Mian-Chowla sequence

From Rosetta Code
Task
Mian-Chowla sequence
You are encouraged to solve this task according to the task description, using any language you may know.

The Mian–Chowla sequence is an integer sequence defined recursively.


Mian–Chowla is an infinite instance of a Sidon sequence, and belongs to the class known as B₂ sequences.


The sequence starts with:

a1 = 1

then for n > 1, an is the smallest positive integer such that every pairwise sum

ai + aj

is distinct, for all i and j less than or equal to n.

The Task
  • Find and display, here, on this page the first 30 terms of the Mian–Chowla sequence.
  • Find and display, here, on this page the 91st through 100th terms of the Mian–Chowla sequence.


Demonstrating working through the first few terms longhand:

a1 = 1
1 + 1 = 2

Speculatively try a2 = 2

1 + 1 = 2
1 + 2 = 3
2 + 2 = 4

There are no repeated sums so 2 is the next number in the sequence.

Speculatively try a3 = 3

1 + 1 = 2
1 + 2 = 3
1 + 3 = 4
2 + 2 = 4
2 + 3 = 5
3 + 3 = 6

Sum of 4 is repeated so 3 is rejected.

Speculatively try a3 = 4

1 + 1 = 2
1 + 2 = 3
1 + 4 = 5
2 + 2 = 4
2 + 4 = 6
4 + 4 = 8

There are no repeated sums so 4 is the next number in the sequence.

And so on...

See also

11l

Translation of: C++
F contains(sums, s, ss)
   L(i) 0 .< ss
      I sums[i] == s
         R 1B
   R 0B

F mian_chowla()
   V n = 100
   V mc = [0] * n
   mc[0] = 1
   V sums = [0] * ((n * (n + 1)) >> 1)
   sums[0] = 2
   V ss = 1
   L(i) 1 .< n
      V le = ss
      V j = mc[i - 1] + 1
      L
         mc[i] = j
         V nxtJ = 0B
         L(k) 0 .. i
            V sum = mc[k] + j
            I contains(sums, sum, ss)
               ss = le
               nxtJ = 1B
               L.break
            sums[ss] = sum
            ss++
         I !nxtJ
            L.break
         j++
   R mc

print(‘The first 30 terms of the Mian-Chowla sequence are:’)
V mc = mian_chowla()
print_elements(mc[0.<30])
print()
print(‘Terms 91 to 100 of the Mian-Chowla sequence are:’)
print_elements(mc[90..])
Output:
The first 30 terms of the Mian-Chowla sequence are:
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312

Terms 91 to 100 of the Mian-Chowla sequence are:
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219

Ada

Works with: Ada version 2012
with Ada.Text_IO;
with Ada.Containers.Hashed_Sets;

procedure Mian_Chowla_Sequence
is
   type Natural_Array is array(Positive range <>) of Natural;

   function Hash(P : in Positive) return Ada.Containers.Hash_Type is
   begin
      return Ada.Containers.Hash_Type(P);
   end Hash;

   package Positive_Sets is new Ada.Containers.Hashed_Sets(Positive, Hash, "=");

   function Mian_Chowla(N : in Positive) return Natural_Array
   is
      return_array : Natural_Array(1 .. N) := (others => 0);
      nth : Positive := 1;
      candidate : Positive := 1;
      seen : Positive_Sets.Set;
   begin
      while nth <= N loop
         declare
            sums : Positive_Sets.Set;
            terms : constant Natural_Array := return_array(1 .. nth-1) & candidate;
            found : Boolean := False;
         begin
            for term of terms loop
               if seen.Contains(term + candidate) then
                  found := True;
                  exit;
               else
                  sums.Insert(term + candidate);
               end if;
            end loop;

            if not found then
               return_array(nth) := candidate;
               seen.Union(sums);
               nth := nth + 1;
            end if;
            candidate := candidate + 1;
         end;
      end loop;
      return return_array;
   end Mian_Chowla;

   length : constant Positive := 100;
   sequence : constant Natural_Array(1 .. length) := Mian_Chowla(length);
begin
   Ada.Text_IO.Put_Line("Mian Chowla sequence first 30 terms :");
   for term of sequence(1 .. 30) loop
      Ada.Text_IO.Put(term'Img);
   end loop;
   Ada.Text_IO.New_Line;
   Ada.Text_IO.Put_Line("Mian Chowla sequence terms 91 to 100 :");
   for term of sequence(91 .. 100) loop
      Ada.Text_IO.Put(term'Img);
   end loop;
end Mian_Chowla_Sequence;
Output:
Mian Chowla sequence first 30 terms :
 1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312
Mian Chowla sequence terms 91 to 100 :
 22526 23291 23564 23881 24596 24768 25631 26037 26255 27219

ALGOL 68

Works with: ALGOL 68G version Any - tested with release 2.8.3.win32

Allocating a large-enough array initially would gain some performance but might be considered cheating - 60 000 elements would be enough for the task.

# Find Mian-Chowla numbers: an
                     where: ai = 1,
                       and: an = smallest integer such that ai + aj is unique
                                             for all i, j in 1 .. n && i <= j
#
BEGIN
    INT max mc           = 100;
    [ max mc ]INT mc;
    INT curr size       :=      0; # initial size of the array     #
    INT size increment   = 10 000; # size to increase the array by #
    REF[]BOOL is sum    := HEAP[ 1 : 0 ]BOOL;
    INT mc count        := 1;
    FOR i WHILE mc count <= max mc DO
        # assume i will be part of the sequence                    #
        mc[ mc count ]  := i;
        # check the sums                                           #
        IF  ( 2 * i ) > curr size THEN
            # the is sum array is too small - make a larger one    #
            REF[]BOOL new sum = HEAP[ curr size + size increment ]BOOL;
            new sum[ 1 : curr size ] := is sum;
            FOR n TO size increment DO new sum[ curr size + n ] := FALSE OD;
            curr size  +:= size increment;
            is sum      := new sum
        FI;
        BOOL is unique  := TRUE;
        FOR mc pos TO mc count WHILE is unique := NOT is sum[ i + mc[ mc pos ] ] DO SKIP OD;
        IF is unique THEN
            # i is a sequence element - store the sums             #
            FOR k TO mc count DO is sum[ i + mc[ k ] ] := TRUE OD;
            mc count +:= 1
        FI
    OD;

    # print parts of the sequence                                  #
    print( ( "Mian Chowla sequence elements 1..30:", newline ) );
    FOR i TO 30 DO print( ( " ", whole( mc[ i ], 0 ) ) ) OD;
    print( ( newline ) );
    print( ( "Mian Chowla sequence elements 91..100:", newline ) );
    FOR i FROM 91 TO 100 DO print( ( " ", whole( mc[ i ], 0 ) ) ) OD

END
Output:
Mian Chowla sequence elements 1..30:
 1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312
Mian Chowla sequence elements 91..100:
 22526 23291 23564 23881 24596 24768 25631 26037 26255 27219

elapsed time approx 0.25 seconds on my Windows 7 system (note under Windows, A68G runs as an interpreter only).

Arturo

mianChowla: function [n][
    result: new [1]
    sums: new [2]
    candidate: 1

    while [n > size result][
        fit: false
        'result ++ 0
        while [not? fit][
            candidate: candidate + 1
            fit: true
            result\[dec size result]: candidate
            loop result 'val [
                if contains? sums val + candidate [
                    fit: false
                    break
                ]
            ]
        ]

        loop result 'val [
            'sums ++ val + candidate
            unique 'sums
        ]
    ]
    return result
]

seq100: mianChowla 100

print "The first 30 terms of the Mian-Chowla sequence are:"
print slice seq100 0 29
print ""

print "Terms 91 to 100 of the sequence are:"
print slice seq100 90 99
Output:
The first 30 terms of the Mian-Chowla sequence are:
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312 

Terms 91 to 100 of the sequence are:
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219

AWK

Translation of the ALGOL 68 - largely implements the "by hand" method in the task.

# Find Mian-Chowla numbers: an
#                    where: ai = 1,
#                      and: an = smallest integer such that ai + aj is unique
#                                            for all i, j in 1 .. n && i <= j
#
BEGIN \
{

    FALSE      = 0;
    TRUE       = 1;

    mcCount    = 1;

    for( i = 1; mcCount <= 100; i ++ )
    {
        # assume i will be part of the sequence
        mc[ mcCount ] = i;
        # check the sums
        isUnique = TRUE;
        for( mcPos = 1; mcPos <= mcCount && isUnique; mcPos ++ )
        {
            isUnique = ! ( ( i + mc[ mcPos ] ) in isSum );
        } # for j
        if( isUnique )
        {
            # i is a sequence element - store the sums
            for( k = 1; k <= mcCount; k ++ )
            {
                isSum[ i + mc[ k ] ] = TRUE;
            } # for k
            mcCount ++;
        } # if isUnique
    } # for i
    # print the sequence
    printf( "Mian Chowla sequence elements 1..30:\n" );
    for( i = 1; i <= 30; i ++ )
    {
        printf( " %d", mc[ i ] );
    } # for i
    printf( "\n" );
    printf( "Mian Chowla sequence elements 91..100:\n" );
    for( i = 91; i <= 100; i ++ )
    {
        printf( " %d", mc[ i ] );
    } # for i
    printf( "\n" );

} # BEGIN
Output:
Mian Chowla sequence elements 1..30:
 1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312
Mian Chowla sequence elements 91..100:
 22526 23291 23564 23881 24596 24768 25631 26037 26255 27219

elapsed time approx 0.20 seconds on my Windows 7 system.

