# Inventory sequence

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

To build the inventory sequence, first take inventory of what numbers are already in the sequence, then, starting with 0 add the count of each number in turn to the end of the sequence. When you reach a number that has a count of 0, stop, add the 0 to the end of the sequence, and then restart the inventory from 0.

E.G.
Start taking inventory; how many 0s are there? 0. Add a 0 to the end of the sequence and restart the inventory. (0)
How many 0s are there? 1. Add a 1 to the end of the sequence. How many 1s are there? 1. Add a 1 to the end of the sequence. How many 2s are there? 0. Add a 0 to the end of the sequence and restart the inventory. (0 1 1 0)
and so on.

• Find and display the first 100 elements of the sequence.
• Find and display the position and value of the first element greater than or equal to 1000.

Stretch
• Find and display the position and value of the first element greater than or equal to 2000, 3000 ... 10,000.
• Plot a graph of the first 10,000 elements of the sequence.

## 8080 Assembly

```PRINT   equ     9       ; CP/M call to print a string
org     100h
lxi     h,tally         ; Zero out the array
lxi     d,11000 * 2     ; Two bytes each
mvi     b,0
zero:   mov     m,b
inx     h
dcx     d
mov     a,d
ora     e
jnz     zero
restrt: lxi     b,0             ; BC = current number

takinv: mov     d,b             ; Look up count of current number
mov     e,c
call    talidx
mov     e,m             ; DE = current element
inx     h
mov     d,m

lxi     h,tabct         ; Any items left to print for the table?
mov     a,m
ora     a
jz      cktrsh
dcr     m               ; In any case there's now one less
call    atoi
lxi     h,numbuf        ; Print the current number
call    prstr
lxi     h,tabcol        ; Done with the row yet?
dcr     m
jnz     cktrsh
mvi     m,10            ; Yes, reset counter and print a newline
lxi     h,nl
call    prstr

cktrsh: lhld    trshld          ; Has the threshold been reached?
call    cdehl
jc      increm
xchg                    ; Yes, fill in the numbers
call    atoi            ; Threshold (from HL)
xchg
push    d               ; And add 1000 to it while we've got it here
lxi     d,1000
pop     d
shld    trshld
lxi     h,trsout
call    numat
call    atoi            ; Our current element
lxi     h,numout
call    numat
lxi     h,trsfmt        ; Print the element and its index
call    prstr

increm: call    talidx          ; HL = address of count of element
call    inxm            ; Increment count
call    incidx          ; Increment index
inx     b               ; Increment the search number
mov     a,d             ; Reached zero?
ora     e
jnz     takinv          ; If not, keep going
lhld    trshld          ; Once we've done enough, stop
lxi     d,-10001
jnc     restrt
ret

tabcol: db      10              ; Column counter for table
tabct:  db      100             ; First 100 items are to be printed
trshld: dw      1000            ; Initial threshold

; 16-bit compare DE and HL
cdehl:  mov     a,d
cmp     h
rnz
mov     a,e
cmp     l
ret

; 16-bit increment of [HL]
inxm:   inr     m
rnz
inx     h
inr     m
ret

; Get HL = &tally[DE]
talidx: push    d
xchg
lxi     d,tally
pop     d
ret

; print string at HL
prstr:  push    h
push    d
push    b
xchg
mvi     c,PRINT
call    5
pop     b
pop     d
pop     h
ret

; copy number buffer to HL
numat:  push    d
push    b
lxi     d,numbuf
mvi     b,5
numcpy: ldax    d
mov     m,a
inx     d
inx     h
dcr     b
jnz     numcpy
pop     b
pop     d
ret

; set numbuf to DE as decimal number
atoi:   push    h
push    d
push    b
mvi     a,5             ; Space out the buffer
lxi     h,nbufe         ; End of number buffer
push    h               ; Keep it on the stack
aspc:   dcx     h
mvi     m,' '
dcr     a
jnz     aspc
lxi     b,-10
adgt:   xchg                    ; Number in HL
lxi     d,-1
adgtlp: inx     d               ; Extract digit
mvi     a,'0'+10        ; ASCII digit in A
pop     h               ; Get buffer pointer back
dcx     h
mov     m,a
push    h
mov     a,d             ; Any digits left?
ora     e
jnz     adgt            ; If so, next digits
pop     b               ; Otherwise we're done
pop     b
pop     d
pop     h
ret
numbuf: db      '*****'
nbufe:  db      '\$'

; Format to print first number above threshold, ends in index
trsfmt: db      'First > '
trsout: db      '*****: '
numout: db      '***** at '

; Index counter, kept as ASCII decimal so we don't need 32-bit math
idx:    db      '     0'
nl:     db      13,10,'\$'       ; Newline string which also ends index

; Increment index
incidx: lxi     h,nl-1
inclp:  mov     a,m
cpi     ' '             ; Space should be set to 1
jz      one
inr     m
cpi     '9'
rnz
mvi     m,'0'
dcx     h
jmp     inclp
one:    mvi     m,'1'
ret

tally   equ     \$               ; Array stored at the end of the program
```
Output:
```    0    1    1    0    2    2    2    0    3    2
4    1    1    0    4    4    4    1    4    0
5    5    4    1    6    2    1    0    6    7
5    1    6    3    3    1    0    7    9    5
3    6    4    4    2    0    8    9    6    4
9    4    5    2    1    3    0    9   10    7
5   10    6    6    3    1    4    2    0   10
11    8    6   11    6    9    3    2    5    3
2    0   11   11   10    8   11    7    9    4
3    6    4    5    0   12   11   10    9   13
First >  1000:  1001 at  24255
First >  2000:  2009 at  43301
First >  3000:  3001 at  61708
First >  4000:  4003 at  81456
First >  5000:  5021 at  98704
First >  6000:  6009 at 121342
First >  7000:  7035 at 151756
First >  8000:  8036 at 168804
First >  9000:  9014 at 184428
First > 10000: 10007 at 201788```

## 8086 Assembly

```        cpu     8086
org     100h
section .text
xor     ax,ax           ; Zero out the array
mov     di,tally
mov     cx,11000
rep     stosw
mov     bp,1000         ; BP = next threshold
mov     cx,640Ah        ; CH = table counter, CL = column counter

restrt: xor     bx,bx           ; BX = current number
takinv: shl     bx,1
mov     si,[tally+bx]   ; SI = current element
shr     bx,1
test    ch,ch           ; Any items left to print for the table?
jz      cktrsh          ; If not check threshold
dec     ch
mov     di,format.cell  ; Print cell
mov     dx,di
mov     ax,si
call    atoi
mov     ah,9
int     21h
dec     cl              ; Row done?
jnz     cktrsh
mov     dx,format.nl
mov     cl,10
int     21h

cktrsh: cmp     si,bp           ; Has the threshold been reached?
jna     increm
mov     ax,bp           ; Yes, fill in the numbers
mov     di,format.trs
call    atoi
mov     ax,si
mov     di,format.num
call    atoi
mov     dx,format       ; Print the string
mov     ah,9
int     21h

increm: mov     di,si           ; Increment count of current element
shl     di,1
inc     word [tally+di]
call    incidx          ; Increment current index
inc     bx              ; Increment search number
test    si,si           ; Reached zero?
jnz     takinv          ; If not, keep going
cmp     bp,10000        ; Last threshold reached?
jna     restrt          ; If not, start over
int     20h             ; Otherwise we're done

; Convert AX to decimal and store at DI (must be 5-char buffer)
atoi:   push    ax
push    bx
push    cx
push    dx
mov     cx,5
mov     bx,10
.digit: xor     dx,dx
div     bx
dec     di
dec     cx
mov     [di],dl
test    ax,ax
jnz     .digit
jcxz    .out
mov     al,' '
dec     di
std
rep     stosb
cld
.out:   pop     dx
pop     cx
pop     bx
pop     ax
ret

; Increment index as ASCII in place
incidx: mov     di,format.nl-1
.loop:  cmp     byte [di],' '   ; Space -> 1
je      .one
inc     byte [di]
cmp     byte [di],'9'
ja      .carry
ret
.carry: mov     byte [di],'0'
dec     di
jmp     .loop
.one:   mov     byte [di],'1'
ret

