
int a, b, c, d, e, f, g; int[] h = new int[g = 1000];
for (a = 0, b = 1, c = 2; c < g; a = b, b = c++)
for (d = a, e = b - d, f = h[b]; e <= b; e++)
if (f == h[d--]) { h[c] = e; break; }
void sho(int i) { System.Console.WriteLine(string.Join(" ",
h.Skip(i).Take(10))); } sho(0); sho(990); } }
Output:
0 0 1 0 2 0 2 2 1 6
4 7 30 25 67 225 488 0 10 136

C++[edit]

#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[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)

COBOL[edit]

        IDENTIFICATION DIVISION.
PROGRAM-ID. VAN-ECK.
 
DATA DIVISION.
WORKING-STORAGE SECTION.
01 CALCULATION.
02 ECK PIC 999 OCCURS 1000 TIMES.
02 I PIC 9999.
02 J PIC 9999.
01 OUTPUT-FORMAT.
02 ITEM PIC ZZ9.
02 IDX PIC ZZZ9.
 
PROCEDURE DIVISION.
B. PERFORM GENERATE-ECK.
PERFORM SHOW VARYING I FROM 1 BY 1 UNTIL I = 11.
PERFORM SHOW VARYING I FROM 991 BY 1 UNTIL I = 1001.
STOP RUN.
 
SHOW.
MOVE I TO IDX.
MOVE ECK(I) TO ITEM.
DISPLAY 'ECK(' IDX ') = ' ITEM.
 
GENERATE-ECK SECTION.
B. SET ECK(1) TO 0.
SET I TO 1.
PERFORM GENERATE-TERM
VARYING I FROM 2 BY 1 UNTIL I = 1001.
 
GENERATE-TERM SECTION.
B. SUBTRACT 2 FROM I GIVING J.
LOOP.
IF J IS LESS THAN 1 GO TO TERM-IS-NEW.
IF ECK(J) = ECK(I - 1) GO TO TERM-IS-OLD.
SUBTRACT 1 FROM J.
GO TO LOOP.
 
TERM-IS-NEW.
SET ECK(I) TO 0.
GO TO DONE.
 
TERM-IS-OLD.
COMPUTE ECK(I) = (I - J) - 1.
 
DONE. EXIT.
Output:
ECK(   1) =   0
ECK(   2) =   0
ECK(   3) =   1
ECK(   4) =   0
ECK(   5) =   2
ECK(   6) =   0
ECK(   7) =   2
ECK(   8) =   2
ECK(   9) =   1
ECK(  10) =   6
ECK( 991) =   4
ECK( 992) =   7
ECK( 993) =  30
ECK( 994) =  25
ECK( 995) =  67
ECK( 996) = 225
ECK( 997) = 488
ECK( 998) =   0
ECK( 999) =  10
ECK(1000) = 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)
 
(defun van-eck-nm-sequence (n m)
(loop with ac repeat m
for i = (position (car ac) (cdr ac)) do
(push (if i (1+ i) 0) ac)
finally (return (nthcdr (1- n) (nreverse ac)))))
 
(format t "The first 10 elements are: ~{~a ~}~%" (van-eck-nm-sequence 1 10))
(format t "The 991-1000th elements are: ~{~a ~}" (van-eck-nm-sequence 991 1000))
 
Output:
The first 10 elements are: 0 0 1 0 2 0 2 2 1 6 
The 991-1000th elements are: 4 7 30 25 67 225 488 0 10 136 

Cowgol[edit]

include "cowgol.coh";
 
sub print_list(ptr: [uint16], n: uint8) is
while n > 0 loop
print_i16([ptr]);
print_char(' ');
n := n - 1;
ptr := @next ptr;
end loop;
print_nl();
end sub;
 
const LIMIT := 1000;
var eck: uint16[LIMIT];
MemZero(&eck as [uint8], @bytesof eck);
var i: @indexof eck;
var j: @indexof eck;
 
i := 0;
while i < LIMIT-1 loop
j := i-1;
while j != -1 loop
if eck[i] == eck[j] then
eck[i+1] := i-j;
break;
end if;
j := j - 1;
end loop;
i := i + 1;
end loop;
 
print_list(&eck[0], 10);
print_list(&eck[LIMIT-10], 10);
Output:
0 0 1 0 2 0 2 2 1 6
4 7 30 25 67 225 488 0 10 136