Alternate

Translation of: Go

Hopefully the comments help explain the algorithm.

# helper functions
#
#    determine if a list is empty or not
function isEmpty(a) { for (ii in a) return 0; return 1 }
#    list concatination
function concat(a, b) { for (cc in b) a[cc] = cc }

BEGIN \
{
    mc[0] = 1; sums[2] = 0;      # initialize lists
    for ( i = 1; i < 100; i ++ ) # iterate for each item in result
    {
        for ( j = mc[i-1]+1; ; j ++ ) # iterate thru trial values
        {
            mc[i] = j;           # set trial value into result
            for ( k = 0; k <= i; k ++ ) # test new iteration of sums
            {
                # test trial sum against old sums list
                if ((sum = mc[k] + j) in sums) 
                {                # collision, so
                    delete ts;   # toss out any accumulated items,
                    break;       #  and break out to the next j
                }
                ts[sum] = sum;   # (else) accumulate to new sum list
            } # for k
            if ( isEmpty( ts ) ) # nothing to add, 
                continue;        #  so try next j
            concat( sums, ts );  # combine new sums to old,
            delete ts;           #  clear out the new,
            break;               #  break out to next i
        } # for j
    } # for i
    # print the sequence
    ps = "Mian Chowla sequence elements %d..%d:\n";
    for ( i = 0; i < 100; i ++ )
    {
        if ( i == 0 )  printf ps, 1, 30;
        if ( i == 90 ) printf "\n\n" ps, 91, 100;
        if ( i < 30 || i >= 90 ) printf "%d ", mc[ i ];
    } # for i
    print "\n"
} # BEGIN
Output:
Mian Chowla sequence elements 1..30:
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312 

Mian Chowla sequence elements 91..100:
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219

Computation time is about 110 ms on tio.run

C

Translation of: Go
#include <stdio.h>
#include <stdbool.h>
#include <time.h>
 
#define n 100
#define nn ((n * (n + 1)) >> 1)

bool Contains(int lst[], int item, int size) {
	for (int i = size - 1; i >= 0; i--)
 		if (item == lst[i]) return true;
	return false;
}
 
int * MianChowla()
{
	static int mc[n]; mc[0] = 1;
	int sums[nn];	sums[0] = 2;
	int sum, le, ss = 1;
	for (int i = 1; i < n; i++) {
		le = ss;
		for (int j = mc[i - 1] + 1; ; j++) {
			mc[i] = j;
			for (int k = 0; k <= i; k++) {
				sum = mc[k] + j;
				if (Contains(sums, sum, ss)) {
					ss = le; goto nxtJ;
				}
				sums[ss++] = sum;
			}
			break;
		nxtJ:;
		}
	}
	return mc;
}
 
int main() {
	clock_t st = clock(); int * mc; mc = MianChowla();
        double et = ((double)(clock() - st)) / CLOCKS_PER_SEC;
	printf("The first 30 terms of the Mian-Chowla sequence are:\n");
	for (int i = 0; i < 30; i++) printf("%d ", mc[i]);
	printf("\n\nTerms 91 to 100 of the Mian-Chowla sequence are:\n");
	for (int i = 90; i < 100; i++) printf("%d ", mc[i]); 
	printf("\n\nComputation time was %f seconds.", et);
}
Output:
The first 30 terms of the Mian-Chowla sequence are:
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312 

Terms 91 to 100 of the Mian-Chowla sequence are:
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219 

Computation time was 1.575556 seconds.

Quick, but...

...is memory hungry. This will allocate a bigger buffer as needed to keep track of the sums involved. Based on the ALGOL 68 version. The minimum memory needed is double of the highest entry calculated. This program doubles the buffer size each time needed, so it will use more than the minimum. The ALGOL 68 increments by a fixed increment size. Which could be just as wasteful if the increment is too large and slower if the increment is too small).

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <time.h>

// helper function for indicating memory used.
void approx(char* buf, double count)
{
    const char* suffixes[] = { "Bytes", "KiB", "MiB" };
    uint s = 0; 
    while (count >= 1024 && s < 3) { s++; count /= 1024; }
    if (count - (double)((int)count) == 0.0)
        sprintf(buf, "%d %s", (int)count, suffixes[s]);
    else
        sprintf(buf, "%.1f %s", count, suffixes[s]);
}

int main() {
    int i, j, k, c = 0, n = 100, nn = 110;
    int* mc = (int*) malloc((n) * sizeof(int));
    bool* isSum = (bool*) calloc(nn, sizeof(bool));
    char em[] = "unable to increase isSum array to %ld.";
    if (n > 100)  printf("Computing terms 1 to %d...\n", n);
    clock_t st = clock();
    for (i = 1; c < n; i++) {
        mc[c] = i;
        if (i + i > nn) {
            bool* newIs = (bool*)realloc(isSum, (nn <<= 1) * sizeof(bool));
            if (newIs == NULL) { printf(em, nn); return -1; }
            isSum = newIs;
            for (j = (nn >> 1); j < nn; j++) isSum[j] = false;
        }
        bool isUnique = true;
        for (j = 0; (j < c) && isUnique; j++) isUnique = !isSum[i + mc[j]];
        if (isUnique) {
            for (k = 1; k <= c; k++) isSum[i + mc[k]] = true;
            c++;
        }
    }
    double et = 1e3 * ((double)(clock() - st)) / CLOCKS_PER_SEC;
    free(isSum);
    printf("The first 30 terms of the Mian-Chowla sequence are:\n");
    for (i = 0; i < 30; i++) printf("%d ", mc[i]);
    printf("\n\nTerms 91 to 100 of the Mian-Chowla sequence are:\n");
    for (i = 90; i < 100; i++) printf("%d ", mc[i]);
    if (c > 100) printf("\nTerm %d is: %d" ,c , mc[c - 1]);
    free(mc);
    char buf[100]; approx(buf, nn * sizeof(bool));
    printf("\n\nComputation time was %6.3f ms.  Allocation was %s.", et, buf);
}
Output:
The first 30 terms of the Mian-Chowla sequence are:
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312 

Terms 91 to 100 of the Mian-Chowla sequence are:
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219 

Computation time was  1.773 ms.  Allocation was 55 KiB.

Here is the output for a larger calculation:

Computing terms 1 to 1300...
The first 30 terms of the Mian-Chowla sequence are:
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312 

Terms 91 to 100 of the Mian-Chowla sequence are:
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219 
Term 1300 is: 29079927

Computation time was 7979.042 ms.  Allocation was 110 MiB.

C#

Translation of: Go
using System;
using System.Collections.Generic;
using System.Diagnostics;
using System.Linq;

static class Program {
    static int[] MianChowla(int n) {
        int[] mc = new int[n - 1 + 1];
        HashSet<int> sums = new HashSet<int>(), ts = new HashSet<int>();
        int sum; mc[0] = 1; sums.Add(2);
        for (int i = 1; i <= n - 1; i++) {
            for (int j = mc[i - 1] + 1; ; j++) {
                mc[i] = j;
                for (int k = 0; k <= i; k++) {
                    sum = mc[k] + j;
                    if (sums.Contains(sum)) { ts.Clear(); break; }
                    ts.Add(sum);
                }
                if (ts.Count > 0) { sums.UnionWith(ts); break; }
            }
        }
        return mc;
    }

    static void Main(string[] args)
    {
        const int n = 100; Stopwatch sw = new Stopwatch();
        string str = " of the Mian-Chowla sequence are:\n";
        sw.Start(); int[] mc = MianChowla(n); sw.Stop();
        Console.Write("The first 30 terms{1}{2}{0}{0}Terms 91 to 100{1}{3}{0}{0}" +
            "Computation time was {4}ms.{0}", '\n', str, string.Join(" ", mc.Take(30)),
            string.Join(" ", mc.Skip(n - 10)), sw.ElapsedMilliseconds);
    }
}
Output:
The first 30 terms of the Mian-Chowla sequence are:
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312

Terms 91 to 100 of the Mian-Chowla sequence are:
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219

Computation time was 17ms.

