4 and 9 in either order.
1
2
3
5
6
7
10
11
14
15
19
23 cannot be produced at all.
4
8
9
12
13
16
17
18
20
21
22
24
25
26
27
28
29
30
31
32
33
34
35
37
38
39
41
42
43
46
47
50
51
55
59 can be produced in exactly one way each. For example 2*9 + 1*4 = 22, 9*9 + 5*4 = 20, and 3*9 + 8*4 = 59.
36
40
44
45
48
49
52
53
54
56
57
58 can all be produced in two ways each. For example, 45 = 1*9 + 9*4 or 5*9 + 0 * 4, 57 = 1*9 + 12*4 or 5*9 + 3*4.
As seen, alternative ways are found by offsetting 4 9's with 9 4's, so long as this doesn't lead to a negative number of either.
All numbers above 59 can be achieved in at least two different ways.
The following program determined the values of 9 and 4:
DEFDBL AZ
CLS
FOR a = 2 TO 59
FOR b = 2 TO 59
IF a <> b THEN
dnd = a: coefa1 = 1: coefb1 = 0
dvr = b: coefa0 = 0: coefb0 = 1
DO
quot = INT(dnd / dvr)
rmnd = dnd  quot * dvr
coefar = coefa1  quot * coefa0
coefbr = coefb1  quot * coefb0
dnd = dvr: dvr = rmnd
coefa1 = coefa0: coefa0 = coefar
coefb1 = coefb0: coefb0 = coefbr
LOOP UNTIL dvr = 0
gcd = dnd
IF gcd = 1 THEN
had0 = 0: had1 = 0: had2 = 0: had59 = 0
FOR goal = 1 TO 100
coefa = goal * coefa1
coefb = goal * coefb1
IF coefa < 0 THEN neg$ = "a": ELSE neg$ = "b"
DO
coefa = coefa + coefa0
coefb = coefb + coefb0
LOOP UNTIL neg$ = "a" AND coefa >= 0 OR neg$ = "b" AND coefb >= 0
ct = 0
WHILE coefb >= 0 AND coefa >= 0
ct = ct + 1
coefa = coefa + coefa0
coefb = coefb + coefb0
WEND
IF ct = 0 THEN had0 = 1
IF ct = 1 AND goal < 59 THEN had1 = 1
IF ct > 1 THEN had2 = 1
IF ct = 1 AND goal = 59 AND had0 = 1 AND had1 = 1 AND had2 = 1 THEN
had59 = 1
saveca = coefa  coefa0: savecb = coefb  coefb0
END IF
IF goal > 59 AND ct < 2 THEN had59 = 0
NEXT goal
IF had59 = 1 THEN
PRINT a; b, saveca; savecb, coefa0; coefb0
sct = sct + 1: IF sct MOD 40 = 0 THEN DO: LOOP UNTIL INKEY$ > "": PRINT
END IF
END IF
END IF
NEXT
NEXT
