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# Ulam numbers

Ulam numbers is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

With the following initial conditions:

```      u1 = 1
u2 = 2
```

We define the   nth   Ulam number   un   as the smallest number that is greater than   un-1     and
is the unique sum of two different Ulam numbers   ui (<n)   and   uj (<n).

Write a function to generate the   nth   Ulam number.

References

## 11l

Translation of: Python
`F ulam(n)   I n <= 2      R n   V mx = 1352000   V lst = [1, 2] [+] [0] * mx   V sums = [0] * (mx * 2 + 1)   sums[3] = 1   V size = 2   Int query   L size < n      query = lst[size - 1] + 1      L         I sums[query] == 1            L(i) 0 .< size               V sum = query + lst[i]               V t = sums[sum] + 1               I t <= 2                  sums[sum] = t            (lst[size], size) = (query, size + 1)            L.break         query++   R query L(p) 5   V n = 10 ^ p   print(‘The #.#. Ulam number is #.’.format(n, I n != 1 {‘th’} E ‘st’, ulam(n)))`
Output:
```The 1st Ulam number is 1
The 10th Ulam number is 18
The 100th Ulam number is 690
The 1000th Ulam number is 12294
The 10000th Ulam number is 132788
```

## Action!

Calculations on a real Atari 8-bit computer take quite long time. It is recommended to use an emulator capable with increasing speed of Atari CPU.

`PROC Main()  DEFINE MAX="1000"  INT ARRAY ulam(MAX)  INT uCount,n,count,x,y  BYTE flag   ulam(0)=1 ulam(1)=2 uCount=2  PrintI(ulam(0)) Put(32) PrintI(ulam(1))  n=3  WHILE uCount<MAX  DO    flag=0 count=0    FOR x=0 TO uCount-2    DO      FOR y=x+1 TO uCount-1      DO        IF ulam(x)+ulam(y)=n THEN          flag=1 count==+1        FI      OD    OD    IF flag=1 AND count=1 THEN      ulam(uCount)=n uCount==+1      Put(32) PrintI(n)    FI    n==+1  ODRETURN`
Output:
```1 2 3 4 6 8 11 13 16 18 26 28 36 38 47 48 53 57 62 69 72 77 82 87 97 99 102 106 114 126 131 138 145 148 155 175 ...
```

## AWK

` # syntax: GAWK -f ULAM_NUMBERS.AWKBEGIN {    u = split("1,2",ulam,",")    for (n=3; ; n++) {      count = 0      for (x=1; x<=u-1; x++) {        for (y=x+1; y<=u; y++) {          if (ulam[x] + ulam[y] == n) {            count++          }        }      }      if (count == 1) {        ulam[++u] = n        if (u ~ /^(10|50|100|500|1000)\$/) {          printf("%6d %6d\n",u,n)          if (++shown >= 5) { break }        }      }    }    exit(0)} `
Output:
```    10     18
50    253
100    690
500   5685
1000  12294
```

## C

Translation of: Phix
`#include <stdio.h>#include <stdlib.h>#include <string.h>#include <time.h> void fatal(const char* message) {    fprintf(stderr, "%s\n", message);    exit(1);} void* xmalloc(size_t n) {    void* ptr = malloc(n);    if (ptr == NULL)        fatal("Out of memory");    return ptr;} void* xrealloc(void* p, size_t n) {    void* ptr = realloc(p, n);    if (ptr == NULL)        fatal("Out of memory");    return ptr;} int* extend(int* array, int min_length, int* capacity) {    int new_capacity = *capacity;    if (new_capacity >= min_length)        return array;    while (new_capacity < min_length)        new_capacity *= 2;    array = xrealloc(array, new_capacity * sizeof(int));    memset(array + *capacity, 0, (new_capacity - *capacity) * sizeof(int));    *capacity = new_capacity;    return array;} int ulam(int n) {    int* ulams = xmalloc((n < 2 ? 2 : n) * sizeof(int));    ulams[0] = 1;    ulams[1] = 2;    int sieve_length = 2;    int sieve_capacity = 2;    int* sieve = xmalloc(sieve_capacity * sizeof(int));    sieve[0] = sieve[1] = 1;    for (int u = 2, ulen = 2; ulen < n; ) {        sieve_length = u + ulams[ulen - 2];        sieve = extend(sieve, sieve_length, &sieve_capacity);        for (int i = 0; i < ulen - 1; ++i)            ++sieve[u + ulams[i] - 1];        for (int i = u; i < sieve_length; ++i) {            if (sieve[i] == 1) {                u = i + 1;                ulams[ulen++] = u;                break;            }        }    }    int result = ulams[n - 1];    free(ulams);    free(sieve);    return result;} int main() {    clock_t start = clock();    for (int n = 1; n <= 100000; n *= 10)        printf("Ulam(%d) = %d\n", n, ulam(n));    clock_t finish = clock();    printf("Elapsed time: %.3f seconds\n", (finish - start + 0.0)/CLOCKS_PER_SEC);    return 0;}`
Output:
```Ulam(1) = 1
Ulam(10) = 18
Ulam(100) = 690
Ulam(1000) = 12294
Ulam(10000) = 132788
Ulam(100000) = 1351223
Elapsed time: 8.630 seconds
```

