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Sep 5, 2011 at 20:51 comment added Will Jagy Right, this is a slowest growing where each new odd number is the largest that does get you a representation of the first missing even number, what Manjul Bhargava called the "truant." It seems, if you take instead the smallest that gives you the truant, you do get all odd numbers (you have all odd numbers up to $2k-1,$ the truant is $4k,$ the smallest new odd number that works is indeed $2k+1,$ as $2k-1 + 2k +1=4k$). So all that is at stake is the definition of "greedy" algorithm, which I interpret, in this setting, as taking the largest odd number that gets the truant.
Sep 5, 2011 at 20:36 comment added Greg Martin well, the slowest growing set is the set of $all$ odd numbers....
Sep 5, 2011 at 20:29 history answered Will Jagy CC BY-SA 3.0