; Format to print first number above threshold
format: db      "First > "
.trs:   db      "*****: "
.num:   db      "***** at "
.idx:   db      "     0"
.nl:    db      13,10,'\$'
.cell:  db      "*****\$"         ; cell placeholder

section .bss
align   2
tally:  resw    11000
```
Output:
```    0    1    1    0    2    2    2    0    3    2
4    1    1    0    4    4    4    1    4    0
5    5    4    1    6    2    1    0    6    7
5    1    6    3    3    1    0    7    9    5
3    6    4    4    2    0    8    9    6    4
9    4    5    2    1    3    0    9   10    7
5   10    6    6    3    1    4    2    0   10
11    8    6   11    6    9    3    2    5    3
2    0   11   11   10    8   11    7    9    4
3    6    4    5    0   12   11   10    9   13
First >  1000:  1001 at  24255
First >  2000:  2009 at  43301
First >  3000:  3001 at  61708
First >  4000:  4003 at  81456
First >  5000:  5021 at  98704
First >  6000:  6009 at 121342
First >  7000:  7035 at 151756
First >  8000:  8036 at 168804
First >  9000:  9014 at 184428
First > 10000: 10007 at 201788```

## ALGOL 68

Calculates the sequence elements without storing them.

```BEGIN # find elements of the inventory sequence                                #

INT next to show :=  1 000;           # next value to show first element > #
INT max to show   = 10 000;           # last value to show first element > #

INT max number    = max to show + 1 000;  # max. element value to consider #
[ 0 : max number ]INT occurs;         # number of times each number occurs #
FOR i FROM LWB occurs TO UPB occurs DO occurs[ i ] := 0 OD;
INT seq pos      := 0;                       # current end of the sequence #
WHILE next to show <= max to show DO
FOR n FROM 0 WHILE next to show <= max to show
AND BEGIN
INT element := occurs[ n ];
seq pos    +:= 1;
IF seq pos <= 100 THEN
print( ( " ", whole( element, -4 ) ) );
IF seq pos MOD 10 = 0 THEN print( ( newline ) ) FI
ELIF element > next to show THEN
print( ( "Element ", whole( seq pos, -8 )
, " (", whole( element, -8 )
, ") is first > ", whole( next to show, -6 )
, newline
)
);
next to show +:= 1 000
FI;
IF element < max number THEN
occurs[ element ] +:= 1
FI;
element /= 0
END
DO SKIP OD
OD

END```
Output:
```    0    1    1    0    2    2    2    0    3    2
4    1    1    0    4    4    4    1    4    0
5    5    4    1    6    2    1    0    6    7
5    1    6    3    3    1    0    7    9    5
3    6    4    4    2    0    8    9    6    4
9    4    5    2    1    3    0    9   10    7
5   10    6    6    3    1    4    2    0   10
11    8    6   11    6    9    3    2    5    3
2    0   11   11   10    8   11    7    9    4
3    6    4    5    0   12   11   10    9   13
Element    24256 (    1001) is first >   1000
Element    43302 (    2009) is first >   2000
Element    61709 (    3001) is first >   3000
Element    81457 (    4003) is first >   4000
Element    98705 (    5021) is first >   5000
Element   121343 (    6009) is first >   6000
Element   151757 (    7035) is first >   7000
Element   168805 (    8036) is first >   8000
Element   184429 (    9014) is first >   9000
Element   201789 (   10007) is first >  10000
```

## APL

Works with: Dyalog APL
```invseq←{
t←1000×⍳10
seq←0{
(⍺+1)∇⍣(n>0)⊢⍵,n←+/⍺=⍵
}⍣{
⍵∨.>⊃⌽t
}⊢⍬
⎕←'First 100 elements:'
⎕←10 10⍴seq
loc←{⊃⍸seq>⍵}¨t
⎕←'⊂First > ⊃,I5,⊂: ⊃,I5,⊂ at ⊃,I6'⎕FMT t,seq[loc],[1.5]loc
}
```
Output:

Note that APL arrays are 1-indexed by default.

```First 100 elements:
0  1  1  0  2  2  2  0  3  2
4  1  1  0  4  4  4  1  4  0
5  5  4  1  6  2  1  0  6  7
5  1  6  3  3  1  0  7  9  5
3  6  4  4  2  0  8  9  6  4
9  4  5  2  1  3  0  9 10  7
5 10  6  6  3  1  4  2  0 10
11  8  6 11  6  9  3  2  5  3
2  0 11 11 10  8 11  7  9  4
3  6  4  5  0 12 11 10  9 13
First >  1000:  1001 at  24256
First >  2000:  2009 at  43302
First >  3000:  3001 at  61709
First >  4000:  4003 at  81457
First >  5000:  5021 at  98705
First >  6000:  6009 at 121343
First >  7000:  7035 at 151757
First >  8000:  8036 at 168805
First >  9000:  9014 at 184429
First > 10000: 10007 at 201789```

## BASIC

```10 DEFINT A-Z
20 DIM I(11000)
30 T=1000
40 N=0
50 E=I(N)
60 IF S#<100 THEN PRINT USING "####";E;
70 IF E>T THEN PRINT USING "First > #####: ##### at ######";T;E;S#: T=T+1000
80 S#=S#+1
90 I(E)=I(E)+1
100 N=N+1
110 IF E>0 THEN 50
120 IF T<=10000 THEN 40
```
Output:
```   0   1   1   0   2   2   2   0   3   2   4   1   1   0   4   4   4   1   4   0
5   5   4   1   6   2   1   0   6   7   5   1   6   3   3   1   0   7   9   5
3   6   4   4   2   0   8   9   6   4   9   4   5   2   1   3   0   9  10   7
5  10   6   6   3   1   4   2   0  10  11   8   6  11   6   9   3   2   5   3
2   0  11  11  10   8  11   7   9   4   3   6   4   5   0  12  11  10   9  13
First >  1000:  1001 at  24255
First >  2000:  2009 at  43301
First >  3000:  3001 at  61708
First >  4000:  4003 at  81456
First >  5000:  5021 at  98704
First >  6000:  6009 at 121342
First >  7000:  7035 at 151756
First >  8000:  8036 at 168804
First >  9000:  9014 at 184428
First > 10000: 10007 at 201788```

## C++

```#include <cstdint>
#include <iomanip>
#include <iostream>
#include <map>
#include <vector>

std::vector<uint32_t> inventory_sequence(uint32_t max_term) {
uint32_t term = 0;
std::vector<uint32_t> result = { term };
std::map<uint32_t, uint32_t> inventory = { { term, 1 } };
while ( result.back() < max_term ) {
inventory.insert({ term, 0 });
const uint32_t count = inventory[term];
term = ( count == 0 ) ? 0 : term + 1;
if ( inventory.find(count) == inventory.end() ) {
inventory.emplace(count, 1);
} else {
inventory[count]++;
}
result.emplace_back(count);
}
return result;
}

int main() {
std::vector<uint32_t> sequence = inventory_sequence(10'000);

uint32_t thousands = 1'000;
std::cout << "The first 100 numbers of the inventory sequence:" << "\n";
for ( uint64_t i = 0; i < sequence.size(); ++i ) {
const uint32_t number = sequence[i];
if ( i < 100 ) {
std::cout << std::setw(2) << number << ( i % 20 == 19 ? "\n" : " " );
} else if ( i == 100 ) {
std::cout << "\n";
} else if ( number >= thousands ) {
std::cout << "The first element ≥ " << std::setw(5) << thousands << " is "
<< std::setw(5) << number << " which occurs at index " << std::setw(6) << i << "\n";
thousands += 1'000;
}
}
}
```
Output:
```The first 100 numbers of the inventory sequence:
0  1  1  0  2  2  2  0  3  2  4  1  1  0  4  4  4  1  4  0
5  5  4  1  6  2  1  0  6  7  5  1  6  3  3  1  0  7  9  5
3  6  4  4  2  0  8  9  6  4  9  4  5  2  1  3  0  9 10  7
5 10  6  6  3  1  4  2  0 10 11  8  6 11  6  9  3  2  5  3
2  0 11 11 10  8 11  7  9  4  3  6  4  5  0 12 11 10  9 13

The first element ≥  1000 is  1001 which occurs at index  24255
The first element ≥  2000 is  2009 which occurs at index  43301
The first element ≥  3000 is  3001 which occurs at index  61708
The first element ≥  4000 is  4003 which occurs at index  81456
The first element ≥  5000 is  5021 which occurs at index  98704
The first element ≥  6000 is  6009 which occurs at index 121342
The first element ≥  7000 is  7035 which occurs at index 151756
The first element ≥  8000 is  8036 which occurs at index 168804
The first element ≥  9000 is  9014 which occurs at index 184428
The first element ≥ 10000 is 10007 which occurs at index 201788
```

## CLU