## C++

Translation of: Phix
`#include <algorithm>#include <chrono>#include <iostream>#include <vector> int ulam(int n) {    std::vector<int> ulams{1, 2};    std::vector<int> sieve{1, 1};    for (int u = 2; ulams.size() < n; ) {        sieve.resize(u + ulams[ulams.size() - 2], 0);        for (int i = 0; i < ulams.size() - 1; ++i)            ++sieve[u + ulams[i] - 1];        auto it = std::find(sieve.begin() + u, sieve.end(), 1);        if (it == sieve.end())            return -1;        u = (it - sieve.begin()) + 1;        ulams.push_back(u);    }    return ulams[n - 1];} int main() {    auto start = std::chrono::high_resolution_clock::now();    for (int n = 1; n <= 100000; n *= 10)        std::cout << "Ulam(" << n << ") = " << ulam(n) << '\n';    auto end = std::chrono::high_resolution_clock::now();    std::chrono::duration<double> duration(end - start);    std::cout << "Elapsed time: " << duration.count() << " seconds\n";}`
Output:
```Ulam(1) = 1
Ulam(10) = 18
Ulam(100) = 690
Ulam(1000) = 12294
Ulam(10000) = 132788
Ulam(100000) = 1351223
Elapsed time: 9.09242 seconds
```

## FreeBASIC

`redim as uinteger ulam(1 to 2)ulam(1) = 1 : ulam(2) = 2 function get_ulam( n as uinteger, ulam() as uinteger ) as uinteger    dim as uinteger nu = ubound(ulam), c, r, s, t, i, usr    if n <= nu then return ulam(n) 'if we have already calculated this one, just return it    'otherwise, calculate it and all intermediate terms    redim preserve ulam(1 to n)    for t = nu+1 to n        i = ulam(t-1)        while true            i += 1 : c = 0            for r = 1 to t-2                for s = r+1 to t-1                    usr = ulam(s)+ulam(r)                    if usr > i then exit for  'efficiency                    if usr = i then                        if c = 1 then continue while                        c += 1                    end if                next s            next r            if c = 0 then continue while 'I'm not 100% sure this is even possible...            ulam(t) = i            exit while        wend    next t    return ulam(n)end function  for i as uinteger = 1 to 4    print 10^i, get_ulam(10^i, ulam())next i `
Output:
``` 10          18
100          690
1000        12294
10000        132788```

## Go

### Version 1

Translation of: Wren
`package main import "fmt" func ulam(n int) int {    ulams := []int{1, 2}    set := map[int]bool{1: true, 2: true}    i := 3    for {        count := 0        for j := 0; j < len(ulams); j++ {            _, ok := set[i-ulams[j]]            if ok && ulams[j] != (i-ulams[j]) {                count++                if count > 2 {                    break                }            }        }        if count == 2 {            ulams = append(ulams, i)            set[i] = true            if len(ulams) == n {                break            }        }        i++    }    return ulams[n-1]} func main() {    for n := 10; n <= 10000; n *= 10 {        fmt.Println("The", n, "\bth Ulam number is", ulam(n))    }}`
Output:
```The 10th Ulam number is 18
The 100th Ulam number is 690
The 1000th Ulam number is 12294
The 10000th Ulam number is 132788
```

### Version 2

Translation of: Phix

The above version is reasonably efficient and runs in about 2.9 seconds on my machine (Intel Core i7-8565U). The following version, which builds up a sieve as it goes along, is (astonishingly) about 40 times faster!

Although not shown here, the 100,000th Ulam number (1,351,223) is computed in about 13.5 seconds.

`package main import (    "fmt"    "time") func ulam(n int) int {    ulams := []int{1, 2}    sieve := []int{1, 1}    u := 2    for len(ulams) < n {        s := u + ulams[len(ulams)-2]        t := s - len(sieve)        for i := 0; i < t; i++ {            sieve = append(sieve, 0)        }        for i := 1; i <= len(ulams)-1; i++ {            v := u + ulams[i-1] - 1            sieve[v]++        }        index := -1        for i, e := range sieve[u:] {            if e == 1 {                index = u + i                break            }        }        u = index + 1        ulams = append(ulams, u)    }    return ulams[n-1]} func commatize(n int) string {    s := fmt.Sprintf("%d", n)    if n < 0 {        s = s[1:]    }    le := len(s)    for i := le - 3; i >= 1; i -= 3 {        s = s[0:i] + "," + s[i:]    }    if n >= 0 {        return s    }    return "-" + s} func main() {    start := time.Now()    for n := 1; n <= 10000; n *= 10 {        s := "th"        if n == 1 {            s = "st"        }        fmt.Println("The", commatize(n), "\b"+s+" Ulam number is", commatize(ulam(n)))    }    fmt.Println("\nTook", time.Since(start))}`
Output:
```The 1st Ulam number is 1
The 10th Ulam number is 18
The 100th Ulam number is 690
The 1,000th Ulam number is 12,294
The 10,000th Ulam number is 132,788

