It's perhaps even clearer if I do it with 5 and 6. Obviously 6 is one 
greater than 5, which means that each row is 1 step out of sync with the 
previous one.

   0000000000111111111122222222223333333333444444444455555555556666666666
   0123456789012345678901234567890123456789012345678901234567890123456789
   X    X    X    X    X    X    X    X    X    X    X    X    X    X
         X    X    X    X    X    X    X    X    X    X    X    X    X
               X    X    X    X    X    X    X    X    X    X    X    X
                     X    X    X    X    X    X    X    X    X    X    X
                           X    X    X    X    X    X    X    X    X    X
   X    XX   XXX  XXXX XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX

By the time we get to the right, the pattern /obviously/ covers all 
possible numbers. Actually we only need to go 5 steps right for this to 
happen.

At each repartition the number 5n-1 is unmakable. This stops being the 
case on the final repartition. So the last number which /is/ unmakable 
exists on the previous repartition. That is,

   5*4 - 1

which is 19. If you check the table above, you see that this is correct.

Going back to 5 and 8, it's not quite clear how this number comes up. 
Wikipedia asserts that the largest unmakable number is called the 
"Frobenius number", and that for a system with two initial integers A 
and B, the Frobenius number is always

   (A - 1)(B - 1) - 1

In the case of 5 and 8, that gives us 4*7-1 which is indeed 27.

Wikipedia also claims that the /total/ number of unmakable numbers is
   (A - 1)(B - 1) / 2
For the <5, 6> case, that's 4*5/2 = 10, which is indeed correct.

Apparently there is no known formula for the case of more than two 
integers...

Post a reply to this message