```grow = proc (arr: array[int], n: int)
while array[int]\$high(arr) < n do
end
end grow

invseq = iter () yields (int)
tally: array[int] := array[int]\$[0:0]
num: int := 0
while true do
grow(tally, num)
el: int := tally[num]
yield(el)
grow(tally, el)
tally[el] := tally[el] + 1
if el=0
then num := 0
else num := num+1
end
end
end invseq

start_up = proc ()
po: stream := stream\$primary_output()
index: int := 0
threshold: int := 1000
for n: int in invseq() do
if index<100 then
stream\$putright(po, int\$unparse(n), 5)
if index // 10 = 9 then stream\$putl(po, "") end
end
if n > threshold then
stream\$puts(po, "First > ")
stream\$putright(po, int\$unparse(threshold), 5)
stream\$puts(po, ": ")
stream\$putright(po, int\$unparse(n), 5)
stream\$puts(po, " at ")
stream\$putright(po, int\$unparse(index), 6)
stream\$putl(po, "")
threshold := threshold + 1000
end
index := index + 1
if threshold > 10000 then break end
end
end start_up```
Output:
```    0    1    1    0    2    2    2    0    3    2
4    1    1    0    4    4    4    1    4    0
5    5    4    1    6    2    1    0    6    7
5    1    6    3    3    1    0    7    9    5
3    6    4    4    2    0    8    9    6    4
9    4    5    2    1    3    0    9   10    7
5   10    6    6    3    1    4    2    0   10
11    8    6   11    6    9    3    2    5    3
2    0   11   11   10    8   11    7    9    4
3    6    4    5    0   12   11   10    9   13
First >  1000:  1001 at  24255
First >  2000:  2009 at  43301
First >  3000:  3001 at  61708
First >  4000:  4003 at  81456
First >  5000:  5021 at  98704
First >  6000:  6009 at 121342
First >  7000:  7035 at 151756
First >  8000:  8036 at 168804
First >  9000:  9014 at 184428
First > 10000: 10007 at 201788```

## Cowgol

```include "cowgol.coh";

sub len(num: uint32): (length: uint8) is
length := 1;
while num>=10 loop
length := length + 1;
num := num / 10;
end loop;
end sub;

sub printtab32(num: uint32, size: uint8) is
var spaces := size - len(num);
while spaces > 0 loop
print_char(' ');
spaces := spaces - 1;
end loop;
print_i32(num);
end sub;

sub printtab16(num: uint16, size: uint8) is
printtab32(num as uint32, size);
end sub;

var tally: uint16[11000];
var threshold: uint16 := 1000;
var index: uint32 := 0;

MemZero(&tally[0] as [uint8], @bytesof tally);

while threshold <= 10000 loop
var number: uint16 := 0;
loop
var element := tally[number];
if index < 100 then
printtab16(element, 5);
if index % 10 == 9 then
print_nl();
end if;
end if;
if element > threshold then
print("First > ");
printtab16(threshold, 5);
print(": ");
printtab16(element, 5);
print(" at ");
printtab32(index, 6);
print_nl();
threshold := threshold + 1000;
end if;
index := index + 1;
number := number + 1;
tally[element] := tally[element] + 1;
if element == 0 then
break;
end if;
end loop;
end loop;```
Output:
```    0    1    1    0    2    2    2    0    3    2
4    1    1    0    4    4    4    1    4    0
5    5    4    1    6    2    1    0    6    7
5    1    6    3    3    1    0    7    9    5
3    6    4    4    2    0    8    9    6    4
9    4    5    2    1    3    0    9   10    7
5   10    6    6    3    1    4    2    0   10
11    8    6   11    6    9    3    2    5    3
2    0   11   11   10    8   11    7    9    4
3    6    4    5    0   12   11   10    9   13
First >  1000:  1001 at  24255
First >  2000:  2009 at  43301
First >  3000:  3001 at  61708
First >  4000:  4003 at  81456
First >  5000:  5021 at  98704
First >  6000:  6009 at 121342
First >  7000:  7035 at 151756
First >  8000:  8036 at 168804
First >  9000:  9014 at 184428
First > 10000: 10007 at 201788```

## Draco

```proc main() void:
[11000]word tally;
word threshold, number, element;
ulong index;

for number from 0 upto 10999 do tally[number] := 0 od;

index := 0;
threshold := 1000;

while threshold <= 10000 do
number := 0;
while
element := tally[number];
if index < 100 then
write(element:5);
if index % 10 = 9 then writeln() fi
fi;
if element > threshold then
writeln("First > ",threshold:5,": ",element:5," at ",index:6);
threshold := threshold + 1000
fi;
index := index + 1;
number := number + 1;
tally[element] := tally[element] + 1;
element /= 0
do od
od
corp```
Output:
```    0    1    1    0    2    2    2    0    3    2
4    1    1    0    4    4    4    1    4    0
5    5    4    1    6    2    1    0    6    7
5    1    6    3    3    1    0    7    9    5
3    6    4    4    2    0    8    9    6    4
9    4    5    2    1    3    0    9   10    7
5   10    6    6    3    1    4    2    0   10
11    8    6   11    6    9    3    2    5    3
2    0   11   11   10    8   11    7    9    4
3    6    4    5    0   12   11   10    9   13
First >  1000:  1001 at  24255
First >  2000:  2009 at  43301
First >  3000:  3001 at  61708
First >  4000:  4003 at  81456
First >  5000:  5021 at  98704
First >  6000:  6009 at 121342
First >  7000:  7035 at 151756
First >  8000:  8036 at 168804
First >  9000:  9014 at 184428
First > 10000: 10007 at 201788```

## EasyLang

```sysconf zero_based
repeat
cnts[] &= 0
n = cnts[i]
cnts[n] += 1
if len cnts[] <= 100
write n & " "
.
i += 1
if n = 0
i = 0
.
until n > 1000
.
print ""
print len cnts[] & " " & n
```
Output:
```0 1 1 0 2 2 2 0 3 2 4 1 1 0 4 4 4 1 4 0 5 5 4 1 6 2 1 0 6 7 5 1 6 3 3 1 0 7 9 5 3 6 4 4 2 0 8 9 6 4 9 4 5 2 1 3 0 9 10 7 5 10 6 6 3 1 4 2 0 10 11 8 6 11 6 9 3 2 5 3 2 0 11 11 10 8 11 7 9 4 3 6 4 5 0 12 11 10 9 13
24256 1001
```

## FreeBASIC

```Dim As Integer max = 10000
Dim As Integer inv()
Dim As Integer counts(max + 100)
counts(0) = 1
Dim As Integer lower = 100
Dim As Integer upper = 1000
Dim As Boolean done = False
Dim As Integer ix = 0
While Not done
Dim As Integer i = 0, c = 0
Do
Dim As Integer j = counts(i)
If Ubound(inv) < max Then
Redim Preserve inv(ix+1)
inv(ix+1) = j
End If
counts(j) += 1
ix += 1
If Ubound(inv) >= lower Then
Print "Inventory sequence, first 100 elements:"
For c = 0 To 99
Print Using "###"; inv(c);
If (c+1) Mod 20 = 0 Then Print
Next
lower = max + 1
End If
If j = 0 Then Exit Do
If j >= upper Then
Print Using !"\nFirst element >= ##,### is ##,### at index ###,###"; upper; j; ix;
If j >= max Then done = True: Exit Do
upper += 1000
End If
i += 1
Loop
Wend

Sleep
```

## J

```   nextinv=:  ((*@] * 1+{.@[), }.@[ , ]) +/@({. = }.)