D[edit]

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[1] = 0
VanEck[2] = 0
VanEck[3] = 1
VanEck[4] = 0
VanEck[5] = 2
VanEck[6] = 0
VanEck[7] = 2
VanEck[8] = 2
VanEck[9] = 1
VanEck[10] = 6

Terms 991 to 1000 of Van Eck's sequence:
VanEck[991] = 4
VanEck[992] = 7
VanEck[993] = 30
VanEck[994] = 25
VanEck[995] = 67
VanEck[996] = 225
VanEck[997] = 488
VanEck[998] = 0
VanEck[999] = 10
VanEck[1000] = 136

Delphi[edit]

See Pascal.

Dyalect[edit]

Translation of: Go
let 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..999])")
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#[edit]

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

Alternative recursive procedure[edit]

 
open System.Collections.Generic
let 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[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 }

Fortran[edit]

      program VanEck
implicit none
integer eck(1000), i, j
 
eck(1) = 0
do 20 i=1, 999
do 10 j=i-1, 1, -1
if (eck(i) .eq. eck(j)) then
eck(i+1) = i-j
go to 20
end if
10 continue
eck(i+1) = 0
20 continue
 
do 30 i=1, 10
30 write (*,'(I4)',advance='no') eck(i)
write (*,*)
 
do 40 i=991, 1000
40 write (*,'(I4)',advance='no') eck(i)
write (*,*)
 
end program
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

Fōrmulæ[edit]

Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. 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 storage and transfer purposes more than visualization and edition.

Programs in Fōrmulæ are created/edited online in its website, However they run on execution servers. By default remote servers are used, but they are limited in memory and processing power, since they are intended for demonstration and casual use. A local server can be downloaded and installed, it has no limitations (it runs in your own computer). Because of that, example programs can be fully visualized and edited, but some of them will not run if they require a moderate or heavy computation/memory resources, and no local server is being used.

In this page you can see the program(s) related to this task and their results.

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:

{-# 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[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

Java[edit]

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[1] = 0
VanEck[2] = 0
VanEck[3] = 1
VanEck[4] = 0
VanEck[5] = 2
VanEck[6] = 0
VanEck[7] = 2
VanEck[8] = 2
VanEck[9] = 1
VanEck[10] = 6

Terms 991 to 1000 of Van Eck's sequence:
VanEck[991] = 4
VanEck[992] = 7
VanEck[993] = 30
VanEck[994] = 25
VanEck[995] = 67
VanEck[996] = 225
VanEck[997] = 488
VanEck[998] = 0
VanEck[999] = 10
VanEck[1000] = 136

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 = Boolean(prev) ? (
i - prev
) : 0;
return [
[v, (seen[x] = i, seen)], v
];
}, [0, {}],
 
enumFromTo(1, n - 1)
)[1]);
 
// ----------------------- TEST ------------------------
const main = () =>
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]

jq[edit]

# Input: an array
# If the rightmost element, .[-1], does not occur elsewhere, return 0;
# otherwise return the "depth" of its rightmost occurrence in .[0:-2]
def depth:
.[-1] as $x
| length as $length
| first(range($length-2; -1; -1) as $i
| select(.[$i] == $x) | $length - 1 - $i)
// 0 ;
 
# Generate a stream of the first $n van Eck integers:
def vanEck($n):
def v:
recurse( if length == $n then empty
else . + [depth] end );
[0] | v | .[-1];
 
# The task:
[vanEck(10)], [vanEck(1000)][990:1001]
 

Output[edit]

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

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
 

Kotlin[edit]

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[1] = 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[1] = 0
VanEck[2] = 0
VanEck[3] = 1
VanEck[4] = 0
VanEck[5] = 2
VanEck[6] = 0
VanEck[7] = 2
VanEck[8] = 2
VanEck[9] = 1
VanEck[10] = 6

Terms 991 to 1000 of Van Eck's sequence:
VanEck[991] = 4
VanEck[992] = 7
VanEck[993] = 30
VanEck[994] = 25
VanEck[995] = 67
VanEck[996] = 225
VanEck[997] = 488
VanEck[998] = 0
VanEck[999] = 10
VanEck[1000] = 136

