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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.

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    $\begingroup$ Note that the first and third would differ by $2$. The linked question claims there is only the one case of numbers that differ by $2$. but the proof is only for $r=2,3$ $\endgroup$
    – Ross Millikan
    Commented Nov 22, 2023 at 22:18
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    $\begingroup$ Something can be found in literature by searching binomial near collisions, for example Conjecture 3.1 in Binomial Collisions and Near Collisions states that given list is a complete list of binomials that differ by $1$, which if true implies there are no three consecutive binomials. $\endgroup$
    – Sil
    Commented Nov 22, 2023 at 23:43
  • $\begingroup$ Sequence of interest oeis.org/A214292 $\endgroup$
    – vallev
    Commented Nov 28, 2023 at 16:59

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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 $$\binom{n}{r}+1=\binom{n}{r+2}$$

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    $\begingroup$ The question does not assume any relation among the rows in which the coefficients lie, only that the coefficients themselves form three consecutive integers, and that none of the coefficients are first, second, next-to-last or last in their individual rows. You have for some reason assumed the three coefficients are from the same two rows, and that some are even adjacent in their row. $\endgroup$
    – coffeemath
    Commented Nov 28, 2023 at 17:36
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    $\begingroup$ you are correct, it did not say adjacent consecutive integers, I must have assumed this due to symmetry of rows $\endgroup$
    – vallev
    Commented Nov 28, 2023 at 17:56

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