10 10\$}.nextinv^:100]0   NB. first 100 elements of inventory sequence
0  1  1  0  2  2  2  0  3  2
4  1  1  0  4  4  4  1  4  0
5  5  4  1  6  2  1  0  6  7
5  1  6  3  3  1  0  7  9  5
3  6  4  4  2  0  8  9  6  4
9  4  5  2  1  3  0  9 10  7
5 10  6  6  3  1  4  2  0 10
11  8  6 11  6  9  3  2  5  3
2  0 11 11 10  8 11  7  9  4
3  6  4  5  0 12 11 10  9 13
({:,_2+#)nextinv^:(1000>{:)^:_]0   NB. first value of at least 1000 and its index
1001 24255
```

The inventory sequence has a "hidden value" which is the number that we are searching for, and counting. So, here, we include it as the first element of the representation of an inventory subsequence. And `nextinv` calculates both the next "hidden value" as well as the corresponding subsequence which incorporates a count of how many times the current "hidden value" appeared.

For the task, we iterate inductively from the initial state (either a 100 times or until we find a value which is greater than 1000).

It would be more efficient to maintain counts of each integer so far encountered, but that efficiency is not necessary for this task (and would require more code).

That said, here's a faster implementation:

```invseq=: {{
cnt=. 0, seq=. i. nxt=. 0
while. -. u seq do.
k=. nxt{cnt
nxt=. (*k)*nxt+1
cnt=. (1+k{cnt) k} cnt=. cnt {.~ (2+k)>.#cnt
seq=. seq, k
end.
}}
```
```   stretch=: }.nextinv^:(10000>{:)^:_]0    NB. or
stretch=: (1e4<:{:) invseq   NB. equivalent, faster approach

(,~ {&stretch) {.I.2000<stretch   NB. first value greater than 2000 and its index
2009 43301
(,~ {&stretch) {.I.3000<stretch
3001 61708
(,~ {&stretch) {.I.4000<stretch
4003 81456
(,~ {&stretch) {.I.5000<stretch
5021 98704
(,~ {&stretch) {.I.6000<stretch
6009 121342
(,~ {&stretch) {.I.7000<stretch
7035 151756
(,~ {&stretch) {.I.8000<stretch
8036 168804
(,~ {&stretch) {.I.9000<stretch
9014 184428
(,~ {&stretch) {.I.10000<stretch
10007 201788
require'plot'
plot 1e4{.stretch
```

## Java

```
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public final class InventorySequence {

public static void main(String[] args) {
List<Integer> inventorySequence = inventorySequence(10_000);

int thousands = 1_000;
System.out.println("The first 100 numbers of the inventory sequence:");
for ( int i = 0; i < inventorySequence.size(); i++ ) {
final int number = inventorySequence.get(i);
if ( i < 100 ) {
System.out.print(String.format("%2d%s", number, ( i % 20 == 19 ? "\n" : " " )));
} else if ( i == 100 ) {
System.out.println();
} else if ( number >= thousands ) {
System.out.println(String.format("%s%5d%s%5d%s%6d",
"The first element ≥ ", thousands, " is ", number, " which occurs at index ", i));
thousands += 1_000;
}
}
}

private static List<Integer> inventorySequence(int maxTerm) {
int term = 0;
List<Integer> result = new ArrayList<Integer>(List.of(term ));
Map<Integer, Integer> inventory = new HashMap<Integer, Integer>(Map.of( 0, 1 ));
while ( result.getLast() < maxTerm ) {
final int count = inventory.computeIfAbsent(term, n -> 0 );
term = ( count == 0 ) ? 0 : term + 1;
inventory.merge(count, 1, Integer::sum);
}
return result;
}

}
```
Output:
```The first 100 numbers of the inventory sequence:
0  1  1  0  2  2  2  0  3  2  4  1  1  0  4  4  4  1  4  0
5  5  4  1  6  2  1  0  6  7  5  1  6  3  3  1  0  7  9  5
3  6  4  4  2  0  8  9  6  4  9  4  5  2  1  3  0  9 10  7
5 10  6  6  3  1  4  2  0 10 11  8  6 11  6  9  3  2  5  3
2  0 11 11 10  8 11  7  9  4  3  6  4  5  0 12 11 10  9 13

The first element ≥  1000 is  1001 which occurs at index  24255
The first element ≥  2000 is  2009 which occurs at index  43301
The first element ≥  3000 is  3001 which occurs at index  61708
The first element ≥  4000 is  4003 which occurs at index  81456
The first element ≥  5000 is  5021 which occurs at index  98704
The first element ≥  6000 is  6009 which occurs at index 121342
The first element ≥  7000 is  7035 which occurs at index 151756
The first element ≥  8000 is  8036 which occurs at index 168804
The first element ≥  9000 is  9014 which occurs at index 184428
The first element ≥ 10000 is 10007 which occurs at index 201788
```

## jq

Works with both jq and gojq, the C and Go implementations of jq

... but note that gojq takes about 5 times longer (and requires much more memory) to complete the task.

With minor modifications, the program below also works quite snappily with jaq, the Rust implementation of jq.

The definition of `_nwise` can be omitted if using the C implementation of jq.

```def _nwise(\$n):
def n: if length <= \$n then . else .[0:\$n] , (.[\$n:] | n) end;
n;

# Emit the inventory sequence ad infinitum
def inventory_sequence:
{num: 0,
emit: 0,
inventory: {} }
| foreach range(0; infinite) as \$n (.;
.emit = (.inventory[.num|tostring] // 0)
| if .emit == 0 then .num = 0 else .num += 1 end
| .inventory[.emit|tostring] += 1 )
| .emit ;

# Report on the progress of an arbitrary sequence, indefinitely
# Emit [.next, \$x, .n]
def probe(s; \$gap):
foreach s as \$x ({n: 0, next: \$gap};
.n += 1
| if \$x >= .next then .emit = {next, \$x, n} | .next += \$gap
else .emit = null
end)
| select(.emit).emit;

def lpad(\$len): tostring | (\$len - length) as \$l | (" " * \$l)[:\$l] + .;

[limit(\$n; inventory_sequence)] | _nwise(10) | map(lpad(3)) | join(" ");

"",
(limit(10; probe(inventory_sequence; 1000))
| "First element >= \(.next) is \(.x) at index \(.n - 1)")```
Output:
```  0   1   1   0   2   2   2   0   3   2
4   1   1   0   4   4   4   1   4   0
5   5   4   1   6   2   1   0   6   7
5   1   6   3   3   1   0   7   9   5
3   6   4   4   2   0   8   9   6   4
9   4   5   2   1   3   0   9  10   7
5  10   6   6   3   1   4   2   0  10
11   8   6  11   6   9   3   2   5   3
2   0  11  11  10   8  11   7   9   4
3   6   4   5   0  12  11  10   9  13

First element >= 1000 is 1001 at index 24255
First element >= 2000 is 2009 at index 43301
First element >= 3000 is 3001 at index 61708
First element >= 4000 is 4003 at index 81456
First element >= 5000 is 5021 at index 98704
First element >= 6000 is 6009 at index 121342
First element >= 7000 is 7035 at index 151756
First element >= 8000 is 8036 at index 168804
First element >= 9000 is 9014 at index 184428
First element >= 10000 is 10007 at index 201788
```

## Julia

```""" rosettacode.org/wiki/Inventory_sequence """

using Printf
using Counters
using Plots

struct InventorySequence
inventory::Counter{Int}
InventorySequence() = new(counter([0]))
end

function Base.iterate(i::InventorySequence, num = 0)
nextval = i.inventory[num]
num = nextval == 0 ? 0 : num + 1
i.inventory[nextval] += 1
return nextval, num
end

const thresholds = [1000 * j for j in 1:10]
const toplot = Int[]

for (i, n) in enumerate(InventorySequence())
if i <= 100
print(rpad(n, 4), i % 20 == 0 ? "\n" : "")
elseif n >= thresholds[begin]
@printf("First element >= %d5: %d5 in position %d.\n", popfirst!(thresholds), n, i)
length(thresholds) == 0 && break
end
length(toplot) < 10000 && push!(toplot, n)
end

plot(toplot)
```
Output:

Similar to Python output.

```            NORMAL MODE IS INTEGER
DIMENSION TALLY(11000)
DIMENSION ROW(10)

VECTOR VALUES ROWF = \$10(I5)*\$
VECTOR VALUES TRSHF = \$8HFIRST > ,I5,2H: ,I5,S1,3HAT ,I6*\$

RWIX = 0
INTERNAL FUNCTION(X)
ENTRY TO TBLOUT.
ROW(RWIX) = X
RWIX = RWIX + 1
WHENEVER RWIX.L.10, FUNCTION RETURN 0
PRINT FORMAT ROWF,ROW(0),ROW(1),ROW(2),ROW(3),ROW(4),
0    ROW(5),ROW(6),ROW(7),ROW(8),ROW(9)
RWIX = 0
END OF FUNCTION

TRSHLD = 1000
IDX = 0
RESTRT      NUM = 0
TAKINV      ELEM = TALLY(NUM)
WHENEVER IDX.L.100, TBLOUT.(ELEM)
WHENEVER ELEM.G.TRSHLD
PRINT FORMAT TRSHF,TRSHLD,ELEM,IDX
TRSHLD = TRSHLD + 1000
END OF CONDITIONAL
IDX = IDX+1
NUM = NUM+1
TALLY(ELEM) = TALLY(ELEM)+1
WHENEVER ELEM.G.0, TRANSFER TO TAKINV
WHENEVER TRSHLD.LE.10000, TRANSFER TO RESTRT
END OF PROGRAM```
Output:
```    0    1    1    0    2    2    2    0    3    2
4    1    1    0    4    4    4    1    4    0
5    5    4    1    6    2    1    0    6    7
5    1    6    3    3    1    0    7    9    5
3    6    4    4    2    0    8    9    6    4
9    4    5    2    1    3    0    9   10    7
5   10    6    6    3    1    4    2    0   10
11    8    6   11    6    9    3    2    5    3
2    0   11   11   10    8   11    7    9    4
3    6    4    5    0   12   11   10    9   13
FIRST >  1000:  1001 AT  24255
FIRST >  2000:  2009 AT  43301
FIRST >  3000:  3001 AT  61708
FIRST >  4000:  4003 AT  81456
FIRST >  5000:  5021 AT  98704
FIRST >  6000:  6009 AT 121342
FIRST >  7000:  7035 AT 151756
FIRST >  8000:  8036 AT 168804
FIRST >  9000:  9014 AT 184428
FIRST > 10000: 10007 AT 201788```