C++

Translation of: Go

The sums array expands by "i" on each iteration from 1 to n, so the max array length can be pre-calculated to the nth triangular number (n * (n + 1) / 2).

using namespace std;

#include <iostream>
#include <ctime>

#define n 100
#define nn ((n * (n + 1)) >> 1)

bool Contains(int lst[], int item, int size) {
	for (int i = 0; i < size; i++) if (item == lst[i]) return true;
	return false;
}

int * MianChowla()
{
	static int mc[n]; mc[0] = 1;
	int sums[nn];	sums[0] = 2;
	int sum, le, ss = 1;
	for (int i = 1; i < n; i++) {
		le = ss;
		for (int j = mc[i - 1] + 1; ; j++) {
			mc[i] = j;
			for (int k = 0; k <= i; k++) {
				sum = mc[k] + j;
				if (Contains(sums, sum, ss)) {
					ss = le; goto nxtJ;
				}
				sums[ss++] = sum;
			}
			break;
		nxtJ:;
		}
	}
	return mc;
}

int main() {
	clock_t st = clock(); int * mc; mc = MianChowla();
	double et = ((double)(clock() - st)) / CLOCKS_PER_SEC;
	cout << "The first 30 terms of the Mian-Chowla sequence are:\n";
	for (int i = 0; i < 30; i++) { cout << mc[i] << ' '; }
	cout << "\n\nTerms 91 to 100 of the Mian-Chowla sequence are:\n";
	for (int i = 90; i < 100; i++) { cout << mc[i] << ' '; }
	cout << "\n\nComputation time was " << et << " seconds.";
}
Output:
The first 30 terms of the Mian-Chowla sequence are:
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312 

Terms 91 to 100 of the Mian-Chowla sequence are:
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219 

Computation time was 1.92958 seconds.

F#

The function

// Generate Mian-Chowla sequence. Nigel Galloway: March 23rd., 2019
let mC=let rec fN i g l=seq{
         let a=(l*2)::[for i in i do yield i+l]@g
         let b=[l+1..l*2]|>Seq.find(fun e->Seq.forall(fun g->(Seq.contains (g-e)>>not) i) a)
         yield b; yield! fN (l::i) (a|>List.filter(fun n->n>b)) b}
       seq{yield 1; yield! fN [] [] 1}

The Tasks

First 30
mC |> Seq.take 30 |> Seq.iter(printf "%d ");printfn ""
Output:
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312
91 to 100
mC |> Seq.skip 90 |> Seq.take 10 |> Seq.iter(printf "%d ");printfn ""
Output:
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219

Factor

USING: fry hash-sets io kernel math prettyprint sequences sets ;

: next ( seq sums speculative -- seq' sums' speculative' )
    dup reach [ + ] with map over dup + suffix! >hash-set pick
    over intersect null?
    [ swapd union [ [ suffix! ] keep ] dip swap ] [ drop ] if
    1 + ;

: mian-chowla ( n -- seq )
    [ V{ 1 } HS{ 2 } [ clone ] bi@ 2 ] dip
    '[ pick length _ < ] [ next ] while 2drop ;

100 mian-chowla
[ 30 head "First 30 terms of the Mian-Chowla sequence:" ]
[ 10 tail* "Terms 91-100 of the Mian-Chowla sequence:" ] bi
[ print [ pprint bl ] each nl nl ] 2bi@
Output:
First 30 terms of the Mian-Chowla sequence:
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312 

Terms 91-100 of the Mian-Chowla sequence:
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219

FreeBASIC

redim as uinteger mian(0 to 1)
redim as uinteger sums(0 to 2)
mian(0) = 1 : mian(1) = 2
sums(0) = 2 : sums(1) = 3 : sums(2) = 4
dim as uinteger n_mc = 2, n_sm = 3, curr = 3, tempsum
while n_mc < 101
    for i as uinteger = 0 to n_mc - 1
        tempsum = curr + mian(i)
        for j as uinteger = 0 to n_sm - 1
            if tempsum = sums(j) then goto loopend
        next j
    next i
    redim preserve as uinteger mian(0 to n_mc)
    mian(n_mc) = curr
    redim preserve as uinteger sums(0 to n_sm + n_mc)
    for j as uinteger = 0 to n_mc - 1
        sums(n_sm + j) = mian(j) + mian(n_mc)
    next j
    n_mc += 1
    n_sm += n_mc
    sums(n_sm-1) = 2*curr
    loopend:
    curr += 1
wend

print "Mian-Chowla numbers 1 through 30: ",
for i as uinteger = 0 to 29
    print mian(i),
next i
print
print "Mian-Chowla numbers 91 through 100: ",
for i as uinteger = 90 to 99
    print mian(i),
next i
print
Output:
Mian-Chowla numbers 1 through 30:         1             2             4             8             13            21            31            45
66            81            97            123           148           182           204           252           290           361           401
475           565           593           662           775           822           916           970           1016          1159          1312

Mian-Chowla numbers 91 through 100:       22526         23291         23564         23881         24596         24768         25631         26037              26255         27219

Go

package main

import "fmt"

func contains(is []int, s int) bool {
    for _, i := range is {
        if s == i {
            return true
        }
    }
    return false
}

func mianChowla(n int) []int {
    mc := make([]int, n)
    mc[0] = 1
    is := []int{2}
    var sum int
    for i := 1; i < n; i++ {
        le := len(is)
    jloop:
        for j := mc[i-1] + 1; ; j++ {
            mc[i] = j
            for k := 0; k <= i; k++ {
                sum = mc[k] + j
                if contains(is, sum) {
                    is = is[0:le]
                    continue jloop
                }
                is = append(is, sum)
            }
            break
        }
    }
    return mc
}

func main() {
    mc := mianChowla(100)
    fmt.Println("The first 30 terms of the Mian-Chowla sequence are:")
    fmt.Println(mc[0:30])
    fmt.Println("\nTerms 91 to 100 of the Mian-Chowla sequence are:")
    fmt.Println(mc[90:100])
}
Output:
The first 30 terms of the Mian-Chowla sequence are:
[1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312]

Terms 91 to 100 of the Mian-Chowla sequence are:
[22526 23291 23564 23881 24596 24768 25631 26037 26255 27219]


Quicker version (runs in less than 0.02 seconds on Celeron N3050 @1.6 GHz), output as before:

package main

import "fmt"

type set map[int]bool

func mianChowla(n int) []int {
    mc := make([]int, n)
    mc[0] = 1
    is := make(set, n*(n+1)/2)
    is[2] = true
    var sum int
    isx := make([]int, 0, n)
    for i := 1; i < n; i++ {
        isx = isx[:0]
    jloop:
        for j := mc[i-1] + 1; ; j++ {
            mc[i] = j
            for k := 0; k <= i; k++ {
                sum = mc[k] + j
                if is[sum] {                   
                    isx = isx[:0]
                    continue jloop
                }
                isx = append(isx, sum)
            }
            for _, x := range isx {
                is[x] = true
            }
            break
        }
    }
    return mc
}

func main() {
    mc := mianChowla(100)
    fmt.Println("The first 30 terms of the Mian-Chowla sequence are:")
    fmt.Println(mc[0:30])
    fmt.Println("\nTerms 91 to 100 of the Mian-Chowla sequence are:")
    fmt.Println(mc[90:100])
}

Haskell

Translation of: Python
Translation of: JavaScript
import Data.Set (Set, fromList, insert, member)

------------------- MIAN-CHOWLA SEQUENCE -----------------
mianChowlas :: Int -> [Int]
mianChowlas =
  reverse . snd . (iterate nextMC (fromList [2], [1]) !!) . subtract 1

nextMC :: (Set Int, [Int]) -> (Set Int, [Int])
nextMC (sumSet, mcs@(n:_)) =
  (foldr insert sumSet ((2 * m) : fmap (m +) mcs), m : mcs)
  where
    valid x = not $ any (flip member sumSet . (x +)) mcs
    m = until valid succ n

--------------------------- TEST -------------------------
main :: IO ()
main =
  (putStrLn . unlines)
    [ "First 30 terms of the Mian-Chowla series:"
    , show (mianChowlas 30)
    , []
    , "Terms 91 to 100 of the Mian-Chowla series:"
    , show $ drop 90 (mianChowlas 100)
    ]
Output:
First 30 terms of the Mian-Chowla series:
[1,2,4,8,13,21,31,45,66,81,97,123,148,182,204,252,290,361,401,475,565,593,662,775,822,916,970,1016,1159,1312]

Terms 91 to 100 of the Mian-Chowla series:
[22526,23291,23564,23881,24596,24768,25631,26037,26255,27219]

J

NB. http://rosettacode.org/wiki/Mian-Chowla_sequence

NB. Dreadfully inefficient implementation recomputes all the sums to n-1
NB. and computes the full addition table rather than just a triangular region
NB. However, this implementation is sufficiently quick to meet the requirements.

NB. The vector head is the next speculative value
NB. Beheaded, the vector is Mian-Chowla sequence.