Lua[edit]

-- Return a table of the first n values of the Van Eck sequence
function 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 seq
end
 
-- Show the set of values in table t from key numbers lo to hi
function showValues (t, lo, hi)
for i = lo, hi do
io.write(t[i] .. " ")
end
print()
end
 
-- Main procedure
local 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

MAD[edit]

            NORMAL MODE IS INTEGER
DIMENSION E(1000)
E(0)=0
THROUGH L1, FOR I=0, 1, I.GE.1000
THROUGH L2, FOR J=I-1, -1, J.L.0
WHENEVER E(J).E.E(I)
E(I+1) = I-J
TRANSFER TO L1
END OF CONDITIONAL
L2 CONTINUE
E(I+1)=0
L1 CONTINUE
THROUGH S, FOR I=0, 1, I.GE.10
S PRINT FORMAT FMT, I, E(I), I+990, E(I+990)
VECTOR VALUES FMT = $2HE(,I3,2H)=,I3,S5,2HE(,I3,2H)=,I3*$
END OF PROGRAM
Output:
E(  0)=  0     E(990)=  4
E(  1)=  0     E(991)=  7
E(  2)=  1     E(992)= 30
E(  3)=  0     E(993)= 25
E(  4)=  2     E(994)= 67
E(  5)=  0     E(995)=225
E(  6)=  2     E(996)=488
E(  7)=  2     E(997)=  0
E(  8)=  1     E(998)= 10
E(  9)=  6     E(999)=136


Mathematica[edit]

 
TakeList[Nest[If[MemberQ[#//Most, #//Last], Join[#, Length[#] - [email protected][#//Most, #//Last]], Append[#, 0]]&, {0}, 999], {10, -10}] // Column
 
Output:
{0,0,1,0,2,0,2,2,1,6}
{4,7,30,25,67,225,488,0,10,136}

Nim[edit]

const max = 1000
var 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[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;
{$IFDEF FPC}
pPosBef, pSeen: pUint32;
{$ELSE}
pPosBef, pSeen: array of UInt32;
{$ENDIF}
begin
HaveSeen := 0;
if MaxTestCnt > MAXNUM then
EXIT;
 
Fillchar(LastSeen, SizeOf(LastSeen), #0);
//setlength and clear
setlength(PosBefore, 0);
setlength(PosBefore, MaxTestCnt);
 
{$IFDEF FPC}
pPosBef := @PosBefore[0];
pSeen := @LastSeen[0];
{$ELSE}
SetLength(pSeen, SizeOf(LastSeen));
setlength(pPosBef, MaxTestCnt);
move(PosBefore[0], pPosBef[0], length(pPosBef));
move(LastSeen[0], pSeen[0], length(pSeen));
{$ENDIF}
 
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;
 
{$IFNDEF FPC}
 
move(pPosBef[0], PosBefore[0], length(pPosBef));
move(pSeen[0], LastSeen[0], length(pSeen));
{$ENDIF}
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[edit]

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[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}

PL/M[edit]

100H:
BDOS: PROCEDURE (FN, ARG); DECLARE FN BYTE, ARG ADDRESS; GO TO 5; END BDOS;
EXIT: PROCEDURE; CALL BDOS(0,0); END EXIT;
PRINT: PROCEDURE (S); DECLARE S ADDRESS; CALL BDOS(9,S); END PRINT;
 
PRINT$NUMBER: PROCEDURE (N);
DECLARE S (7) BYTE INITIAL ('..... $');
DECLARE (N, P) ADDRESS, C BASED P BYTE;
P = .S(5);
DIGIT:
P = P - 1;
C = N MOD 10 + '0';
N = N / 10;
IF N > 0 THEN GO TO DIGIT;
CALL PRINT(P);
END PRINT$NUMBER;
 
PRINT$SLICE: PROCEDURE (LIST, N);
DECLARE (I, N, LIST, L BASED LIST) ADDRESS;
DO I=0 TO N-1;
CALL PRINT$NUMBER(L(I));
END;
CALL PRINT(.(13,10,'$'));
END PRINT$SLICE;
 