## Mathematica /Wolfram Language

```(*Function to generate the inventory sequence*)
InventorySequence[terms_] :=
Module[{num = 0, alst = {0}, inventory, c}, inventory = <|0 -> 1|>;
Table[c = Lookup[inventory, num, 0];
num = If[c == 0, 0, num + 1];
alst = Append[alst, c];
inventory[c] = Lookup[inventory, c, 0] + 1;, {n, 2, terms}];
alst]

(*Generate the inventory sequence*)
biglist = InventorySequence[201790];

(*Print first 100 elements of the sequence*)
partitioned = Partition[Take[biglist, 100], 10];
Do[Print[Row[partitioned[[i]], " "]], {i, Length[partitioned]}]

(*Find and print the first occurrences of elements>=thresholds*)
thresholds = 1000 Range[1, 10];
firstOccurrences =
Reap[Do[If[biglist[[i]] >= thresholds[[1]],
Sow[{thresholds[[1]], biglist[[i]], i}];
thresholds = Rest[thresholds];
If[Length[thresholds] == 0, Break[]];], {i, Length[biglist]}]][[
2, 1]];

(*Print the formatted results*)
Do[Print["First element \[GreaterEqual] ", firstOccurrence[[1]],
" is ", firstOccurrence[[2]], " at index ",
firstOccurrence[[3]]], {firstOccurrence, firstOccurrences}]

(*Plot the first 10,000 elements of the sequence*)
ListPlot[biglist[[1 ;; 10000]], Joined -> True,
PlotStyle -> {Thin, Blue}, PlotRange -> Full]
```
Output:
```  0   1   1   0   2   2   2   0   3   2
4   1   1   0   4   4   4   1   4   0
5   5   4   1   6   2   1   0   6   7
5   1   6   3   3   1   0   7   9   5
3   6   4   4   2   0   8   9   6   4
9   4   5   2   1   3   0   9  10   7
5  10   6   6   3   1   4   2   0  10
11   8   6  11   6   9   3   2   5   3
2   0  11  11  10   8  11   7   9   4
3   6   4   5   0  12  11  10   9  13
First element >= 1000 is 1001 at index 24256
First element >= 2000 is 2009 at index 43302
First element >= 3000 is 3001 at index 61709
First element >= 4000 is 4003 at index 81457
First element >= 5000 is 5021 at index 98705
First element >= 6000 is 6009 at index 121343
First element >= 7000 is 7035 at index 151757
First element >= 8000 is 8036 at index 168805
First element >= 9000 is 9014 at index 184429
First element >= 10000 is 10007 at index 201789
```

## Miranda

```main :: [sys_message]
main = [Stdout "First 100 elements:\n",
Stdout (table 10 5 (take 100 invseq)),
Stdout (lay (map disp (zip2 thresholds firsts)))
]
where thresholds = [1000, 2000..10000]
firsts = first [(>x) | x <- thresholds] invseq
disp (t,(i,x)) = concat ["First > ",
rjustify 5 (shownum t),
": ",
rjustify 5 (shownum x),
" at ",
rjustify 6 (shownum i)]

table :: num->num->[num]->[char]
table w cw nums = lay [ concat (map (rjustify cw . show) row)
| row <- group w nums]

group :: num->[*]->[[*]]
group sz [] = []
group sz ls = take sz ls : group sz (drop sz ls)

invseq :: [num]
invseq = f [] 0
where f acc i = el : f (inc el acc) i'
where el = get i acc
i' = 0, if el=0
i' = i+1

get :: num->[(num,num)]->num
get n []         = 0
get n ((n,x):ns) = x
get n (m:ns)     = get n ns

inc :: num->[(num,num)]->[(num,num)]
inc n []         = [(n, 1)]
inc n ((n,x):ns) = (n,x+1):ns
inc n (m:ns)     = m:inc n ns

first :: [(*->bool)]->[*]->[(num,*)]
first ps = f ps 0
where f []     n xs     = []
f (p:ps) n (x:xs) = (n, x) : f ps (n+1) xs, if p x
f (p:ps) n (x:xs) = f (p:ps) (n+1) xs```
Output:
```First 100 elements:
0    1    1    0    2    2    2    0    3    2
4    1    1    0    4    4    4    1    4    0
5    5    4    1    6    2    1    0    6    7
5    1    6    3    3    1    0    7    9    5
3    6    4    4    2    0    8    9    6    4
9    4    5    2    1    3    0    9   10    7
5   10    6    6    3    1    4    2    0   10
11    8    6   11    6    9    3    2    5    3
2    0   11   11   10    8   11    7    9    4
3    6    4    5    0   12   11   10    9   13
First >  1000:  1001 at  24255
First >  2000:  2009 at  43301
First >  3000:  3001 at  61708
First >  4000:  4003 at  81456
First >  5000:  5021 at  98704
First >  6000:  6009 at 121342
First >  7000:  7035 at 151756
First >  8000:  8036 at 168804
First >  9000:  9014 at 184428
First > 10000: 10007 at 201788```

## Nim

```import std/[strformat, tables]
import gnuplot

iterator inventorySequence(): (int, int) =
var counts: CountTable[int]
var idx = -1
while true:
var i = 0
while true:
let n = counts[i]
inc idx
counts.inc(n)
yield (idx, n)
if n == 0: break
inc i

echo "First 100 elements:"
var x, y: seq[int]
var lim = 1000
for idx, n in inventorySequence():
if idx < 10000:
if idx <= 100:
stdout.write &"{n:>2}"
stdout.write if idx mod 10 == 0: '\n' else: ' '
if idx == 100: echo()
elif n >= lim:
echo &"First element ⩾ {lim:>5} is {n:>5} at index {idx:>6}"
lim += 1000
if lim > 10000: break

withGnuPlot:
plot(x, y, "Inventory sequence", "with impulses lw 0.5")
png("inventory_sequence.png")
```
Output:
```First 100 elements:
0
1  1  0  2  2  2  0  3  2  4
1  1  0  4  4  4  1  4  0  5
5  4  1  6  2  1  0  6  7  5
1  6  3  3  1  0  7  9  5  3
6  4  4  2  0  8  9  6  4  9
4  5  2  1  3  0  9 10  7  5
10  6  6  3  1  4  2  0 10 11
8  6 11  6  9  3  2  5  3  2
0 11 11 10  8 11  7  9  4  3
6  4  5  0 12 11 10  9 13  8

First element ⩾  1000 is  1001 at index  24255
First element ⩾  2000 is  2009 at index  43301
First element ⩾  3000 is  3001 at index  61708
First element ⩾  4000 is  4003 at index  81456
First element ⩾  5000 is  5021 at index  98704
First element ⩾  6000 is  6009 at index 121342
First element ⩾  7000 is  7035 at index 151756
First element ⩾  8000 is  8036 at index 168804
First element ⩾  9000 is  9014 at index 184428
First element ⩾ 10000 is 10007 at index 201788
```