Until =: conjunction def 'u^:(0 = v)^:_'
unique =: -:&# ~.   NB. tally of list matches that of set

next_mc =: [: (, {.) (>:@:{. , }.)Until(unique@:((<:/~@i.@# #&, +/~)@:(}. , {.)))


prime_q =: 1&p:   NB. for fun look at prime generation suitability
   NB. generate sufficient terms of sequence

   A =: (next_mc^:108) 1 1

   NB. first 30 terms
   (,:prime_q)30{.}.A
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312
0 1 0 0  1  0  1  0  0  0  1   0   0   0   0   0   0   0   1   0   0   1   0   0   0   0   0    0    0    0

   NB. terms 91 through 100
   (,: prime_q) A {~ 91+i.10
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219
    0     1     0     0     0     0     0     0     0     0

Java

import java.util.Arrays;

public class MianChowlaSequence {

    public static void main(String[] args) {
        long start = System.currentTimeMillis();
        System.out.println("First 30 terms of the Mian–Chowla sequence.");
        mianChowla(1, 30);
        System.out.println("Terms 91 through 100 of the Mian–Chowla sequence.");
        mianChowla(91, 100);
        long end = System.currentTimeMillis();
        System.out.printf("Elapsed = %d ms%n", (end-start));
    }

    private static void mianChowla(int minIndex, int maxIndex) {
        int [] sums = new int[1];
        int [] chowla = new int[maxIndex+1];
        sums[0] = 2;
        chowla[0] = 0;
        chowla[1] = 1;
        if ( minIndex == 1 ) {
            System.out.printf("%d ", 1);
        }
        int chowlaLength = 1;
        for ( int n = 2 ; n <= maxIndex ; n++ ) {

            //  Sequence is strictly increasing.
            int test = chowla[n - 1];
            //  Bookkeeping.  Generate only new sums.
            int[] sumsNew = Arrays.copyOf(sums, sums.length + n);
            int sumNewLength = sums.length;
            int savedsSumNewLength = sumNewLength;
            
            //  Generate test candidates for the next value of the sequence.
            boolean found = false;
            while ( ! found ) {
                test++;
                found = true;
                sumNewLength = savedsSumNewLength;
                //  Generate test sums
                for ( int j = 0 ; j <= chowlaLength ; j++ ) {
                    int testSum = (j == 0 ? test : chowla[j]) + test;
                    boolean duplicate = false;
                    
                    //  Check if test Sum in array
                    for ( int k = 0 ; k < sumNewLength ; k++ ) {
                        if ( sumsNew[k] == testSum ) {
                            duplicate = true;
                            break;
                        }
                    }
                    if ( ! duplicate ) {
                        //  Add to array 
                        sumsNew[sumNewLength] = testSum;
                        sumNewLength++;
                    }
                    else {
                        //  Duplicate found.  Therefore, test candidate of the next value of the sequence is not OK.
                        found = false;
                        break;
                    }
                }
            }
            
            //  Bingo!  Now update bookkeeping.
            chowla[n] = test;
            chowlaLength++;            
            sums = sumsNew;
            if ( n >= minIndex ) {
                System.out.printf("%d %s", chowla[n], (n==maxIndex ? "\n" : ""));
            }
        }
    }

}
Output:
First 30 terms of the Mian–Chowla sequence.
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312 
Terms 91 through 100 of the Mian–Chowla sequence.
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219 
Elapsed = 220 ms

JavaScript

Translation of: Python

(Functional Python version)

(() => {
    'use strict';

    // main :: IO ()
    const main = () => {
        const genMianChowla = mianChowlas();
        console.log([
            'Mian-Chowla terms 1-30:',
            take(30)(
                genMianChowla
            ),

            '\nMian-Chowla terms 91-100:',
            (
                drop(60)(genMianChowla),
                take(10)(
                    genMianChowla
                )
            )
        ].join('\n') + '\n');
    };

    // mianChowlas :: Gen [Int]
    function* mianChowlas() {
        let
            mcs = [1],
            sumSet = new Set([2]),
            x = 1;
        while (true) {
            yield x;
            [sumSet, mcs, x] = nextMC(sumSet, mcs, x);
        }
    }

    // nextMC :: Set Int -> [Int] -> Int -> (Set Int, [Int], Int)
    const nextMC = (setSums, mcs, n) => {
        // Set of sums -> Series up to n -> Next term in series
        const valid = x => {
            for (const m of mcs) {
                if (setSums.has(x + m)) return false;
            }
            return true;
        };
        const x = until(valid)(x => 1 + x)(n);
        return [
            sumList(mcs)(x)
            .reduce(
                (a, n) => (a.add(n), a),
                setSums
            ),
            mcs.concat(x),
            x
        ];
    };

    // sumList :: [Int] -> Int -> [Int]
    const sumList = xs =>
        // Series so far -> additional term -> new sums
        n => [2 * n].concat(xs.map(x => n + x));


    // ---------------- GENERIC FUNCTIONS ----------------

    // drop :: Int -> [a] -> [a]
    // drop :: Int -> Generator [a] -> Generator [a]
    // drop :: Int -> String -> String
    const drop = n =>
        xs => Infinity > length(xs) ? (
            xs.slice(n)
        ) : (take(n)(xs), xs);


    // length :: [a] -> Int
    const length = xs =>
        // Returns Infinity over objects without finite
        // length. This enables zip and zipWith to choose
        // the shorter argument when one is non-finite,
        // like cycle, repeat etc
        'GeneratorFunction' !== xs.constructor
        .constructor.name ? (
            xs.length
        ) : Infinity;


    // take :: Int -> [a] -> [a]
    // take :: Int -> String -> String
    const take = n =>
        // The first n elements of a list,
        // string of characters, or stream.
        xs => 'GeneratorFunction' !== xs
        .constructor.constructor.name ? (
            xs.slice(0, n)
        ) : [].concat.apply([], Array.from({
            length: n
        }, () => {
            const x = xs.next();
            return x.done ? [] : [x.value];
        }));


    // until :: (a -> Bool) -> (a -> a) -> a -> a
    const until = p =>
        f => x => {
            let v = x;
            while (!p(v)) v = f(v);
            return v;
        };

    // MAIN ---
    return main();
})();
Output:
Mian-Chowla terms 1-30:
1,2,4,8,13,21,31,45,66,81,97,123,148,182,204,252,290,361,401,475,565,593,662,775,822,916,970,1016,1159,1312

Mian-Chowla terms 91-100:
22526,23291,23564,23881,24596,24768,25631,26037,26255,27219

jq

Works with: jq

Works with gojq, the Go implementation of jq

# Input: a bag-of-words (bow)
# Output: either an augmented bow, or nothing if a duplicate is found
def augment_while_unique(stream):
  label $out
  | foreach ((stream|tostring), null) as $word (.;
       if $word == null then .
       elif has($word) then break $out
       else .[$word] = 1 
       end;
      select($word == null) );

# For speedup, store "sums" as a hash
def mian_chowlas:
  {m:[1], sums: {"1":1}}
  | recurse(
      .m as $m
      | .sums as $sums
      | first(range(1+$m[-1]; infinite) as $i
	      | $sums
	      | augment_while_unique( ($m[] | (.+$i)), (2*$i))
	      | [$i, .] ) as [$i, $sums]
      | {m: ($m + [$i]), $sums} )
  | .m[-1] ;

The Tasks

[limit(100; mian_chowlas)]
| "First thirty: \(.[:30]);",
  "91st through 100th: \(.[90:])."
Output:
First thirty: [1,2,4,8,13,21,31,45,66,81,97,123,148,182,204,252,290,361,401,475,565,593,662,775,822,916,970,1016,1159,1312];
91st through 100th: [22526,23291,23564,23881,24596,24768,25631,26037,26255,27219].

Julia

Optimization in Julia can be an incremental process. The first version of this program ran in over 2 seconds. Using a hash table for lookup of sums and avoiding reallocation of arrays helps considerably.

function mianchowla(n)
    seq = ones(Int, n)
    sums = Dict{Int,Int}()
    tempsums = Dict{Int,Int}()
    for i in 2:n
        seq[i] = seq[i - 1] + 1
        incrementing = true
        while incrementing
            for j in 1:i
                tsum = seq[j] + seq[i]
                if haskey(sums, tsum)
                    seq[i] += 1
                    empty!(tempsums)
                    break
                else
                    tempsums[tsum] = 0
                    if j == i
                        merge!(sums, tempsums)
                        empty!(tempsums)
                        incrementing = false
                    end
                end
            end
        end
    end
    seq
end

function testmianchowla()
    println("The first 30 terms of the Mian-Chowla sequence are $(mianchowla(30)).")
    println("The 91st through 100th terms of the Mian-Chowla sequence are $(mianchowla(100)[91:100]).")
end

testmianchowla()
@time testmianchowla()
Output:
...
The first 30 terms of the Mian-Chowla sequence are [1, 2, 4, 8, 13, 21, 31, 45, 66, 81, 97, 123, 148, 182, 204, 252, 290, 361, 401, 475, 565, 593, 662, 775, 822, 916, 970, 1016, 1159, 1312].
The 91st through 100th terms of the Mian-Chowla sequence are [22526, 23291, 23564, 23881, 24596, 24768, 25631, 26037, 26255, 27219].
  0.007524 seconds (168 allocations: 404.031 KiB)

