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    <p><code>[2*n + 3*m +1 | m &lt;- [0..], n &lt;- [0..]]</code> won't work because it starts with <code>m = 0</code> and goes through all the <code>n</code>, and then has <code>m = 1</code> and goes through all the n, etc. But just the <code>m = 0</code> part is infinite, so you will never get to m = 1 or 2 or 3, etc. So <code>[2*n + 3*m +1 | m &lt;- [0..], n &lt;- [0..]]</code> is exactly the same as <code>[2*n + 3*0 +1 | n &lt;- [0..]]</code>.</p> <p>To generate all of them, you either need to realize, like users vartec and Hynek -Pichi- Vychodil, that the set of numbers you want is just the natural numbers - {0,2}. Or you need to somehow enumerate all the pairs (m,n) such that m,n are nonnegative. One way to do that is to go along each of the "diagonals" where <code>m + n</code> is the same. So we start with the numbers where <code>m + n = 0</code>, and then the ones where <code>m + n = 1</code>, etc. Each of these diagonals has a finite number of pairs, so you will always go on to the next one, and all the pairs (m,n) will eventually be counted.</p> <p>If we let <code>i = m + n</code> and <code>j = m</code>, then <code>[(m, n) | m &lt;- [0..], n &lt;- [0..]]</code> becomes <code>[(j, i - j) | i &lt;- [0..], j &lt;- [0..i]]</code></p> <p>So for you, you can just do</p> <pre><code>[2*(i-j) + 3*j +1 | i &lt;- [0..], j &lt;- [0..i]] </code></pre> <p>(Of course this method will also produce duplicates for you because there are multiple (m,n) pairs that generate the same number in your expression.)</p>
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    1. POhow to express {2n+3m+1|n,m∈N} in list comprehension form? ( N is the set of natural numbers including 0)
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    1. COdunno what i was drinking when i posted that answer :P thanks for the explanation though :) and i will delete my useless post
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    2. COYou could always `nub` the list in order to eliminate duplicates, although then it's not strictly a list comprehension only, and it'll use inordinate amounts of memory.
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