Took 74.45373ms
```

### Version 3

Translation of: XPL0

This version is even quicker than Version 2 and reduces the time needed to calculate the 10,000th and 100,000th Ulam numbers to about 40 milliseconds and 3.25 seconds respectively.

As mentioned in the Wren version 3 example, you need to know how much memory to allocate in advance.

`package main import (    "fmt"    "time") func ulam(n int) int {    if n <= 2 {        return n    }    const MAX = 1_352_000    list := make([]int, MAX+1)    list[0], list[1] = 1, 2    sums := make([]byte, 2*MAX+1)    sums[3] = 1    size := 2    var query int    for {        query = list[size-1] + 1        for {            if sums[query] == 1 {                for i := 0; i < size; i++ {                    sum := query + list[i]                    t := sums[sum] + 1                    if t <= 2 {                        sums[sum] = t                    }                }                list[size] = query                size++                break            }            query++        }        if size >= n {            break        }    }    return query} func commatize(n int) string {    s := fmt.Sprintf("%d", n)    if n < 0 {        s = s[1:]    }    le := len(s)    for i := le - 3; i >= 1; i -= 3 {        s = s[0:i] + "," + s[i:]    }    if n >= 0 {        return s    }    return "-" + s} func main() {    start := time.Now()    for n := 10; n <= 100000; n *= 10 {        fmt.Println("The", commatize(n), "\bth Ulam number is", commatize(ulam(n)))    }    fmt.Println("\nTook", time.Since(start))}`
Output:
```The 10th Ulam number is 18
The 100th Ulam number is 690
The 1,000th Ulam number is 12,294
The 10,000th Ulam number is 132,788
The 100,000th Ulam number is 1,351,223

Took 3.226255944s
```

### Lazy List

` import Data.List ulam   :: Integral i =>     Int -> iulam 1 = 1ulam 2 = 2ulam n  | n > 2 = ulams !! (n-1) ulams  :: Integral n =>     [n]ulams = 1:2:(nexts [2,1])nexts us = u: (nexts (u:us))  where    n = length us    [u] = head . filter isSingleton . group . sort  \$             [v | i <- [0 .. n-2], j <- [i+1 .. n-1]                , let s = us !! i               , let t = us !! j               , let v = s+t               , v > head us               ] isSingleton :: [a] -> BoolisSingleton as  | length as == 1 = True  | otherwise      = False`

## Java

Translation of: Phix
`public class UlamNumbers {    public static void main(String[] args) {        long start = System.currentTimeMillis();        for (int n = 1; n <= 100000; n *= 10) {            System.out.printf("Ulam(%d) = %d\n", n, ulam(n));        }        long finish = System.currentTimeMillis();        System.out.printf("Elapsed time: %.3f seconds\n", (finish - start)/1000.0);    }     private static int ulam(int n) {        int[] ulams = new int[Math.max(n, 2)];        ulams[0] = 1;        ulams[1] = 2;        int sieveLength = 2;        int[] sieve = new int[sieveLength];        sieve[0] = sieve[1] = 1;        for (int u = 2, ulen = 2; ulen < n; ) {            sieveLength = u + ulams[ulen - 2];            sieve = extend(sieve, sieveLength);            for (int i = 0; i < ulen - 1; ++i)                ++sieve[u + ulams[i] - 1];            for (int i = u; i < sieveLength; ++i) {                if (sieve[i] == 1) {                    u = i + 1;                    ulams[ulen++] = u;                    break;                }            }        }        return ulams[n - 1];    }     private static int[] extend(int[] array, int minLength) {        if (minLength <= array.length)            return array;        int newLength = 2 * array.length;        while (newLength < minLength)            newLength *= 2;        int[] newArray = new int[newLength];        System.arraycopy(array, 0, newArray, 0, array.length);        return newArray;    }}`
Output:
```Ulam(1) = 1
Ulam(10) = 18
Ulam(100) = 690
Ulam(1000) = 12294
Ulam(10000) = 132788
Ulam(100000) = 1351223
Elapsed time: 9.098 seconds
```

## jq

`# Input: the target number of Ulam numbers to generate# Output: an array of Ulam numbersdef ulams:  . as \$target  | label \$done  | {ulam: [1, 2],     nulams: 2}  | foreach range(3; infinite) as \$n (.;      .count = 0      | .x = 0      | until( .x == .nulams or .count > 1;          .y = .x+1	  | until( .y >= .nulams or .count > 1;              if (.ulam[.x] + .ulam[.y] == \$n) then .count += 1 else . end	      | .y += 1)	   | .x += 1)        | if .count == 1 then .nulams += 1 | .ulam[.nulams-1] = \$n else . end;       select(.nulams >= \$target) | .ulam, break \$done); def nth_ulam: ulams[.-1];`

Illustration

`(5 | nth_ulam) | "5 => \(.)","",([5, 10, 100] as \$in | (100|ulams) as \$u | \$in[] | "\(.) => \(\$u[. - 1])" ) `
Output:
```5 => 6