Kotlin

Translation of Go

// Version 1.3.21

fun mianChowla(n: Int): List<Int> {
    val mc = MutableList(n) { 0 }
    mc[0] = 1
    val hs = HashSet<Int>(n * (n + 1) / 2)
    hs.add(2)
    val hsx = mutableListOf<Int>()
    for (i in 1 until n) {
        hsx.clear()
        var j = mc[i - 1]
        outer@ while (true) {
            j++
            mc[i] = j
            for (k in 0..i) {
                val sum = mc[k] + j
                if (hs.contains(sum)) {
                    hsx.clear()
                    continue@outer
                }
                hsx.add(sum)
            }
            hs.addAll(hsx)
            break
        }
    }
    return mc
}

fun main() {
    val mc = mianChowla(100)
    println("The first 30 terms of the Mian-Chowla sequence are:")
    println(mc.subList(0, 30))
    println("\nTerms 91 to 100 of the Mian-Chowla sequence are:")
    println(mc.subList(90, 100))
}
Output:
The first 30 terms of the Mian-Chowla sequence are:
[1, 2, 4, 8, 13, 21, 31, 45, 66, 81, 97, 123, 148, 182, 204, 252, 290, 361, 401, 475, 565, 593, 662, 775, 822, 916, 970, 1016, 1159, 1312]

Terms 91 to 100 of the Mian-Chowla sequence are:
[22526, 23291, 23564, 23881, 24596, 24768, 25631, 26037, 26255, 27219]

Idiomatic

fun sumsRemainDistinct(candidate: Int, seq: Iterable<Int>, sums: MutableSet<Int>): Boolean {
    val candidateSums = mutableListOf<Int>()

    for (s in seq) {
        when ((candidate + s) !in sums) {
            true -> candidateSums.add(candidate + s)
            false -> return false
        }
    }
    with(sums) {
        addAll(candidateSums)
        add(candidate + candidate)
    }
    return true
}

fun mianChowla(n: Int): List<Int> {
    val bufferSeq = linkedSetOf<Int>()
    val bufferSums = linkedSetOf<Int>()

    val sequence = generateSequence(1) { it + 1 } // [1,2,3,..]
        .filter { sumsRemainDistinct(it, bufferSeq, bufferSums) }
        .onEach { bufferSeq.add(it) }

    return sequence.take(n).toList()
}

fun main() {
    mianChowla(100).also {
        println("Mian-Chowla[1..30] = ${it.take(30)}")
        println("Mian-Chowla[91..100] = ${it.drop(90)}")
    }
}
Output:
Mian-Chowla[1..30] = [1, 2, 4, 8, 13, 21, 31, 45, 66, 81, 97, 123, 148, 182, 204, 252, 290, 
361, 401, 475, 565, 593, 662, 775, 822, 916, 970, 1016, 1159, 1312]

Mian-Chowla[91..100] = [22526, 23291, 23564, 23881, 24596, 24768, 25631, 26037, 26255, 27219]

Mathematica /Wolfram Language

n = {m} = {1};
tmp = {2};
Do[
 m++;
 While[ContainsAny[tmp, m + n],
  m++
  ];
 tmp = Join[tmp, n + m];
 AppendTo[tmp, 2 m];
 AppendTo[n, m]
 ,
 {99}
 ]
Row[Take[n, 30], ","]
Row[Take[n, {91, 100}], ","]
Output:
1,2,4,8,13,21,31,45,66,81,97,123,148,182,204,252,290,361,401,475,565,593,662,775,822,916,970,1016,1159,1312
22526,23291,23564,23881,24596,24768,25631,26037,26255,27219

Nim

import intsets, strutils, times

proc mianchowla(n: Positive): seq[int] =
  result = @[1]
  var sums = [2].toIntSet()
  var candidate = 1

  while result.len < n:
    # Test successive candidates.
    var fit = false
    result.add 0    # Make room for next value.
    while not fit:
      inc candidate
      fit = true
      result[^1] = candidate
      # Check the sums.
      for val in result:
        if val + candidate in sums:
          # Failed to satisfy criterium.
          fit = false
          break
    # Add the new sums to the set of sums.
    for val in result:
      sums.incl val + candidate

let t0 = now()
let seq100 = mianchowla(100)
echo "The first 30 terms of the Mian-Chowla sequence are:"
echo seq100[0..29].join(" ")
echo ""
echo "Terms 91 to 100 of the sequence are:"
echo seq100[90..99].join(" ")

echo ""
echo "Computation time: ", (now() - t0).inMilliseconds, " ms"
Output:
The first 30 terms of the Mian-Chowla sequence are:
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312

Terms 91 to 100 of the sequence are:
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219

Computation time: 2 ms

Pascal

Works with: Free Pascal

keep sum of all sorted.Memorizing the compare positions speeds up.

const
  deltaK = 250;
  maxCnt = 25000;
 Using 
  tElem = Uint64;
  t_n_sum_all = array of tElem; //dynamic array
    n          mian-chowla[n]  average dist    runtime
   250                317739           1270       429 ms// runtime setlength of 2.35 GB ~ 400ms
   500               2085045           7055       589 ms
   750               6265086          16632      1053 ms
..
  1500              43205712          67697      6669 ms
..
  3000             303314913         264489     65040 ms //2xn -> runtime x9,75
..
  6000            2189067236        1019161    719208 ms //2xn -> runtime x11,0
  6250            2451223363        1047116    825486 ms
..
 12000           15799915996        3589137   8180177 ms //2xn -> runtime x11,3
 12250           16737557137        3742360   8783711 ms
 12500           17758426186        4051041   9455371 ms
..
 24000          115709049568       13738671  99959526 ms  //2xn -> runtime x12
 24250          119117015697       13492623 103691559 ms
 24500          122795614247       14644721 107758962 ms
 24750          126491059919       14708578 111875949 ms
 25000          130098289096       14414457 115954691 ms //dt = 4078s ->16s/per number
 
 real  1932m34,698s => 1d8h12m35
program MianChowla;
//compiling with /usr/lib/fpc/3.2.0/ppcx64.2 -MDelphi -O4 -al "%f"
{$CODEALIGN proc=8,loop=4 }
uses
  sysutils;
const
  deltaK = 100;
  maxCnt = 1000;
type
  tElem  = Uint32;
  tpElem = pUint32;
  t_n = array[0..maxCnt+1] of tElem;
  t_n_sum_all = array[0..(maxCnt+1)*(maxCnt+2) DIV 2] of tElem;

var
  n_LastPos,
  n : t_n;

  n_sum_all : t_n_sum_all;

  maxIdx,
  maxN,
  max_SumIdx : NativeUInt;

procedure Init;
var
  i : NativeInt;
begin
  maxIdx := 1;
  maxN   := 1;
  n[maxIdx] := maxN;
  max_SumIdx := 1;
  n_sum_all[max_SumIdx] := 2*maxN;

  For i := 0 to maxCnt do
    n_LastPos[i] := 1;
end;

procedure InsertNew_sum(NewValue:NativeUint);
//insertion already knowning the positions
var
  pElem :tpElem;
  InsIdx,chkIdx,oldIdx,newIdx : nativeInt;
Begin
  newIdx := maxIdx;
  oldIdx := max_SumIdx;
  //append new value
  inc(maxIdx);
  n[maxIdx] := NewValue;
  //extend sum_
  inc(max_SumIdx,maxIdx);
  //heighest value already known
  InsIdx := max_SumIdx;
  n_sum_all[InsIdx] := 2*NewValue;
  //stop mark
  n_sum_all[InsIdx+1] := High(tElem);
  pElem := @n_sum_all[0];
  dec(InsIdx);
  //n_LastPos[newIdx]+newIdx-1 == InsIdx
  repeat
    //move old bigger values
    chkIdx := n_LastPos[newIdx]+newIdx-1;
    while InsIdx > chkIdx do
    Begin
      pElem[InsIdx] := pElem[oldIdx];
      dec(InsIdx);
      dec(oldIdx);
    end;
    //insert new value
    pElem[InsIdx] := NewValue+n[newIdx];
    dec(InsIdx);
    dec(newIdx);
    //all inserted
  until newIdx <= 0;
  //new minimum search position one behind, oldidx is one to small
  inc(oldidx,2);
  For newIdx := 1 to maxIdx do
    n_LastPos[newIdx] := oldIdx;
end;
procedure FindNew;
var
  pSumAll,pn : tpElem;
  i,LastCheckPos,newValue,newSum : NativeUint;
  TestRes : boolean;
begin
  //start value = last inserted value
  newValue := n[maxIdx];
  pSumAll := @n_sum_all[0];
  pn := @n[0];
  repeat
    //try next number
    inc(newValue);
    LastCheckPos := n_LastPos[1];
    i := 1;
    //check if sum = new is already n all_sum
    repeat
      newSum := newValue+pn[i];
      IF LastCheckPos < n_LastPos[i] then
        LastCheckPos := n_LastPos[i];
      while pSumAll[LastCheckPos] < newSum do
        inc(LastCheckPos);
      //memorize LastCheckPos;
      n_LastPos[i] := LastCheckPos;
      TestRes:= pSumAll[LastCheckPos] = newSum;
      IF TestRes then
        BREAK;
      inc(i);
    until i>maxIdx;
    //found?
    If not(TestRes) then
      BREAK;
  until false;
  InsertNew_sum(newValue);
end;

var
  T1,T0: Int64;
  i,k : NativeInt;

procedure Out_num(k:NativeInt);
Begin
  T1 := GetTickCount64;
  //     k      n[k]     average dist last deltak          total time
  writeln(k:6,n[k]:12,(n[k]-n[k-deltaK+1]) DIV deltaK:8,T1-T0:8,' ms');
end;

