Consider the interior of Pascal's triangle: the triangle without numbers of the form $\binom{n}{0},\binom{n}{1},\binom{n}{n-1},\binom{n}{n}$.
A006987 (which is described here) is a list of the smallest $10000$ numbers in the interior of Pascal's triangle, without repeats.
It contains $15$ instances of two consecutive integers. It contains only one instance of integers separated by $2$. But it does not contain three consecutive integers.
So I wonder:
Does the interior of Pascal's triangle contain three consecutive integers?
Looking at the formula for the binomial coefficients, $\binom{n}{r}=\dfrac{n!}{r!(n-r)!}$, I see no reason why there could not be three consecutive integers, but I cannot find any examples either.
This question seems open questiony, but I cannot find any references.
EDIT: In the comments, @Sil notes that it is conjectured that a certain list is a complete list of binomial coefficients that differ by $1$, which if true would imply that the answer to the OP is no. But regardless of the outcome of that conjecture, perhaps there is a way to show that there can be no three consecutive binomial coefficients. Maybe proof by contradiction, or a combinatorial argument?
EDIT 2: Posted at MO.
I don't think this answers the question yet but it may help.
Consider the well-known identity that any entry in Pascal's triangle is the sum of the two above it : $$\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$$ .
Let $a,b,c,d$ be the entries above consecutive integer entries $x,x+1,x+2$.
This yields the system
$x=a+b$
$x=b+c-1$
$x=c+d-2$
Which yields
$c=a+1$ and $d=b+1$
Therefore the entries above the consecutive integers $x,x+1,x+2$ must be of the form $a,b,a+1,b+1$
This is equivalent to saying (since $a=\binom{n}{k}$ for some $n,r$ ) that there exist $n$ and $r$ for which $$\binom{n}{r}+1=\binom{n}{r+2}$$