5 => 6
10 => 18
100 => 690
```

## Julia

Translation of: Wren
`function nthUlam(n)    ulams = [1, 2]    memoized = Set([1, 2])    i = 3    while true        count = 0        for j in 1:length(ulams)            if i - ulams[j] in memoized && ulams[j] != i - ulams[j]                (count += 1) > 2 && break            end        end        if count == 2            push!(ulams, i)            push!(memoized, i)            length(ulams) == n && break        end        i += 1    end    return ulams[n]end nthUlam(5) for n in [10, 100, 1000, 10000]    @time println("The ", n, "th Ulam number is: ", nthUlam(n))end `
Output:
```The 10th Ulam number is: 18
0.000657 seconds (27 allocations: 1.422 KiB)
The 100th Ulam number is: 690
0.000959 seconds (39 allocations: 7.094 KiB)
The 1000th Ulam number is: 12294
0.027564 seconds (52 allocations: 72.188 KiB)
The 10000th Ulam number is: 132788
3.076024 seconds (63 allocations: 473.125 KiB)
```

## Lua

Implemented from scratch, but algorithmically equivalent to other solutions where a running count of number-of-ways-to-reach-sum is maintained in order to sieve candidate values.

`function ulam(n)  local ulams, nways, i = { 1,2 }, { 0,0,1 }, 3  repeat    if nways[i] == 1 then      for j = 1, #ulams do        local sum = i + ulams[j]        nways[sum] = (nways[sum] or 0) + 1      end      ulams[#ulams+1] = i    end    i = i + 1  until #ulams == n  return ulams[#ulams]end for _,n in ipairs({10,100,1000,10000,100000}) do  local s, u, e = os.clock(), ulam(n), os.clock()  print(string.format("%dth is %d (%f seconds elapsed)", n, u, e-s))end`
Output:

Times are Lua 5.4 on [email protected]

```10th is 18 (0.000000 seconds elapsed)
100th is 690 (0.000000 seconds elapsed)
1000th is 12294 (0.020000 seconds elapsed)
10000th is 132788 (1.724000 seconds elapsed)
100000th is 1351223 (277.824000 seconds elapsed)```

## Nim

### Phix algorithm

Translation of: Wren

This is a translation of Wren second solution which uses Phix algorithm.

It has been compiled with option `-d:release` which means that all runtime checks are done but debugging data is limited.

`import strformat, times func ulam(n: Positive): int =  var    ulams = @[1, 2]    sieve = @[1, 1]    u = 2  while ulams.len < n:    let s = u + ulams[^2]    sieve.setLen(s)    for i in 0..<ulams.high:      let v = u + ulams[i] - 1      inc sieve[v]    for i in u..sieve.high:      if sieve[i] == 1:        u = i + 1        break    ulams.add u  result = ulams[^1] let t0 = cpuTime()var n = 1for _ in 0..4:  let suffix = if n == 1: "st" else: "th"  echo &"The {n}{suffix} Ulam number is {ulam(n)}."  n *= 10echo &"\nTook {cpuTime() - t0:.3f} s."`
Output:

The program runs in about 150 ms on my laptop (i5-8250U with 8GB of RAM and running Linux Manjaro). It allows to get the 100_000th Ulam number in about 30 seconds.

```The 1st Ulam number is 2.
The 10th Ulam number is 18.
The 100th Ulam number is 690.
The 1000th Ulam number is 12294.
The 10000th Ulam number is 132788.

Took 0.148 s.```

### XPL0 algorithm

Translation of: Wren

This is a translation of Wren third solution which uses XPL0 algorithm.

It has been compiled with the same option as the other version.

` import strformat, times func ulam(n: Positive): int =  if n <= 2: return n  const Max = 1352000  var list = newSeq[int](Max)  list[0] = 1  list[1] = 2  var sums = newSeq[byte](2 * Max + 1)  sums[3] = 1  var size = 2  var query: int  while size < n:    query = list[size-1] + 1    while true:      if sums[query] == 1:        for i in 0..<size:          let sum = query + list[i]          let t = sums[sum] + 1          if t <= 2: sums[sum] = t        list[size] = query        inc size        break      inc query  result = query let t0 = cpuTime()var n = 10while n <= 100_000:  echo &"The {n}th Ulam number is {ulam(n)}."  n *= 10echo &"\nTook {cpuTime() - t0:.3f} s."`
Output:

In a first translation, we got the result in about 24 seconds which is the time obtained by the XPL0 version on a less powerful machine. We found that declaring "sums" as a sequence of bytes (8 bits) instead of a sequence of integers (64 bits) makes a big difference. We then got the result in 5 seconds. Note than there are also some ways to improve the other algorithm by using 32 bits integers rather than 64 bits integers, but it will still remain at least 50 % less efficient.

```The 10th Ulam number is 18.
The 10th Ulam number is 18.
The 100th Ulam number is 690.
The 1000th Ulam number is 12294.
The 10000th Ulam number is 132788.
The 100000th Ulam number is 1351223.

