Let's start with a triplet of integers, say (1, 2, 5) and a set of mathematical operations (+, -, *, /, ^, sqrt, fact!, concatenation, brackets).
Our task will be to represent all (or almost all - as explained below) integers from 1 to n using some or all of the initial triplet and any quantity of operations defined above.
So:
1=1
6=1+5
9=5*2-1
13=15-2
27=51-4!
60=12*5 etc
Let's define n as the first occurrence of not being able to find a valid representation for n+1 and for n+2. I believe that in our case n=17 (15+2), since neither 18 nor 19 get valid solutions.
You are requested to find a triplet of integers (a,b,c) enabling a maximal n.
Ch, the triplet chosen by you is definitely better than mine - i tried 9,9,9 and your 4,9,9 can be easily transformed into mine since !4=9.
However I have squeezed out of nine significantly more single digits:
sqrt9=3 3!=6 6!=720 !3=2 !2=1
plus !4=9 and !(4+1)=44 and 4!!=8 ETC
Just by adding those transformations we can fill in some holes in your list 35 =36-1; 37=36+1; 41=44-3; 44=!(4+1): 47=44+3: 53=44+9;.....59 not solved yet, 62=64-2; 65=64+1; 89=81+8=9*9+4!!
Did not extend your list, but still leave it to you,
if 92 or 93 can be represented as the rules permit ----we have an improved answer.
6!!!=18
some hints & remarks @ Ch
