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There are 
total possible ways to pick any three integers from the set.
Of the total, there are 18 consisting of consecutive triplets
.
Now, of the total, suppose you fix two integers to be consecutive. There would be 19 possible pairs
, and for each pair 18 possible choices for the third integer (for instance, 
can be taken with 3, 4, ..., 20), to a total of 
. To avoid double-counting (e.g. can't go with 3; 
can't go with 1 or 4), we subtract 1 from the extreme pairs and 
(twice), and 2 from the rest (17 times).
So, the number of triplets that don't consist of pairwise consecutive integers is
I don't know how useful this would be to you, but I've verified the count in Mathematica:
In[8]:= DeleteCases[Subsets[Range[1, 20], {3}], x_ /; x[[2]] == x[[1]] + 1 || x[[3]] == x[[2]] + 1] // Length
Out[8]= 816
Of the total, there are 18 consisting of consecutive triplets
Now, of the total, suppose you fix two integers to be consecutive. There would be 19 possible pairs
So, the number of triplets that don't consist of pairwise consecutive integers is
I don't know how useful this would be to you, but I've verified the count in Mathematica:
In[8]:= DeleteCases[Subsets[Range[1, 20], {3}], x_ /; x[[2]] == x[[1]] + 1 || x[[3]] == x[[2]] + 1] // Length
Out[8]= 816