Took 4.986 s.```

## Pascal

Works with: Free Pascal
like GO,PHIX who was first
`program UlamNumbers;{\$IFDEF FPC}  {\$MODE DELPHI}  {\$Optimization On,All}{\$ENDIF}uses  sysutils;const  maxUlam = 100000;  Limit = 1351223+4000;type  tCheck  =  Uint16;  tpCheck = pUint16;var  Ulams : array of Uint32;  Check0 : array of tCheck;  Ulam_idx :NativeInt; procedure init;begin  setlength(Ulams,maxUlam);  Ulams[0] := 1;  Ulams[1] := 2;  Ulam_idx := 1;  setlength(check0,Limit);   check0[1]:=1;  check0[2]:=1;end; procedure OutData(idx,num:NativeInt);Begin  writeln(' Ulam(',idx+1,')',#9#9,num:10);end; function findNext(i:nativeInt;pCh0:tpCheck):NativeInt;begin  result := i;  repeat    if pCh0[result] = 1 then      break;    inc(result);  until false;end; procedure SumOne(idx:NativeUint;pCh:tpCheck;pUlams:pUint32);//seperated speeds up a lot by reducing register pressure in mainBegin  For idx := idx downto 0 do    pCh[pUlams[idx]] +=1;end; var  pCh0,pCh : tpCheck;  pUlams   : pUint32;  ul,idx,lmtidx :nativeInt;Begin   Init;  lmtidx := 9;  pCh0:= @Check0[0];  pUlams := @Ulams[0];  OutData(0,pUlams[0]);  ul := pUlams[Ulam_idx];  pCh:= @pCh0[ul];  repeat    SumOne(Ulam_idx-1,pCh,pUlams);    ul := findNext(ul+1,pCh0);    inc(Ulam_idx);    pUlams[Ulam_idx] := ul;    pCh:= @pCh0[ul];    IF ul>Limit DIV 2 then      break;    if Ulam_idx=lmtIdx then    Begin      OutData(Ulam_idx,ul);      lmtidx := lmtidx*10+9;    end;  until Ulam_idx >= maxUlam-1;   idx := Ulam_idx-1;  //now reducing then the highest used summing idx  while Ulam_idx< maxUlam-1 do  begin    while ul+pUlams[idx] > limit do      dec(idx);    SumOne(idx,pCh,pUlams);    ul := findNext(ul+1,pCh0);    inc(Ulam_idx);    pUlams[Ulam_idx] := ul;    pCh:= @pCh0[ul];  end;  OutData(Ulam_idx,ul);  setlength(check0,0);  setlength(Ulams,0);end. `
Output:
``` Ulam(1)                 1
Ulam(10)               18
Ulam(100)             690
Ulam(1000)          12294
Ulam(10000)        132788
Ulam(100000)      1351223

// real  0m4,731s AMD 2200G
//TIO.RUN
Free Pascal Compiler version 3.0.4 [2018/07/13] for x86_64
Copyright (c) 1993-2017 by Florian Klaempfl and others
Target OS: Linux for x86-64
Compiling .code.tio.pp
/usr/bin/ld: warning: link.res contains output sections; did you forget -T?
92 lines compiled, 0.2 sec

Real time: 3.025 s```

## Perl

Translation of: Julia
`use strict;use warnings;use feature <say state>; sub ulam {    my(\$n) = @_;    state %u     = (1 => 1, 2 => 1);    state @ulams = <0 1 2>; # 0 a placeholder to shift indexing up one     return \$ulams[\$n] if \$ulams[\$n];     \$n++;    my \$i = 3;     while () {        my \$count = 0;             \$u{ \$i - \$ulams[\$_] }        and \$ulams[\$_] != \$i - \$ulams[\$_]        and \$count++ > 2        and last            for 0..\$#ulams;             \$count == 2        and push(@ulams,\$i)        and \$u{\$i} = 1        and @ulams == \$n        and last;         \$i++;    }    \$ulams[\$n-1];} printf "The %dth Ulam number is: %d\n", 10**\$_, ulam(10**\$_) for 1..4;`
Output:
```The 10th Ulam number is: 18
The 100th Ulam number is: 690
The 1000th Ulam number is: 12294
The 10000th Ulam number is: 132788
The 10000th Ulam number is: 132788```

## Phix

```with javascript_semantics
function ulam(integer n)
sequence ulams = {1, 2},
sieve = {1, 1}
integer u := 2
while length(ulams)<n do
integer s = u+ulams[\$-1], t
sieve &= repeat(0,s-length(sieve))
for i=1 to length(ulams)-1 do
s = u+ulams[i]
t = sieve[s]+1
if t<=2 then
sieve[s] = t
end if
end for
u = find(1,sieve,u+1)
ulams &= u
end while
return ulams[n]
end function

atom t0 = time()
for p=0 to 4 do
integer n = power(10,p)
printf(1,"The %,d%s Ulam number is %,d\n",{n,ord(n),ulam(n)})
end for
?elapsed(time()-t0)
```
Output:
```The 1st Ulam number is 1
The 10th Ulam number is 18
The 100th Ulam number is 690
The 1,000th Ulam number is 12,294
The 10,000th Ulam number is 132,788
"1.0s"
```

For comparison, Julia took 4.5s (9.3s on repl.it), Go took 4.9s, Wren (on tio) 27s, Ring timed out (>60s) on tio before getting the 1,000th number, REXX (also on tio) got to the 1,000th number in 12.3s but timed out before getting the 10,000th, Raku (on repl.it) 9mins 50s, FreeBASIC 17mins 44s, and I cancelled XPL0 (on EXPL32) after 53 minutes. The Haskell entry does not compile for me on either tio or repl.it (The above timings all predate any {{trans|Phix}})

The above algorithm can also yield "The 100,000th Ulam number is 1,351,223" in 1 minute and 40s, for me. (I fully expect translations of this better algorithm to run even faster, btw)

