[seqfan] More Frobenius numbers

[Eric Angelini]

Hello SeqFans,
"Coin problem" again: 
http://en.wikipedia.org/wiki/Coin_problem

We want here to represent "n" (first column) with the smallest possible
"integer" a(n) -- with a(n)'s digits seen as "coins".

1 -->   23 because with coins 2 and 3 the largest "impossible" amount is 1
2 -->  345 because with coins 3, 4 and 5 the largest "impossible" amount is 2
3 -->   25 because with coins 2 and 5 the largest "impossible" amount is 3
4 -->  357 because with coins 3, 5 and 7 the largest "impossible" amount is 4
5 -->   27 because with coins 2 and 7 the largest "impossible" amount is 5
6 -->  457 because with coins 4, 5 and 7 the largest "impossible" amount is 6
7 -->   29 because with coins 2 and 9 the largest "impossible" amount is 7
8 --> 5679 because with coins 5,6,7 and 9 the largest "impossible" amount is 8
9 -->  467 because with coins 4, 6 and 7 the largest "impossible" amount is 9
10 --> 479 because with coins 4, 7 and 9 the largest "impossible" amount is 10
11 -->  37 because with coins 3 and 7 the largest "impossible" amount is 11
12 --> 589 because with coins 5, 8 and 9 the largest "impossible" amount is 12
...

I'm not sure at all that the beginning of F is correct, though:

F = 23,345,25,357,27,457,29,5679,467,479,37,589,...

As F is finite, we could decide that F's last term is a(55) = 89,
-- "holes" up to a(55) being replaced by zeros.

Best,
É.