## Perl

Library: ntheory
Translation of: Raku
```use strict;
use warnings;
use feature 'say';

use List::AllUtils <max firstidx>;
use GD::Graph::bars;

sub comma { reverse ((reverse shift) =~ s/.{3}\K/,/gr) =~ s/^,//r }
sub table { my \$t = 20 * (my \$c = 1 + length max @_); ( sprintf( ('%'.\$c.'d')x@_, @_) ) =~ s/.{1,\$t}\K/\n/gr }

my(\$i, @inventory, %i) = 0;
do {
my \$count = \$i{\$i} // 0;
\$i = \$count ? \$i+1 : 0;
++\$i{\$count};
push @inventory, \$count
} until \$inventory[-1] > 10_000;

say "Inventory sequence, first 100 elements:\n" .  table @inventory[0..99]; say '';

for my \$n (map { \$_ * 1000 } 1..10) {
my \$i = firstidx { \$_ >= \$n } @inventory;
printf "First element >= %6s is %6s in position: %s\n", comma(\$n), comma(\$inventory[\$i]), comma \$i;
}

# graph
my @data = ( [0..5000], [@inventory[0..5000]] );
my \$graph = GD::Graph::bars->new(800, 600);
\$graph->set(
title          => 'Inventory sequence',
y_max_value    => 250,
x_tick_number  => 5,
r_margin       => 10,
dclrs          => [ 'blue' ],
) or die \$graph->error;
my \$gd = \$graph->plot(\@data) or die \$graph->error;

open my \$fh, '>', 'Perl-inventory-sequence.png';
binmode \$fh;
print \$fh \$gd->png();
close \$fh;
```
Output:
```Inventory sequence, first 100 elements:
0  1  1  0  2  2  2  0  3  2  4  1  1  0  4  4  4  1  4  0
5  5  4  1  6  2  1  0  6  7  5  1  6  3  3  1  0  7  9  5
3  6  4  4  2  0  8  9  6  4  9  4  5  2  1  3  0  9 10  7
5 10  6  6  3  1  4  2  0 10 11  8  6 11  6  9  3  2  5  3
2  0 11 11 10  8 11  7  9  4  3  6  4  5  0 12 11 10  9 13

First element >=  1,000 is  1,001 in position: 24,255
First element >=  2,000 is  2,009 in position: 43,301
First element >=  3,000 is  3,001 in position: 61,708
First element >=  4,000 is  4,003 in position: 81,456
First element >=  5,000 is  5,021 in position: 98,704
First element >=  6,000 is  6,009 in position: 121,342
First element >=  7,000 is  7,035 in position: 151,756
First element >=  8,000 is  8,036 in position: 168,804
First element >=  9,000 is  9,014 in position: 184,428
First element >= 10,000 is 10,007 in position: 201,788
```

## Phix

```-- demo\rosetta\Inventory_sequence.exw
with javascript_semantics
function inventory(integer limit)
sequence inv = {0}, counts = {1}
integer ix = 0, thousands = 1000
while true do
integer i = 0
while true do
integer j = iff(i>=length(counts)?0:counts[i+1])
inv &= j
while j>=length(counts) do counts &= 0 end while
counts[j+1] += 1
ix += 1
if length(inv)=100 then
printf(1,"Inventory sequence, first 100 elements:\n%s\n",
{join_by(inv,1,20,"",fmt:="%3d")})
end if
if j=0 then exit end if
if j>=thousands then
printf(1,"First element >= %,6d is %,6d at index %,7d\n", {thousands, j, ix})
if j>=limit then return inv[1..limit] end if
thousands += 1000
end if
i += 1
end while
end while
end function
constant lim = 1e4
sequence x = tagset(lim),
y = inventory(lim)

include pGUI.e
include IupGraph.e
function get_data(Ihandle graph)
integer {w,h} = IupGetIntInt(graph,"SIZE")
IupSetInt(graph,"XTICK",iff(w<500?iff(w<350?iff(w<250?5000:2500):2000):1000))
IupSetInt(graph,"YTICK",iff(h<350?iff(h<200?iff(h<150? 200: 100):  80):  40))
return {{x,y,CD_BLUE}}
end function

IupOpen()
Ihandle graph = IupGraph(get_data,"XMIN=0,XMAX=10000,YMIN=0,YMAX=400"),
dlg = IupDialog(graph,`TITLE=gGraph,SIZE=320x240,MINSIZE=240x140`)
IupShow(dlg)
if platform()!=JS then
IupMainLoop()
end if
```
Output:
```Inventory sequence, first 100 elements:
0  1  1  0  2  2  2  0  3  2  4  1  1  0  4  4  4  1  4  0
5  5  4  1  6  2  1  0  6  7  5  1  6  3  3  1  0  7  9  5
3  6  4  4  2  0  8  9  6  4  9  4  5  2  1  3  0  9 10  7
5 10  6  6  3  1  4  2  0 10 11  8  6 11  6  9  3  2  5  3
2  0 11 11 10  8 11  7  9  4  3  6  4  5  0 12 11 10  9 13

First element >=  1,000 is  1,001 at index  24,255
First element >=  2,000 is  2,009 at index  43,301
First element >=  3,000 is  3,001 at index  61,708
First element >=  4,000 is  4,003 at index  81,456
First element >=  5,000 is  5,021 at index  98,704
First element >=  6,000 is  6,009 at index 121,342
First element >=  7,000 is  7,035 at index 151,756
First element >=  8,000 is  8,036 at index 168,804
First element >=  9,000 is  9,014 at index 184,428
First element >= 10,000 is 10,007 at index 201,788
```

## PL/M

```100H:
BDOS: PROCEDURE (FN, ARG); DECLARE FN BYTE, ARG ADDRESS; GO TO 5; END BDOS;
EXIT: PROCEDURE; GO TO 0; END EXIT;
PRINT: PROCEDURE (STR); DECLARE STR ADDRESS; CALL BDOS(9, STR); END PRINT;

PRINT\$NUM: PROCEDURE (N, W);
DECLARE BUF (6) BYTE INITIAL ('.....\$');
DECLARE N ADDRESS, (I, W) BYTE;
DO I=0 TO 4; BUF(I) = ' '; END;
I = 5;
DIGIT:
BUF(I := I-1) = '0' + N MOD 10;
IF (N := N/10) > 0 THEN GO TO DIGIT;
CALL PRINT(.BUF(5-W));
END PRINT\$NUM;

CALC\$SEQ: PROCEDURE;

INV\$SEQ(0) = 0;
SEARCH\$NUM = 0;

DO ITEM = 1 TO LAST(INV\$SEQ);
CALL PRINT\$NUM(ITEM, 5);
CALL PRINT(.(13,'\$'));
INV\$SEQ(ITEM) = 0;

DO SEARCH\$IDX = 0 TO ITEM-1;
IF INV\$SEQ(SEARCH\$IDX) = SEARCH\$NUM THEN
INV\$SEQ(ITEM) = INV\$SEQ(ITEM) + 1;
END;

IF INV\$SEQ(ITEM) = 0
THEN SEARCH\$NUM = 0;
ELSE SEARCH\$NUM = SEARCH\$NUM + 1;
END;
END CALC\$SEQ;

DO I = 0 TO LAST(INV\$SEQ);
IF INV\$SEQ(I) >= N THEN RETURN I;
END;
END FIND\$FIRST;

CALL CALC\$SEQ;

CALL PRINT(.('FIRST 100 ITEMS OF INVENTORY SEQUENCE:',13,10,'\$'));
DO I=0 TO 99;
CALL PRINT\$NUM(INV\$SEQ(I), 4);
IF I MOD 10 = 9 THEN CALL PRINT(.(13,10,'\$'));
END;

CALL PRINT(.'FIRST ELEMENT >= 1000: \$');
CALL PRINT\$NUM(INV\$SEQ(I := FIND\$FIRST(1000)), 5);
CALL PRINT(.' AT \$');
CALL PRINT\$NUM(I, 5);
CALL EXIT;
EOF```
Output:
```FIRST 100 ITEMS OF INVENTORY SEQUENCE:
0   1   1   0   2   2   2   0   3   2
4   1   1   0   4   4   4   1   4   0
5   5   4   1   6   2   1   0   6   7
5   1   6   3   3   1   0   7   9   5
3   6   4   4   2   0   8   9   6   4
9   4   5   2   1   3   0   9  10   7
5  10   6   6   3   1   4   2   0  10
11   8   6  11   6   9   3   2   5   3
2   0  11  11  10   8  11   7   9   4
3   6   4   5   0  12  11  10   9  13
FIRST ELEMENT >= 1000:  1001 AT 24255```