## Python

Translation of: XPL0
`import time def ulam(n):    if n <= 2:        return n    mx = 1352000    lst = [1, 2] + [0] * mx    sums = [0] * (mx * 2 + 1)    sums[3] = 1    size = 2    while size < n:        query = lst[size-1] + 1        while True:            if sums[query] == 1:                for i in range(size):                    sum = query + lst[i]                    t = sums[sum] + 1                    if t <= 2:                        sums[sum] = t                lst[size], size = query, size + 1                break            query += 1    return query t0 = time.time()for p in range(5):    n = 10**p    print(f"The {n}{'th' if n!=1 else 'st'} Ulam number is {ulam(n)}") print("\nElapsed time:", time.time() - t0) `
Output:
```The 1st Ulam number is 1
The 10th Ulam number is 18
The 100th Ulam number is 690
The 1_000th Ulam number is 12_294
The 10_000th Ulam number is 132_788

(Elapsed time: 11.470759391784668 seconds)```
Extra
```In [1]: %time ulam(100_000)
Wall time: 23min 58s
Out[1]: 1351223

In [37]:
```

## Raku

`my @ulams = 1, 2, &next-ulam … *; sub next-ulam {    state \$i = 1;    state @sums = 0,1,1;    my \$last = @ulams[\$i];    (^\$i).map: { @sums[@ulams[\$_] + \$last]++ };    ++\$i;    quietly (\$last ^.. *).first: { @sums[\$_] == 1 };} for 1 .. 4 {    say "The {10**\$_}th Ulam number is: ", @ulams[10**\$_ - 1]}`
Output:
```The 10th Ulam number is: 18
The 100th Ulam number is: 690
The 1000th Ulam number is: 12294
The 10000th Ulam number is: 132788```

## REXX

Translation of: Wren

This REXX version has several speed improvements.

`/*REXX program finds & displays the Nth Ulam number (or any number of specified values).*/parse arg \$                                      /*obtain optional argument from the CL.*/if \$='' | \$=","  then \$= 10 100 1000 10000       /*Not specified? Then use the defaults.*/         do k=1  for words(\$)                     /*process each of the specified values.*/        x= Ulam( word(\$, k) )                    /*invoke Ulam to find a  Ulam  number. */        say 'the '        commas(#)th(#)       ' Ulam number is: '           commas(x)        end   /*k*/exit 0                                           /*stick a fork in it,  we're all done. *//*──────────────────────────────────────────────────────────────────────────────────────*/commas: parse arg _;  do jc=length(_)-3  to 1  by -3; _= insert(',', _, jc); end; return _th:     parse arg th; return word('th st nd rd', 1 + (th//10)*(th//100%10\==1)*(th//10<4))/*──────────────────────────────────────────────────────────────────────────────────────*/Ulam: parse arg n;    @.1= 1;   @.2= 2;   #= 2   /*1st two terms;  #:  sequence length. */                      !.= 0;    !.1= 1;   !.2= 1 /*semaphore for each term in sequence. */      z= 3                                       /*value of next possible term in seq.  */                 do until #==n                 cnt= 0                        do j=1  for #;        _= z - @.j    /*_:   short circuit value. */                        if !._  then if @.j\==_  then do;   cnt= cnt + 1                                                            if cnt>2  then leave                                                      end                        end   /*j*/                  if cnt==2  then do;  #= # + 1              /*bump the number of terms. */                                      @.#= z                /*add  Z  to the sequence.  */                                      !.z= 1                /*set the semaphore for  Z. */                                 end                 z= z + 1                                   /*bump next possible term.  */                 end   /*until*/      return @.#`
output   when using the default input of:     10   100   1000   10000
```the  10th  Ulam number is:  18
the  100th  Ulam number is:  690
the  1,000th  Ulam number is:  12,294
the  10,000th  Ulam number is:  132,788
```
output   (courtesy of Paul Kislanko's PC)   when using the input of:     100000
```the  100,000th  Ulam number is:  1,351,223
```

## Ring

` load "stdlib.ring" limit = 12500Ulam = []add(Ulam,1)add(Ulam,2) for n = 3 to limit    flag = 0    count = 0    len = len(Ulam)    for x = 1 to len-1        for y = x+1 to len               if Ulam[x] + Ulam[y] = n               flag = 1               count = count + 1            ok        next     next     if flag = 1 and count = 1        add(Ulam,n)        ln = len(Ulam)        if ln = 10           see "The 10th Ulam number is: " + n + nl        ok        if ln = 100           see "The 100th Ulam number is: " + n + nl        ok        if ln = 1000           see "The 1000th Ulam number is: " + n + nl        ok        if ln = 10000           see "The 10000th Ulam number is: " + n + nl        ok     oknext `

Output:

```The 10th Ulam number is: 18
The 100th Ulam number is: 690
The 1000th Ulam number is: 12294
The 10000th Ulam number is: 132788
```

## Rust

Translation of: Phix
`fn ulam(n: usize) -> usize {    let mut ulams = vec![1, 2];    let mut sieve = vec![1, 1];    let mut u = 2;    while ulams.len() < n {        sieve.resize(u + ulams[ulams.len() - 2], 0);        for i in 0..ulams.len() - 1 {            sieve[u + ulams[i] - 1] += 1;        }        for i in u..sieve.len() {            if sieve[i] == 1 {                u = i + 1;                ulams.push(u);                break;            }        }    }    ulams[n - 1]} fn main() {    use std::time::Instant;    let start = Instant::now();    let mut n = 1;    while n <= 100000 {        println!("Ulam({}) = {}", n, ulam(n));        n *= 10;    }    println!("Elapsed time: {:.2?}", start.elapsed());}`
Output:
```Ulam(1) = 1
Ulam(10) = 18
Ulam(100) = 690
Ulam(1000) = 12294
Ulam(10000) = 132788
Ulam(100000) = 1351223
Elapsed time: 10.68s
```

