Let's say the first N integers divisible by 3 starting with 9.
I'm sure there is some one line solution using lambdas, I just don't know it that area of the language well enough yet.
Just to be different (and to avoid using a where statement) you could also do:
var numbers = Enumerable.Range(0, n).Select(i => i * 3 + 9);
Update This also has the benefit of not running out of numbers.
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