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