## Sidef

Translation of: Perl
`func ulam(n) {     static u     = Set(1,2)    static ulams = [0, 1, 2]     return ulams[n] if (ulams.end >= n)     ++n     for(var i = 3; true; ++i) {        var count = 0         ulams.each {|v|            if (u.has(i - v) && (v != i-v)) {                break if (count++ > 2)            }        }         if (count == 2) {            ulams << i            u << i            break if (ulams.len == n)        }    }     ulams.tail} for k in (1..3) {    say "The 10^#{k}-th Ulam number is: #{ulam(10**k)}"}`
Output:
```The 10^1-th Ulam number is: 18
The 10^2-th Ulam number is: 690
The 10^3-th Ulam number is: 12294
```

## Vlang

### Version 1

Translation of: Go
`import timefn ulam(n int) int {    mut ulams := [1, 2]    mut set := {1: true, 2: true}    mut i := 3    for {        mut count := 0        for j := 0; j < ulams.len; j++ {            ok := set[i-ulams[j]]            if ok && ulams[j] != (i-ulams[j]) {                count++                if count > 2 {                    break                }            }        }        if count == 2 {            ulams << i            set[i] = true            if ulams.len == n {                break            }        }        i++    }    return ulams[n-1]} fn main() {    start := time.now()    for n := 10; n <= 10000; n *= 10 {        println("The \${n}th Ulam number is \${ulam(n)}")    }    println("\nTook \${time.since(start)}")}`
Output:
```The 10th Ulam number is 18
The 100th Ulam number is 690
The 1000th Ulam number is 12294
The 10000th Ulam number is 132788

Took 9.611s
```

### Version 2

Translation of: Go

The following version, which builds up a sieve as it goes along, is (astonishingly) about 40 times faster!

`import timefn ulam(n int) int {    mut ulams := [1, 2]    mut sieve := [1, 1]    mut u := 2    for ulams.len < n {        s := u + ulams[ulams.len-2]        t := s - sieve.len        for i := 0; i < t; i++ {            sieve << 0        }        for i := 1; i <= ulams.len-1; i++ {            v := u + ulams[i-1] - 1            sieve[v]++        }        mut index := -1        for i, e in sieve[u..] {            if e == 1 {                index = u + i                break            }        }        u = index + 1        ulams << u    }    return ulams[n-1]} fn commatize(n int) string {    mut s := '\$n'    if n < 0 {        s = s[1..]    }    le := s.len    for i := le - 3; i >= 1; i -= 3 {        s = '\${s[0..i]},\${s[i..]}'    }    if n >= 0 {        return s    }    return "-\$s"} fn main() {    start := time.now()    for n := 1; n <= 10000; n *= 10 {        mut s := "th"        if n == 1 {            s = "st"        }        println("The \${commatize(n)}\$s Ulam number is \${commatize(ulam(n))}")    }    println("\nTook \${time.since(start)}")}`
Output:
```The 1st Ulam number is 1
The 10th Ulam number is 18
The 100th Ulam number is 690
The 1,000th Ulam number is 12,294
The 10,000th Ulam number is 132,788

Took 415.000ms
```

### Version 3

Translation of: Go

This version is even quicker than Version 2 and reduces the time needed to calculate the 10,000th and 100,000th Ulam numbers to about 40 milliseconds and 3.25 seconds respectively.

As mentioned in the Wren version 3 example, you need to know how much memory to allocate in advance.

`import timefn ulam(n int) int {    if n <= 2 {        return n    }    max := 1_352_000    mut list := []int{len:max+1}    list[0], list[1] = 1, 2    mut sums := []byte{len:2*max+1}    sums[3] = 1    mut size := 2    mut query := 0    for {        query = list[size-1] + 1        for {            if sums[query] == 1 {                for i in 0..size {                    sum := query + list[i]                    t := sums[sum] + 1                    if t <= 2 {                        sums[sum] = t                    }                }                list[size] = query                size++                break            }            query++        }        if size >= n {            break        }    }    return query} fn commatize(n int) string {    mut s := '\$n'    if n < 0 {        s = s[1..]    }    le := s.len    for i := le - 3; i >= 1; i -= 3 {        s = '\${s[0..i]},\${s[i..]}'    }    if n >= 0 {        return s    }    return "-\$s"} fn main() {    start := time.now()    for n := 1; n <= 100000; n *= 10 {        mut s := "th"        if n == 1 {            s = "st"        }        println("The \${commatize(n)}\$s Ulam number is \${commatize(ulam(n))}")    }    println("\nTook \${time.since(start)}")}`
Output:
```The 1st Ulam number is 1
The 10th Ulam number is 18
The 100th Ulam number is 690
The 1,000th Ulam number is 12,294
The 10,000th Ulam number is 132,788
The 100,000th Ulam number is 1,351,223

Took 42.912s
```

