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.

(In reply to re: Record by Dej Mar)

Given the limitation to kth factorials where k <= 3 and with the other imposed restrictions of operations, with the assumption that a member of the set of triplets is a single digit number, the limit may be n=210 using the triplets {4, 4, 5}. 

Thinking "outside the box", I realized that the triplets are not stated to be required to be single digits, and with the SQRT operation, the initial set of integers can begin as larger numbers, which can possibly increase the range and limits. For example, instead of {4, 4, 5}, the set was {16, 16, 25}. One could take the SQRT(x) for each member of the triplet and have the same results but with the increased options now available to extend the limit. The process can be done again with squaring of these numbers, eg., {256. 256, 625}, to further increase one's options as one does not need to take the square root for each integer of the set, but may choose to do so for any.  The limit will definitely be increased as one could now reach 211 with
256 - (!(SQRT(SQRT(256))) × (SQRT(SQRT(625))))
= 256 - (9×5) = 211.     

Posted by Dej Mar on 2020-09-16 14:30:38