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
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
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
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 #
HEAP[ 1 : 0 ]BOOL empty sum;
REF[]BOOL is sum := empty sum;
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. 0.1 seconds on TIO.RUN.
ALGOL W
Based on the Algol 68 sample but using a hash table to store the sequence elements.
% 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
record HashedSum ( integer hSum; reference(HashedSum) hNext );
integer HASH_MOD, MAX_MC;
HASH_MOD := 10000;
MAX_MC := 100;
begin
% hash table of sums of the sequence elements encountered so far %
reference(HashedSum) array sums ( 0 :: HASH_MOD - 1 );
% table of the sequence elements encountered so far %
integer array mc ( 1 :: MAX_MC );
integer mcCount, i;
for i := 0 until HASH_MOD - 1 do sums( i ) := null;
mcCount := 1;
i := 0;
while begin i := i + 1; mcCount <= MAX_MC end do begin
logical isUnique;
integer mcPos;
% assume i will be part of the sequence %
mc( mcCount ) := i;
% check the sums %
isUnique := true;
mcPos := 0;
while begin mcPos := mcPos + 1; mcPos <= mcCount and isUnique end do begin
integer s;
reference(HashedSum) hs;
% attempt to find the sum in the hash table %
s := i + mc( mcPos );
hs := sums( s rem HASH_MOD );
while hs not = null and s not = hSum(hs) do hs := hNext(hs);
isUnique := hs = null
end while_isUnique;
if isUnique then begin
% i is a sequence element - store its sums %
for mcPos := 1 until mcCount do begin
integer newSum, sumHash;
newSum := i + mc( mcPos );
sumHash := newSum rem HASH_MOD;
sums( sumHash ) := HashedSum( newSum, sums( sumHash ) )
end for_mcPos ;
mcCount := mcCount + 1
end if_isUnique
end while_mcCount_le_MAX_MC;
% print parts of the sequence %
write( "Mian Chowla sequence elements 1..30:" );write();
for i := 1 until 30 do writeon( i_w := 1, s_w := 0, " ", mc( i ) );
write( "Mian Chowla sequence elements 91..100:" );write();
for i := 91 until 100 do writeon( i_w := 1, s_w := 0, " ", mc( i ) )
end
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
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
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
#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#
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++
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.
EasyLang
func[] mian_chowla n .
len mc[] n
mc[1] = 1
is[] = [ 0 1 ]
for i = 2 to n
j = mc[i - 1]
repeat
j += 1
mc[i] = j
for k = 1 to i
sum = mc[k] + j
if sum > len is[]
len is[] sum + 10000
.
if is[sum] = 1
isnew[] = [ ]
break 1
.
isnew[] &= sum
.
until len isnew[] > 0
.
for v in isnew[]
is[v] = 1
.
.
return mc[]
.
mc[] = mian_chowla 100
for i to 30
write mc[i] & " "
.
print ""
print ""
for i = 91 to 100
write mc[i] & " "
.
- 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
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
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
(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 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
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
'''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
« { 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
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
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
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
' 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
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
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
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