## Python

```''' rosettacode.org/wiki/Inventory_sequence '''
from collections import Counter
from matplotlib.pyplot import plot

def inventory_sequence(terms):
''' From the code by Branicky at oeis.org/A342585 '''
num, alst, inventory = 0, [0], Counter([0])
for n in range(2, terms+1):
c = inventory[num]
num = 0 if c == 0 else num + 1
alst.append(c)
inventory.update([c])
return alst

biglist = inventory_sequence(201_790)
thresholds = [1000 * j for j in range(1, 11)]

for i, k in enumerate(biglist):
if i < 100:
print(f'{k:<4}', end='\n' if (i + 1) % 20 == 0 else '')
elif k >= thresholds[0]:
print(f'\nFirst element >= {thresholds.pop(0):5}: {k:5} in position {i:6}')
if len(thresholds) == 0:
break

plot(biglist[:10_000], linewidth=0.3)
plt.show()
```
Output:
```0   1   1   0   2   2   2   0   3   2   4   1   1   0   4   4   4   1   4   0
5   5   4   1   6   2   1   0   6   7   5   1   6   3   3   1   0   7   9   5
3   6   4   4   2   0   8   9   6   4   9   4   5   2   1   3   0   9   10  7
5   10  6   6   3   1   4   2   0   10  11  8   6   11  6   9   3   2   5   3
2   0   11  11  10  8   11  7   9   4   3   6   4   5   0   12  11  10  9   13

First element >=  1000:  1001 in position  24255

First element >=  2000:  2009 in position  43301

First element >=  3000:  3001 in position  61708

First element >=  4000:  4003 in position  81456

First element >=  5000:  5021 in position  98704

First element >=  6000:  6009 in position 121342

First element >=  7000:  7035 in position 151756

First element >=  8000:  8036 in position 168804

First element >=  9000:  9014 in position 184428

First element >= 10000: 10007 in position 201788
```

## Quackery

```  [ 0 unrot witheach
[ over =
rot + swap ]
drop ]                  is occurs  ( n [ --> n   )

[ 2dup occurs
tuck join
dip [ 0 != tuck * + ] ] is additem ( n [ --> n [ )

witheach
[ dup 10 < if sp
echo sp
i^ 10 mod 9 = if cr ]
drop
cr
0 []
dup -1 peek 999 >
until ]
say "Element #"
dup size 1 - echo
say " is "
-1 peek echo
say "." cr
drop```
Output:
``` 0  1  1  0  2  2  2  0  3  2
4  1  1  0  4  4  4  1  4  0
5  5  4  1  6  2  1  0  6  7
5  1  6  3  3  1  0  7  9  5
3  6  4  4  2  0  8  9  6  4
9  4  5  2  1  3  0  9 10  7
5 10  6  6  3  1  4  2  0 10
11  8  6 11  6  9  3  2  5  3
2  0 11 11 10  8 11  7  9  4
3  6  4  5  0 12 11 10  9 13

Element #24255 is 1001.```

## Raku

```use Lingua::EN::Numbers;

my (\$i, %i) = 0;

my @inventory = (^∞).map: {
my \$count = %i{\$i} // 0;
\$i = \$count ?? \$i+1 !! 0;
++%i{\$count};
\$count
}

say "Inventory sequence, first 100 elements:\n" ~
@inventory[^100].batch(20)».fmt("%2d").join: "\n";

say '';

for (1..10).map: * × 1000 {
my \$k = @inventory.first: * >= \$_, :k;
printf "First element >= %6s is %6s in position: %s\n",
.&comma, @inventory[\$k].&comma, comma \$k;
}

use SVG;
use SVG::Plot;

my @x = ^10000;

'Inventory-raku.svg'.IO.spurt:
SVG.serialize: SVG::Plot.new(
background  => 'white',
width       => 1000,
height      => 600,
plot-width  => 950,
plot-height => 550,
x           => @x,
values      => [@inventory[@x],],
title       => "Inventory Sequence - First {+@x} values (zero indexed)",
).plot: :lines;
```
Output:
```Inventory sequence, first 100 elements:
0  1  1  0  2  2  2  0  3  2  4  1  1  0  4  4  4  1  4  0
5  5  4  1  6  2  1  0  6  7  5  1  6  3  3  1  0  7  9  5
3  6  4  4  2  0  8  9  6  4  9  4  5  2  1  3  0  9 10  7
5 10  6  6  3  1  4  2  0 10 11  8  6 11  6  9  3  2  5  3
2  0 11 11 10  8 11  7  9  4  3  6  4  5  0 12 11 10  9 13

First element >=  1,000 is  1,001 in position: 24,255
First element >=  2,000 is  2,009 in position: 43,301
First element >=  3,000 is  3,001 in position: 61,708
First element >=  4,000 is  4,003 in position: 81,456
First element >=  5,000 is  5,021 in position: 98,704
First element >=  6,000 is  6,009 in position: 121,342
First element >=  7,000 is  7,035 in position: 151,756
First element >=  8,000 is  8,036 in position: 168,804
First element >=  9,000 is  9,014 in position: 184,428
First element >= 10,000 is 10,007 in position: 201,788```

converted to a .png to reduce size for display here:

## Refal

```\$ENTRY Go {
, <InvSeqTo 210000>: e.Inv
, <First 100 e.Inv>: (e.100) e.1
, <Each (Mul 1000) <Iota 1 10>>: e.Ts
= <Prout 'First 100 elements:'>
<Tbl 10 5 e.100>
<Each (DispFind (e.Inv)) e.Ts>;
};

Iota {
s.E s.E = s.E;
s.S s.E = s.S <Iota <+ 1 s.S> s.E>;
};

DispFind {
t.L s.T, <Find t.L s.T>: s.X s.N =
<Prout 'First > ' <Fmt 5 s.T> ': '
<Fmt 5 s.X> ' at ' <Fmt 6 s.N>>;
};

Find {
(s.X e.Y) s.T = <Find (s.X e.Y) 0 s.T>;
(s.X e.Y) s.N s.T, <Compare s.X s.T>: {
'+' = s.X s.N;
s.1 = <Find (e.Y) <+ 1 s.N> s.T>;
};
};

Tbl {
s.W s.CW = ;
s.W s.CW e.X, <First s.W e.X>: (e.R) e.Y =
<Prout <Each (Fmt s.CW) e.R>>
<Tbl s.W s.CW e.Y>;
};

Fmt {
s.W s.N,
<Rep s.W ' '> <Symb s.N>: e.F,
<Last s.W e.F>: (e.1) e.2 = e.2;
};

Rep {
0 s.C = ;
s.N s.C = s.C <Rep <- s.N 1> s.C>;
};

Each {
(e.F) = ;
(e.F) t.I e.X = <Mu e.F t.I> <Each (e.F) e.X>;
};

InvSeqTo {
s.Num = <InvSeqTo () () s.Num>;
(e.X) (e.L) s.Num,
<SeqStep (e.X) (e.L) 0>: (e.X2) (e.L2),
<Lenw e.L2>: s.Len e.L2,
<Compare s.Len s.Num>: {
'-' = <InvSeqTo (e.X2) (e.L2) s.Num>;
s.1 = e.L2;
};
};

SeqStep {
(e.X) (e.L) s.N,
<GetN (e.X) s.N>: s.C,
<IncN (e.X) s.C>: e.X2,
e.L s.C: e.L2,
s.C: {
0 = (e.X2) (e.L2);
s.C = <SeqStep (e.X2) (e.L2) <+ 1 s.N>>;
};
};

GetN {
(e.X (s.N s.C) e.Y) s.N = s.C;
(e.X) s.N = 0;
};

IncN {
(e.X (s.N s.C) e.Y) s.N = e.X (s.N <+ 1 s.C>) e.Y;
(e.X) s.N = e.X (s.N 1);
};```
Output:
```First 100 elements:
0    1    1    0    2    2    2    0    3    2
4    1    1    0    4    4    4    1    4    0
5    5    4    1    6    2    1    0    6    7
5    1    6    3    3    1    0    7    9    5
3    6    4    4    2    0    8    9    6    4
9    4    5    2    1    3    0    9   10    7
5   10    6    6    3    1    4    2    0   10
11    8    6   11    6    9    3    2    5    3
2    0   11   11   10    8   11    7    9    4
3    6    4    5    0   12   11   10    9   13
First >  1000:  1001 at  24255
First >  2000:  2009 at  43301
First >  3000:  3001 at  61708
First >  4000:  4003 at  81456
First >  5000:  5021 at  98704
First >  6000:  6009 at 121342
First >  7000:  7035 at 151756
First >  8000:  8036 at 168804
First >  9000:  9014 at 184428
First > 10000: 10007 at 201788```

## RPL

For efficiency reasons, two different programs are needed to generate the sequence or search for the first high value.