DECLARE ECK (1000) ADDRESS;
DECLARE (I, J) ADDRESS;
 
ECK(0) = 0;
DO I=0 TO LAST(ECK)-1;
J = I - 1;
DO WHILE J <> 0FFFFH; /* WHAT IS SIGNED MATH */
IF ECK(I) = ECK(J) THEN DO;
ECK(I+1) = I-J;
GO TO NEXT;
END;
J = J - 1;
END;
ECK(I+1) = 0;
NEXT:
END;
 
CALL PRINT$SLICE(.ECK(0), 10);
CALL PRINT$SLICE(.ECK(990), 10);
CALL EXIT;
EOF
Output:
0 0 1 0 2 0 2 2 1 6
4 7 30 25 67 225 488 0 10 136

Prolog[edit]

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[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 go
 
 
# 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
 
 
# 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 by map-accumulation'''
 
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 m: 'N=' + str(m), 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 go
 
 
# ------------------------ 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]

Quackery[edit]

  [ ' [ 0 ] 
swap 1 - times
[ dup behead swap find
1+ 2dup swap found *
swap join ]
reverse ] is van-eck ( n --> [ )
 
10 van-eck echo cr
1000 van-eck -10 split echo drop
Output:
[ 0 0 1 0 2 0 2 2 1 6 ]
[ 4 7 30 25 67 225 488 0 10 136 ]

Racket[edit]

#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[edit]

(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[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   20,000   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[edit]

Ruby: Using an Array[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]

Ruby: Using a Hash[edit]

class VenEch
include Enumerable
 
def initialize()
@i = 0;
@val = 0;
@seen = {};
end
 
def add_num num
@val = @i - @seen.fetch(num, @i)
@seen[num] = @i
@i += 1
end
 
def each(&block)
loop { block.call(@val); add_num @val }
end
end
 
ve = VenEch.new.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[edit]

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[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).

SNOBOL4[edit]

        define('eck(n)i,j')                     :(eck_end)
eck eck = array(n,0)
i = 0
eouter i = lt(i,n - 1) i + 1  :f(return)
j = i
einner j = gt(j,0) j - 1  :f(eouter)
eck<i + 1> = eq(eck<i>,eck<j>) i - j  :s(eouter)f(einner)
eck_end
 
define('list(arr,start,stop)')  :(list_end)
list list = list arr<start> ' '
start = lt(start,stop) start + 1  :s(list)f(return)
list_end
 
ecks = eck(1000)
output = list(ecks, 1, 10)
output = list(ecks, 991, 1000)
end
Output:
0 0 1 0 2 0 2 2 1 6
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]

Swift[edit]

Translation of: Rust
struct VanEckSequence: Sequence, IteratorProtocol {
private var index = 0
private var lastTerm = 0
private var lastPos = Dictionary<Int, Int>()
 
mutating func next() -> Int? {
let result = lastTerm
var nextTerm = 0
if let v = lastPos[lastTerm] {
nextTerm = index - v
}
lastPos[lastTerm] = index
lastTerm = nextTerm
index += 1
return result
}
}
 
let seq = VanEckSequence().prefix(1000)
 
print("First 10 terms of the Van Eck sequence:")
for n in seq.prefix(10) {
print(n, terminator: " ")
}
print("\nTerms 991 to 1000 of the Van Eck sequence:")
for n in seq.dropFirst(990) {
print(n, terminator: " ")
}
print()
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 

Tcl[edit]

## 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 9
show 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

Visual Basic .NET[edit]

Translation of: C#
Imports System.Linq
Module Module1
Dim h() As Integer
Sub sho(i As Integer)
Console.WriteLine(String.Join(" ", h.Skip(i).Take(10)))
End Sub
Sub Main()
Dim a, b, c, d, f, g As Integer : g = 1000
h = new Integer(g){} : a = 0 : b = 1 : For c = 2 To g
f = h(b) : For d = a To 0 Step -1
If f = h(d) Then h(c) = b - d: Exit For
Next : a = b : b = c : Next : sho(0) : sho(990)
End Sub
End Module
Output:

Same as C#.

Wren[edit]

Translation of: Go
var max = 1000
var 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[edit]

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