BEGIN
  writeln('Allocated memory ',2*SizeOf(t_n)+Sizeof(t_n_sum_all));
  T0 := GetTickCount64;
  while t0 = GetTickCount64 do;
  T0 := GetTickCount64;
  Init;

  k := deltaK;
  i := 1;
  repeat
    repeat
      FindNew;
      inc(i);
    until i=k;
    Out_num(k);
    k := k+deltaK;
  until k>maxCnt;
  writeln;
  writeln(#13,'The first 30 terms of the Mian-Chowla sequence are');
  For i := 1 to 30 do
    write(n[i],' ');
  writeln;
  writeln;
  writeln('The terms 91 - 100 of the Mian-Chowla sequence are');
  For i := 91 to 100 do
    write(n[i],' ');
  writeln;
END.
Output:
Allocated memory 2014024
   100       27219     272   0.002 s
   200      172922    1443   0.011 s
   300      514644    3404   0.037 s
   400     1144080    6197   0.090 s
   500     2085045    9398   0.179 s
   600     3375910   12689   0.311 s
   700     5253584   18705   0.520 s
   800     7600544   23438   0.801 s
   900    10441056   28339   1.160 s
  1000    14018951   35611   1.640 s
The first 30 terms of the Mian-Chowla sequence are
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312

The terms 91 - 100 of the Mian-Chowla sequence are
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219

Perl

use strict; 
use warnings;
use feature 'say';

sub generate_mc {
    my($max)  = @_;
    my $index = 0;
    my $test  = 1;
    my %sums  = (2 => 1);
    my @mc    = 1;
    while ($test++) {
        my %these = %sums;
        map { next if ++$these{$_ + $test} > 1 } @mc[0..$index], $test;
        %sums = %these;
        $index++;
        return @mc if (push @mc, $test) > $max-1;
    }
}

my @mian_chowla = generate_mc(100);
say "First 30 terms in the Mian–Chowla sequence:\n", join(' ', @mian_chowla[ 0..29]),
    "\nTerms 91 through 100:\n",                     join(' ', @mian_chowla[90..99]);
Output:
First 30 terms in the Mian–Chowla sequence:
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312

Terms 91 through 100:
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219

Phix

function mian_chowla(integer n)
sequence mc = {1},
         is = {false,true}
    integer len_is = 2, s
    for i=2 to n do
        sequence isx = {}
        integer j = mc[i-1]+1
        mc = append(mc,j)
        while true do
            for k=1 to length(mc) do
                s = mc[k] + j
                if s<=len_is and is[s] then
                    isx = {}
                    exit
                end if
                isx = append(isx,s)
            end for
            if length(isx) then
                s = isx[$]
                if s>len_is then
                    is &= repeat(false,s-len_is)
                    len_is = length(is)
                end if
                for k=1 to length(isx) do
                    is[isx[k]] = true
                end for
                exit
            end if
            j += 1
            mc[i] = j
        end while
    end for
    return mc
end function
 
atom t0 = time()
sequence mc = mian_chowla(100)
printf(1,"The first 30 terms of the Mian-Chowla sequence are:\n %V\n",{mc[1..30]})
printf(1,"Terms 91 to 100 of the Mian-Chowla sequence are:\n %V\n",{mc[91..100]})
printf(1,"completed in %s\n",{elapsed(time()-t0)})
Output:
The first 30 terms of the Mian-Chowla sequence are:
 {1,2,4,8,13,21,31,45,66,81,97,123,148,182,204,252,290,361,401,475,565,593,662,775,822,916,970,1016,1159,1312}
Terms 91 to 100 of the Mian-Chowla sequence are:
 {22526,23291,23564,23881,24596,24768,25631,26037,26255,27219}
completed in 0.1s

Python

Procedural

from itertools import count, islice, chain
import time

def mian_chowla():
    mc = [1]
    yield mc[-1]
    psums = set([2])
    newsums = set([])
    for trial in count(2):
        for n in chain(mc, [trial]):
            sum = n + trial
            if sum in psums:
                newsums.clear()
                break
            newsums.add(sum)
        else:
            psums |= newsums
            newsums.clear()
            mc.append(trial)
            yield trial

def pretty(p, t, s, f):
    print(p, t, " ".join(str(n) for n in (islice(mian_chowla(), s, f))))

if __name__ == '__main__':
    st = time.time()
    ts = "of the Mian-Chowla sequence are:\n"
    pretty("The first 30 terms", ts, 0, 30)
    pretty("\nTerms 91 to 100", ts, 90, 100)
    print("\nComputation time was", (time.time()-st) * 1000, "ms")
Output:
The first 30 terms of the Mian-Chowla sequence are:
 1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312

Terms 91 to 100 of the Mian-Chowla sequence are:
 22526 23291 23564 23881 24596 24768 25631 26037 26255 27219

Computation time was 53.58004570007324 ms

Functional

Works with: Python version 3.7
'''Mian-Chowla series'''

from itertools import (islice)
from time import time


# mianChowlas :: Gen [Int]
def mianChowlas():
    '''Mian-Chowla series - Generator constructor
    '''
    mcs = [1]
    sumSet = set([2])
    x = 1
    while True:
        yield x
        (sumSet, mcs, x) = nextMC(sumSet, mcs, x)


# nextMC :: (Set Int, [Int], Int) -> (Set Int, [Int], Int)
def nextMC(setSums, mcs, n):
    '''(Set of sums, series so far, current term) ->
        (updated sum set, updated series, next term)
    '''
    def valid(x):
        for m in mcs:
            if x + m in setSums:
                return False
        return True

    x = until(valid)(succ)(n)
    setSums.update(
        [x + y for y in mcs] + [2 * x]
    )
    return (setSums, mcs + [x], x)


# TEST ----------------------------------------------------
# main :: IO ()
def main():
    '''Tests'''

    start = time()
    genMianChowlas = mianChowlas()
    print(
        'First 30 terms of the Mian-Chowla series:\n',
        take(30)(genMianChowlas)
    )
    drop(60)(genMianChowlas)
    print(
        '\n\nTerms 91 to 100 of the Mian-Chowla series:\n',
        take(10)(genMianChowlas),
        '\n'
    )
    print(
        '(Computation time c. ' + str(round(
            1000 * (time() - start)
        )) + ' ms)'
    )


# GENERIC -------------------------------------------------

# drop :: Int -> [a] -> [a]
# drop :: Int -> String -> String
def drop(n):
    '''The suffix of xs after the
       first n elements, or [] if n > length xs'''
    def go(xs):
        if isinstance(xs, list):
            return xs[n:]
        else:
            take(n)(xs)
            return xs
    return lambda xs: go(xs)


# succ :: Int -> Int
def succ(x):
    '''The successor of a numeric value (1 +)'''
    return 1 + x


# take :: Int -> [a] -> [a]
# take :: Int -> String -> String
def take(n):
    '''The prefix of xs of length n,
       or xs itself if n > length xs.'''
    return lambda xs: (
        xs[0:n]
        if isinstance(xs, list)
        else list(islice(xs, n))
    )


# until :: (a -> Bool) -> (a -> a) -> a -> a
def until(p):
    '''The result of applying f until p holds.
       The initial seed value is x.'''
    def go(f, x):
        v = x
        while not p(v):
            v = f(v)
        return v
    return lambda f: lambda x: go(f, x)


if __name__ == '__main__':
    main()
Output:
First 30 terms of the Mian-Chowla series:
 [1, 2, 4, 8, 13, 21, 31, 45, 66, 81, 97, 123, 148, 182, 204, 252, 290, 361, 401, 475, 565, 593, 662, 775, 822, 916, 970, 1016, 1159, 1312]

Terms 91 to 100 of the Mian-Chowla series:
 [22526, 23291, 23564, 23881, 24596, 24768, 25631, 26037, 26255, 27219] 

(Computation time c. 27 ms)

Quackery

  [ stack ]                          is makeable    (   --> s )

  [ temp put
    1 bit makeable put
    ' [ 1 ] 1
    [ true temp put
      1+
      over witheach
        [ over + bit
          makeable share & if
            [ false temp replace
              conclude ] ]
      temp take if
        [ dup dip join 
          over witheach
           [ over + bit
             makeable take
             | makeable put ] ]
      over size temp share = until ]
    makeable release
    temp release
    drop ]                           is mian-chowla ( n --> [ )

100 mian-chowla
30 split swap echo cr
-10 split nip echo
Output:
[ 1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312 ]
[ 22526 23291 23564 23881 24596 24768 25631 26037 26255 27219 ]

Raku

(formerly Perl 6)

my @mian-chowla = 1, |(2..Inf).map: -> $test {
    state $index = 1;
    state %sums  = 2 => 1;
    my $next;
    my %these;
    @mian-chowla[^$index].map: { ++$next and last if %sums{$_ + $test}:exists; ++%these{$_ + $test} };
    next if $next;
    ++%sums{$test + $test}; 
    %sums.push: %these;
    ++$index;
    $test
};

put "First 30 terms in the Mian–Chowla sequence:\n", @mian-chowla[^30];
put "\nTerms 91 through 100:\n", @mian-chowla[90..99];
Output:
First 30 terms in the Mian–Chowla sequence:
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312

Terms 91 through 100:
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219

REXX

Programming note:   the   do   loop   (line ten):

      do j=i  for t-i+1;  ···

can be coded as:

      do j=i  to t;       ···

but the 1st version is faster.

