I am looking for an effective way to generate a complete list of integer sequences
{a_1,a_2,...,a_n}
of the length $n$ such that $$0\le a_1\le a_2\le\dots\le a_n< m,$$
with two integer parameters $n$ and $m$.
I can imagine to perform this via
Table[Sort[IntegerDigits[x-1,m,n]],{x,m^n}]
and then delete duplicates, but surely there should exist a much more effective way.
Since we can map such sequence
$$0\leq a_1\leq a_2\leq a_3 \leq \cdots \leq a_{n-1}\leq a_n < m $$
to
$$0 < b_1 = a_1+1 < b_2 = a_2+2 < b_3 =a_3+3 <\cdots < b_n=a_n+n < m+n $$
and
$\{b_1,b_2,\cdots b_n\}$ is the n subsets of Range[m+n-1]
And we can get $\{a_1,a_2,\cdots a_n\}$ from $\{b_1,b_2,\cdots b_n\}-\{1,2,\cdots,n\}$
m = 8;
n = 5;
list = Subsets[Range[m+n-1], {n}]
Subtract[#, Range[n]] & /@ list
With a small trick, we can do this using the Table function. This is necessary because Table has the attribute HoldAll.
For a small example, we first set m and n:
m=4;
n=2;
We then create a list of variables and a list of iterators and join them into the body of Table:
var = Table[x[i], {i, n}];
iter = Table[{x[i], x[i - 1] + 1, m-1}, {i, n}] /. x[0] -> -1;
body = PrependTo[iter, var]
Finally we apply Table to the body and Flatten to get ride of superfluous braces:
Flatten[Table @@ body, 1]
This gives:
{{0, 1}, {0, 2}, {0, 3}, {1, 2}, {1, 3}, {2, 3}}
With[{m = 10, n = 5}, Subtract[1 + #, Range[n]] & /@ Subsets[Range[m + n - 1], {n}]]? (+1) – kglr Dec 02 '20 at 12:53