[seqfan] Minimum quantity of coins

[Eric Angelini]

Hello SeqFans,
To pay 12 with coins labelled 1 and 2 you'll need at least 6 coins (6 x "2" = 12);
To pay 26 with coins labelled 2 and 6 you'll need at least 5 coins (4 x "6" + 1 x "2" = 26);
To pay 65 with coins labelled 6 and 5 you'll need at least 11 coins (10 x "6" + 1 x "5" = 65);
To pay 511 with coins labelled 5 and 11 you'll need at least 47 coins (46 x "11" + 1 x "5" = 511);
To pay 1147 with coins labelled 11 and 47 you'll need at least 29 coins (23 x "47" + 6 x "11" = 1147);
To pay 4729 with coins labelled 11 and 47 you'll need at least 101 coins (100 x "47" + 1 x "29" = 4729);
To pay 29101, etc.
...

In shorthand:

S = 1,2,6,5,11,47,29,101,...

Could someone extend S? Is S finite?
Best,
É.