/*REXX program computes and displays  any  range  of the  Mian─Chowla  integer sequence.*/
parse arg LO HI .                                /*obtain optional arguments from the CL*/
if LO=='' | LO==","  then LO=  1                 /*Not specified?  Then use the default.*/
if HI=='' | HI==","  then HI= 30                 /* "       "        "   "   "     "    */
r.= 0                                            /*initialize the rejects stemmed array.*/
#=  0                                            /*count of numbers in sequence (so far)*/
$=                                               /*the Mian─Chowla sequence  (so far).  */
   do t=1  until #=HI;      @.= r.0              /*process numbers until range is filled*/
     do i=1    for t;       if r.i  then iterate /*I  already rejected?  Then ignore it.*/
       do j=i  for t-i+1;   if r.j  then iterate /*J     "        "        "     "    " */
       _= i + j                                  /*calculate the sum of   I   and   J.  */
       if @._  then do;  r.t= 1; iterate t;  end /*reject  T  from Mian─Chowla sequence.*/
       @._= 1                                    /*mark  _  as one of the sequence sums.*/
       end   /*j*/
     end     /*i*/
   #= # + 1                                      /*bump the counter of terms in the list*/
   if #>=LO  then  if  #<=HI  then $= $ t        /*In the specified range?  Add to list.*/
   end       /*t*/
                                                 /*stick a fork in it,  we're all done. */
say 'The Mian─Chowla sequence for terms '      LO      "──►"       HI      ' (inclusive):'
say strip($)                                     /*ignore the leading superfluous blank.*/
output   when using the default inputs:
The Mian─Chowla sequence for terms  1 ──► 30  (inclusive):
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312
output   when using the input of:     91   100
The Mian─Chowla sequence for terms 91 ──► 100  (inclusive):
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219

RPL

Works with: RPL version HP48-R
« { 2 } → n sums
  « { 1 } 
    WHILE DUP SIZE n < REPEAT 
       DUP DUP SIZE GET 
       1 CF
       DO 1 + 
          DUP2 ADD DUP sums + SORT ΔLIST
          IF 0 POS THEN DROP ELSE 1 SF END
       UNTIL 1 FS? END 
       OVER DUP + + 'sums' STO+ +
    END 
» » 'A5282' STO
30 A5282
Output:
1: { 1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312 }

Sieve version

This approach is 25 times faster and uses less memory, thanks to a dynamic sieve.

« 39 DUP STWS / CEIL
  « # 0b » 'x' 1 4 ROLL 1 SEQ
» 'CSV' STO                @ ( size → { sieve } )

« IF DUP2 EVAL SIZE 39 * > THEN
     DUP2 EVAL SIZE SWAP 39 / CEIL 1 - START DUP #0 STO+ NEXT
  END
  SWAP 1 - 39 MOD LASTARG / IP 1 + 
  ROT SWAP DUP2 GET 2 5 ROLL ^ R→B OR PUT
» 'SSV' STO                @ ( val 'sieve' → )

« IF DUP2 EVAL SIZE 39 * > THEN DROP2 0 ELSE
     SWAP 1 - 39 MOD LASTARG / IP 1 + 
     ROT SWAP GET 2 ROT ^ R→B AND # 0b ≠
  END
» 'SVS?' STO               @ ( val 'sieve' → boolean )

« 500 CSV 'Sums' STO { 1 }
    WHILE DUP2 SIZE > REPEAT 
       DUP DUP SIZE GET 
       DUP 2 * 'Sums' SSV
       DO 1 + 
          1 SF
          DUP2 ADD
          1 OVER SIZE FOR j
             IF DUP j GET 'Sums' SVS? THEN 1 CF SIZE 'j' STO END
          NEXT
       UNTIL 1 FS? END 
       1 
       1 3 PICK SIZE START GETI 'Sums' SSV NEXT
       DROP2 +
   END 
   SWAP DROP 'Sums' PURGE
» 'A5282' STO
100 A5282
DUP 1 30 SUB SWAP 91 100 SUB
Output:
2: { 1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312 }
1: { 22526 23291 23564 23881 24596 24768 25631 26037 26255 27219 }

Ruby

Translation of: Go
require 'set'
n, ts, mc, sums = 100, [], [1], Set.new
sums << 2
st = Time.now
for i in (1 .. (n-1))
   for j in mc[i-1]+1 .. Float::INFINITY
      mc[i] = j
      for k in (0 .. i)
         if (sums.include?(sum = mc[k]+j))
            ts.clear
            break 
         end
         ts << sum
      end
      if (ts.length > 0)
         sums = sums | ts
         break
      end
   end
end
et = (Time.now - st) * 1000
s = " of the Mian-Chowla sequence are:\n"
puts "The first 30 terms#{s}#{mc.slice(0..29).join(' ')}\n\n"
puts "Terms 91 to 100#{s}#{mc.slice(90..99).join(' ')}\n\n"
puts "Computation time was #{et.round(1)}ms."
Output:
The first 30 terms of the Mian-Chowla sequence are:
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312

Terms 91 to 100 of the Mian-Chowla sequence are:
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219

Computation time was 63.0ms.

Or using an Enumerator:

mian_chowla = Enumerator.new do |yielder|
  mc, sums  = [1], {}
  1.step do |n|
    mc << n
    if  mc.none?{|k| sums[k+n] } then
      mc.each{|k| sums[k+n] = true }
      yielder << n
    else 
      mc.pop # n didn't work, get rid of it.
    end
  end
end

res = mian_chowla.take(100).to_a

s = " of the Mian-Chowla sequence are:\n"
puts "The first 30 terms#{s}#{res[0,30].join(' ')}\n
Terms 91 to 100#{s}#{res[90,10].join(' ')}"

Sidef

Translation of: Go
var (n, sums, ts, mc) = (100, Set(2), [], [1])
var st = Time.micro
for i in (1 ..^ n) {
   for j in (mc[i-1]+1 .. Inf) {
      mc[i] = j
      for k in (0 .. i) {
         var sum = mc[k]+j
         if (sums.has(sum)) {
            ts.clear
            break
         }
         ts << sum
      }
      if (ts.len > 0) {
         sums |= Set(ts...)
         break
      }
   }
}
var et = (Time.micro - st)
var s = " of the Mian-Chowla sequence are:\n"
say "The first 30 terms#{s}#{mc.first(30).join(' ')}\n"
say "Terms 91 to 100#{s}#{mc.slice(90).first(10).join(' ')}\n"
say "Computation time was #{et} seconds."
Output:
The first 30 terms of the Mian-Chowla sequence are:
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312

Terms 91 to 100 of the Mian-Chowla sequence are:
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219

Computation time was 2.6288 seconds.

Swift

Translation of: Go
public func mianChowla(n: Int) -> [Int] {
  var mc = Array(repeating: 0, count: n)
  var ls = [2: true]
  var sum = 0

  mc[0] = 1

  for i in 1..<n {
    var lsx = [Int]()

    jLoop: for j in (mc[i-1]+1)... {
      mc[i] = j

      for k in 0...i {
        sum = mc[k] + j

        if ls[sum] ?? false {
          lsx = []
          continue jLoop
        }

        lsx.append(sum)
      }

      for n in lsx {
        ls[n] = true
      }

      break
    }
  }

  return mc
}

let seq = mianChowla(n: 100)

print("First 30 terms in sequence are: \(Array(seq.prefix(30)))")
print("Terms 91 to 100 are: \(Array(seq[90..<100]))")
Output:
First 30 terms in sequence are: [1, 2, 4, 8, 13, 21, 31, 45, 66, 81, 97, 123, 148, 182, 204, 252, 290, 361, 401, 475, 565, 593, 662, 775, 822, 916, 970, 1016, 1159, 1312]
Terms 91 to 100 are: [22526, 23291, 23564, 23881, 24596, 24768, 25631, 26037, 26255, 27219]

