6 would be 1,5,6,6,7,11,12
In this case the two 6's are not actually duplicates bc one is 6+none of the other numbers and the other is 5+1, yet 1+5 would be a duplicate since those numbers had been added already.
Is there a simple way of finding these values for a set of 5 values?
In the ABC definition above, repeated solutions with a different letter orders would count as a distinct solutions but with numbers the order wouldn't matter. I'm not sure if I am over thinking this and making this a more difficult problem than it is? Any advice?…