```« → max
« { 0 1 1 0 }                              @ need to start with a non-null cycle to have ∑LIST work
WHILE DUP SIZE max < REPEAT
0 max FOR j
DUP 1 « j == » DOLIST ∑LIST       @ count occurrences in the list
IF DUP NOT THEN max 'j' STO END
+
NEXT
END
1 max SUB
» 'INVT' STO                                 @ ( n → { a(1)..a(n)} )

« DUP 1 + { } + 0 CON -1 → max counts j
« 2 CF 1
DO 'counts' 'j' INCR 1 +
IFERR GET THEN DROP2 0 END
IF DUP NOT THEN -1 'j' STO END
IF DUP max > THEN
"element" →TAG SWAP "position" →TAG 2 SF
ELSE
'counts' SWAP 1 + DUP2 GET 1 + PUT 1 +
END
UNTIL 2 FS? END
» » 'INVT1ST' STO                            @ ( n → 1st_value_>_n pos )
```
```100 INVT
1000 INVT1ST
```
Output:
```3: { 0 1 1 0 2 2 2 0 3 2 4 1 1 0 4 4 4 1 4 0 5 5 4 1 6 2 1 0 6 7 5 1 6 3 3 1 0 7 9 5 3 6 4 4 2 0 8 9 6 4 9 4 5 2 1 3 0 9 10 7 5 10 6 6 3 1 4 2 0 10 11 8 6 11 6 9 3 2 5 3 2 0 11 11 10 8 11 7 9 4 3 6 4 5 0 12 11 10 9 13 }
2: element: 1001
1: position: 24256
```

## Ruby

Not actually counting but keeping count in a hash:

```n = 0
counter = Hash.new(0)
inventory = loop.lazy.map do
c = counter[n]
counter[c] += 1
c == 0 ? n = 0 : n += 1
c
end
inventory.first(100).each_slice(10){|s| puts "%4d"*s.size % s}
puts

(1000..10000).step(1000).each do |t|
counter.clear
puts "First element >= #{t} : %d index %d" % inventory.with_index.detect{|e,i| e > t}
end
```
Output:
```   0   1   1   0   2   2   2   0   3   2
4   1   1   0   4   4   4   1   4   0
5   5   4   1   6   2   1   0   6   7
5   1   6   3   3   1   0   7   9   5
3   6   4   4   2   0   8   9   6   4
9   4   5   2   1   3   0   9  10   7
5  10   6   6   3   1   4   2   0  10
11   8   6  11   6   9   3   2   5   3
2   0  11  11  10   8  11   7   9   4
3   6   4   5   0  12  11  10   9  13

First element >= 1000 : 1001 index 24255
First element >= 2000 : 2009 index 43301
First element >= 3000 : 3001 index 61708
First element >= 4000 : 4003 index 81456
First element >= 5000 : 5021 index 98704
First element >= 6000 : 6009 index 121342
First element >= 7000 : 7035 index 151756
First element >= 8000 : 8036 index 168804
First element >= 9000 : 9014 index 184428
First element >= 10000 : 10007 index 201788
```

## SETL

```program inventory_sequence;
loop init
inv := {};
next := 1000;
while next <= 10**4 do
loop init
i := 0;
until el = 0 do
el := inv(i) ? 0;
if (seq +:= 1) <= 100 then
if seq mod 10 = 0 then print; end if;
elseif el > next then
print("First > " + lpad(str next, 10) + ": "
next +:= 1000;
end if;
inv(el) +:= 1;
i +:= 1;
end loop;
end loop;
end program;```
Output:
```    0    1    1    0    2    2    2    0    3    2
4    1    1    0    4    4    4    1    4    0
5    5    4    1    6    2    1    0    6    7
5    1    6    3    3    1    0    7    9    5
3    6    4    4    2    0    8    9    6    4
9    4    5    2    1    3    0    9   10    7
5   10    6    6    3    1    4    2    0   10
11    8    6   11    6    9    3    2    5    3
2    0   11   11   10    8   11    7    9    4
3    6    4    5    0   12   11   10    9   13
First >       1000:       1001 at     24255
First >       2000:       2009 at     43301
First >       3000:       3001 at     61708
First >       4000:       4003 at     81456
First >       5000:       5021 at     98704
First >       6000:       6009 at    121342
First >       7000:       7035 at    151756
First >       8000:       8036 at    168804
First >       9000:       9014 at    184428
First >      10000:      10007 at    201788```

## Wren

Library: DOME
Library: Wren-plot
Library: Wren-iterate
Library: Wren-fmt
```import "dome" for Window
import "graphics" for Canvas, Color
import "./plot" for Axes
import "./iterate" for Stepped
import "./fmt" for Fmt

var max = 10000
var inv = [0]
var counts = List.filled(max + 100, 0) // say
counts[0] = 1
var lower = 100
var upper = 1000
var done = false
var ix = 0
while (!done) {
var i = 0
while(true) {
var j = counts[i]
counts[j] = counts[j] + 1
ix = ix + 1
if (inv.count >= lower) {
System.print("Inventory sequence, first 100 elements:")
Fmt.tprint("\$2d", inv[0..99], 20)
System.print()
lower = max + 1
}
if (j == 0) break
if (j >= upper) {
Fmt.print("First element >= \$,6d is \$,6d at index \$,7d", upper, j, ix)
if (j >= max) {
done = true
break
}
upper = upper + 1000
}
i = i + 1
}
}

// generate points for the plot
var Pts = (0...max).map { |i| [i, inv[i]] }.toList

class Main {
construct new() {
Window.title = "Inventory sequence - first 10,000 elements."
Canvas.resize(1000, 600)
Window.resize(1000, 600)
Canvas.cls(Color.white)
var axes = Axes.new(100, 500, 800, 400, 0..10000, 0..450)
axes.draw(Color.black, 2)
var xMarks = Stepped.new(0..10000, 500)
var yMarks = Stepped.new(0..400, 50)
axes.mark(xMarks, yMarks, Color.black, 2)
var xMarks2 = Stepped.new(0..10000, 1000)
var yMarks2 = Stepped.new(0..400, 100)
axes.label(xMarks2, yMarks2, Color.black, 2, Color.black)
axes.lineGraph(Pts, Color.blue, 2)
}

init() {}

update() {}

draw(alpha) {}
}

var Game = Main.new()
```
Output:

Terminal output:

```Inventory sequence, first 100 elements:
0  1  1  0  2  2  2  0  3  2  4  1  1  0  4  4  4  1  4  0
5  5  4  1  6  2  1  0  6  7  5  1  6  3  3  1  0  7  9  5
3  6  4  4  2  0  8  9  6  4  9  4  5  2  1  3  0  9 10  7
5 10  6  6  3  1  4  2  0 10 11  8  6 11  6  9  3  2  5  3
2  0 11 11 10  8 11  7  9  4  3  6  4  5  0 12 11 10  9 13

First element >=  1,000 is  1,001 at index  24,255
First element >=  2,000 is  2,009 at index  43,301
First element >=  3,000 is  3,001 at index  61,708
First element >=  4,000 is  4,003 at index  81,456
First element >=  5,000 is  5,021 at index  98,704
First element >=  6,000 is  6,009 at index 121,342
First element >=  7,000 is  7,035 at index 151,756
First element >=  8,000 is  8,036 at index 168,804
First element >=  9,000 is  9,014 at index 184,428
First element >= 10,000 is 10,007 at index 201,788
```

## XPL0

```include xpllib; \for Print

def Size = 201_790;
int Seq(Size), SeqEnd, Num, Count, N, Thresh;
[SeqEnd:= 0;
loop    [Num:= 0;
repeat  Count:= 0;
for N:= 0 to SeqEnd-1 do
if Num = Seq(N) then Count:= Count+1;
Seq(SeqEnd):= Count;
SeqEnd:= SeqEnd+1;
if SeqEnd >= Size then quit;
Num:= Num+1;
until   Count = 0;
];
Thresh:= 1000;
for N:= 0 to SeqEnd-1 do
if N < 100 then
[Print("%3.0f", float(Seq(N)));
if rem(N/20) = 19 then CrLf(0);
]
else if Seq(N) >= Thresh then
[Print("First element >= %5.0f: %5.0f in position %6.0f\n",
float(Thresh), float(Seq(N)), float(N));
if Thresh >= 10000 then return;
Thresh:= Thresh + 1000;
];
]```
Output:
```  0  1  1  0  2  2  2  0  3  2  4  1  1  0  4  4  4  1  4  0
5  5  4  1  6  2  1  0  6  7  5  1  6  3  3  1  0  7  9  5
3  6  4  4  2  0  8  9  6  4  9  4  5  2  1  3  0  9 10  7
5 10  6  6  3  1  4  2  0 10 11  8  6 11  6  9  3  2  5  3
2  0 11 11 10  8 11  7  9  4  3  6  4  5  0 12 11 10  9 13
First element >=  1000:  1001 in position  24255
First element >=  2000:  2009 in position  43301
First element >=  3000:  3001 in position  61708
First element >=  4000:  4003 in position  81456
First element >=  5000:  5021 in position  98704
First element >=  6000:  6009 in position 121342
First element >=  7000:  7035 in position 151756
First element >=  8000:  8036 in position 168804
First element >=  9000:  9014 in position 184428
First element >= 10000: 10007 in position 201788
```