VBScript

' Mian-Chowla sequence - VBScript - 15/03/2019
    Const m = 100, mm=28000
    ReDim r(mm), v(mm * 2)
    Dim n, t, i, j, l, s1, s2, iterate_t
    ReDim seq(m)
    t0=Timer
    s1 = "1": s2 = ""
    seq(1) = 1: n = 1: t = 1
    Do While n < m
        t = t + 1
        iterate_t = False
        For i = 1 to t * 2
            v(i) = 0
        Next
        i = 1
        Do While i <= t And Not iterate_t
            If r(i) = 0 Then
                j = i
                Do While j <= t And Not iterate_t
                    If r(j) = 0 Then
                        l = i + j
                        If v(l) = 1 Then
                            r(t) = 1
                            iterate_t = True
                        End If
                        If Not iterate_t Then v(l) = 1
                    End If
                    j = j + 1
                Loop
            End If
            i = i + 1
        Loop
        If Not iterate_t Then
            n = n + 1
            seq(n) = t
            if           n<= 30 then s1 = s1 & " " & t
            if n>=91 and n<=100 then s2 = s2 & " " & t
        End If
    Loop
    wscript.echo "t="& t
    wscript.echo "The Mian-Chowla sequence for elements 1 to 30:"
    wscript.echo s1
    wscript.echo "The Mian-Chowla sequence for elements 91 to 100:"
    wscript.echo s2
    wscript.echo "Computation time: "&  Int(Timer-t0) &" sec"
Output:
The Mian-Chowla sequence for elements 1 to 30:
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312
The Mian-Chowla sequence for elements 91 to 100:
 22526 23291 23564 23881 24596 24768 25631 26037 26255 27219
Computation time: 2381 sec

Execution time: 40 min

Shorter execution time

Translation of: Go
' Mian-Chowla sequence - VBScript - March 19th, 2019

    Function Find(x(), val) ' finds val on a pre-sorted list
        Dim l, u, h : l = 0 : u = ubound(x) : Do : h = (l + u) \ 2
            If val = x(h) Then Find = h : Exit Function
            If val > x(h) Then l = h + 1 Else u = h - 1
        Loop Until l > u : Find = -1
    End Function

    ' adds next item from a() to result (r()), adds all remaining items
    ' from b(), once a() is exhausted
    Sub Shuffle(ByRef r(), a(), b(), ByRef i, ByRef ai, ByRef bi, al, bl)
        r(i) = a(ai) : ai = ai + 1 : If ai > al Then Do : i = i + 1 : _
            r(i) = b(bi) : bi = bi + 1 : Loop until bi = bl
    End Sub

    Function Merger(a(), b(), bl) ' merges two pre-sorted lists
        Dim res(), ai, bi, i : ReDim res(ubound(a) + bl) : ai = 0 : bi = 0
        For i = 0 To ubound(res)
            If a(ai) < b(bi) Then Shuffle res, a, b, i, ai, bi, ubound(a), bl _
            Else Shuffle res, b, a, i, bi, ai, bl, ubound(a)
        Next : Merger = res
    End Function

    Const n = 100 : Dim mc(), sums(), ts(), sp, tc : sp = 1 : tc = 0
    ReDim mc(n - 1), sums(0), ts(n - 1) : mc(0) = 1 : sums(sp - 1) = 2
    Dim sum, i, j, k, st : st = Timer
    wscript.echo "The Mian-Chowla sequence for elements 1 to 30:"
    wscript.stdout.write("1 ")
    For i = 1 To n - 1 : j = mc(i - 1) + 1 : Do    
            mc(i) = j : For k = 0 To i
                sum = mc(k) + j : If Find(sums, sum) >= 0 Then _
                    tc = 0 : Exit For Else ts(tc) = sum : tc = tc + 1
            Next : If tc > 0 Then
              nu = Merger(sums, ts, tc) : ReDim sums(ubound(nu)) 
              For e = 0 To ubound(nu) : sums(e) = nu(e) : Next
              tc = 0 : Exit Do 
            End If : j = j + 1 : Loop
        if i = 90 then wscript.echo vblf & vbLf & _
            "The Mian-Chowla sequence for elements 91 to 100:"
        If i < 30 or i >= 90 Then wscript.stdout.write(mc(i) & " ")
    Next
    wscript.echo vblf & vbLf & "Computation time: "& Timer - st &" seconds."
Output:

Hint: save the code to a .vbs file (such as "mc.vbs") and start it with this command Line: "cscript.exe /nologo mc.vbs". This will send the output to the console instead of a series of message boxes.
This goes faster because the cache of sums is maintained throughout the computation instead of being reinitialized at each iteration. Also the sums() array is kept sorted to find any previous values quicker.

The Mian-Chowla sequence for elements 1 to 30:
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312

The Mian-Chowla sequence for elements 91 to 100:
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219

Computation time: 1.328125 seconds.

Visual Basic .NET

Translation of: Go
Module Module1
Function MianChowla(ByVal n As Integer) As Integer()
        Dim mc(n - 1) As Integer, sums, ts As New HashSet(Of Integer),
        sum As Integer : mc(0) = 1 : sums.Add(2)
        For i As Integer = 1 To n - 1
            For j As Integer = mc(i - 1) + 1 To Integer.MaxValue
                mc(i) = j
                For k As Integer = 0 To i
                    sum = mc(k) + j
                    If sums.Contains(sum) Then ts.Clear() : Exit For
                    ts.Add(sum)
                Next
                If ts.Count > 0 Then sums.UnionWith(ts) : Exit For
            Next
        Next
        Return mc
    End Function

    Sub Main(ByVal args As String())
        Const n As Integer = 100
        Dim sw As New Stopwatch(), str As String = " of the Mian-Chowla sequence are:" & vbLf
        sw.Start() : Dim mc As Integer() = MianChowla(n) : sw.Stop()
        Console.Write("The first 30 terms{1}{2}{0}{0}Terms 91 to 100{1}{3}{0}{0}" &
            "Computation time was {4}ms.{0}", vbLf, str,
            String.Join(" ", mc.Take(30)), String.Join(" ", mc.Skip(n - 10)), sw.ElapsedMilliseconds)
    End Sub
End Module
Output:
The first 30 terms of the Mian-Chowla sequence are:
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312

Terms 91 to 100 of the Mian-Chowla sequence are:
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219

Computation time was 18ms.

Wren

Translation of: C#
var mianChowla = Fn.new { |n|
    var mc = List.filled(n, 0)
    var sums = {}
    var ts = {}
    mc[0] = 1
    sums[2] = true
    for (i in 1...n) {
        var j = mc[i-1] + 1
        while (true) {
            mc[i] = j
            for (k in 0..i) {
                var sum = mc[k] + j
                if (sums[sum]) {
                    ts.clear()
                    break
                }
                ts[sum] = true
            }
            if (ts.count > 0) {
                for (key in ts.keys) sums[key] = true
                break
            }
            j = j + 1
        }
    }
    return mc
}

var start = System.clock
var mc = mianChowla.call(100)
System.print("The first 30 terms of the Mian-Chowla sequence are:\n%(mc[0..29].join(" "))")
System.print("\nTerms 91 to 100 of the Mian-Chowla sequence are:\n%(mc[90..99].join(" "))")
System.print("\nTook %(((System.clock - start)*1000).round) milliseconds")
Output:
The first 30 terms of the Mian-Chowla sequence are:
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312

Terms 91 to 100 of the Mian-Chowla sequence are:
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219

Took 32 milliseconds

XPL0

Translation of: C

Takes about 1.5 seconds on RPi-4.

define N = 100;
define NN = (N * (N+1)) >> 1;

func Contains(Lst, Item, Size);
int  Lst, Item, Size, I;
[for I:= Size-1 downto 0 do
    if Item = Lst(I) then return true;
return false;
];
 
int MC(N);

proc MianChowla;
int  Sums(NN), Sum, LE, SS, I, J, K;
[MC(0):= 1;
Sums(0):= 2;
SS:= 1;
for I:= 1 to N-1 do
    [LE:= SS;
    J:= MC(I-1) + 1;
    MC(I):= J;
    K:= 0;
    loop    [Sum:= MC(K) + J;
            if Contains(Sums, Sum, SS) then
                [SS:= LE;
                J:= J+1;
                MC(I):= J;
                K:= 0;
                ]
            else
                [Sums(SS):= Sum;
                SS:= SS+1;
                K:= K+1;
                if K > I then quit;
                ];
            ];
        ];
];
 
int  I;
[MianChowla;
Text(0, "The first 30 terms of the Mian-Chowla sequence are:^m^j");
for I:= 0 to 30-1 do
    [IntOut(0, MC(I));  ChOut(0, ^ )];
Text(0, "^m^j^m^jTerms 91 to 100 of the Mian-Chowla sequence are:^m^j");
for I:= 90 to 100-1 do 
    [IntOut(0, MC(I));  ChOut(0, ^ )];
CrLf(0);
]
Output:
The first 30 terms of the Mian-Chowla sequence are:
1 2 4 8 13 21 31 45 66 81 97 123 148 182 204 252 290 361 401 475 565 593 662 775 822 916 970 1016 1159 1312 

Terms 91 to 100 of the Mian-Chowla sequence are:
22526 23291 23564 23881 24596 24768 25631 26037 26255 27219