## Wren

### Version 1

Library: Wren-set
`import "/set" for Set var ulam = Fn.new() { |n|    var ulams = [1, 2]    var set = Set.new(ulams)    var i = 3    while (true) {        var count = 0        for (j in 0...ulams.count) {            if (set.contains(i - ulams[j]) && ulams[j] != (i - ulams[j])) {                count = count + 1                if (count > 2) break            }        }        if (count == 2) {            ulams.add(i)            set.add(i)            if (ulams.count == n) break        }        i = i + 1    }    return ulams[-1]} var n = 1while (true) {    n = n * 10    System.print("The %(n)th Ulam number is %(ulam.call(n))")    if (n == 10000) break}`
Output:
```The 10th Ulam number is 18
The 100th Ulam number is 690
The 1000th Ulam number is 12294
The 10000th Ulam number is 132788
```

### Version 2

Translation of: Phix
Library: Wren-seq
Library: Wren-fmt

The above version is reasonably efficient and runs in about 21.6 seconds on my machine (Intel Core i7-8565U). The following version, which builds up a sieve as it goes along, is more than 3 times faster.

`import "/seq" for Lstimport "/fmt" for Fmt var ulam = Fn.new { |n|    var ulams = [1, 2]    var sieve = [1, 1]    var u = 2    while (ulams.count < n) {        var s = u + ulams[-2]        sieve = sieve + ([0] * (s - sieve.count))        for (i in 1..ulams.count - 1) {            var v = u + ulams[i-1] - 1            sieve[v] = sieve[v] + 1        }        u = Lst.indexOf(sieve, 1, u) + 1        ulams.add(u)    }    return ulams[n-1]} var start = System.clockfor (p in 0..4) {    var n = 10.pow(p)    Fmt.print("The \$,r Ulam number is \$,d", n, ulam.call(n))}System.print("\nTook %(System.clock - start) seconds.")`
Output:
```The 1st Ulam number is 1
The 10th Ulam number is 18
The 100th Ulam number is 690
The 1,000th Ulam number is 12,294
The 10,000th Ulam number is 132,788

Took 6.366709 seconds.
```

### Version 3

Translation of: XPL0

This version is even quicker than Version 2 and reduces the time needed to calculate the 10,000th Ulam number to about 3.65 seconds. It also makes the 100,000th Ulam number a viable proposition for the Wren interpreter coming in at about 6 minutes 50 seconds.

The only downside with this version is that you need to know how much memory to allocate in advance.

`import "/fmt" for Fmt var ulam = Fn.new { |n|    if (n <= 2) return n    var max = 1352000    var list = List.filled(max+1, 0)    list[0] = 1    list[1] = 2    var sums = List.filled(max*2+1, 0)    sums[3] = 1    var size = 2    var query    while (true) {        query = list[size-1] + 1        while (true) {            if (sums[query] == 1) {                for (i in 0..size-1) {                    var sum = query + list[i]                    var t = sums[sum] + 1                    if (t <= 2) sums[sum] = t                }                list[size] = query                size = size + 1                break            }            query = query + 1        }        if (size >= n) break    }    return query} var start = System.clockvar n = 10while (true) {    Fmt.print("The \$,r Ulam number is \$,d", n, ulam.call(n))    n = n * 10    if (n > 100000) break}System.print("\nTook %(System.clock - start) seconds.")`
Output:
```The 10th Ulam number is 18
The 100th Ulam number is 690
The 1,000th Ulam number is 12,294
The 10,000th Ulam number is 132,788
The 100,000th Ulam number is 1,351,223

Took 409.990502 seconds.
```

## XPL0

Seeing "set" in the Go version and "sieve" in Phix, etc. lit a little light bulb. This version exploits those ideas and finds the 100,000th Ulam number in 24.7 seconds on a Pi4.

`func    Ulam(N);                \Return Nth Ulam numberint     N;def     Max = 1_352_000;        \enough for 100_000th Ulam numberint     List(Max);              \array of Ulam numberschar    Sums(Max*2);            \array: 0, 1, or more ways to sum Ulamsint     I, Size, Query, Sum, T;[if N <= 2 then return N;for I:= 0 to Max*2 do Sums(I):= 0;List(0):= 1;  List(1):= 2;Sums(3):= 1;                    \only one way to sum Ulams: 1+2 = 3Size:= 2;                       \start after first 2 Ulamsrepeat  Query:= List(Size-1)+1;                 \possible next Ulam no.        loop    [if Sums(Query) = 1 then        \sums 1 way so it's next                    [for I:= 0 to Size-1 do     \update Sums array with                        [Sum:= Query + List(I); \ all combos of sums                        T:= Sums(Sum)+1;        \ but limit max count                        if T <= 2 then Sums(Sum):= T;                        ];                    List(Size):= Query;         \add Query to List                    Size:= Size+1;                    quit;                    ];                Query:= Query+1;                \possible next Ulam no.                ];until   Size >= N;return Query;]; int  N;[N:= 10;repeat  Text(0, "The ");  IntOut(0, N);          Text(0, "th Ulam number is ");        IntOut(0, Ulam(N));  CrLf(0);        N:= N*10;until   N > 100_000;]`
Output:
```The 10th Ulam number is 18
The 100th Ulam number is 690
The 1000th Ulam number is 12294
The 10000th Ulam number is 132788
The 100000th Ulam number is 1351223
```