1. We count 4-tuples $(b_1,b_2,b_3,b_4)$ such that:

• each $b_i \in \{1,2,\dots,100\}$,
• all $b_i$ are distinct,
• either $(b_1,b_2,b_3)$ are consecutive integers, or $(b_2,b_3,b_4)$ are consecutive integers.

Let $A=\{(b_1,b_2,b_3,b_4): b_1,b_2,b_3 \text{ are consecutive integers}\}$ and $B=\{(b_1,b_2,b_3,b_4): b_2,b_3,b_4 \text{ are consecutive integers}\}.$ We need $|A\cup B|=|A|+|B|-|A\cap B|.$

2. Count $|A|$.

For $b_1,b_2,b_3$ to be consecutive integers, they must be some permutation of $\{k,k+1,k+2\}, \qquad k=1,2,\dots,98.$ So there are $98$ choices for the set of three consecutive integers.

These three can be arranged in $3!=6$ ways in positions $b_1,b_2,b_3$.

Now $b_4$ can be any number from $1$ to $100$ except those already used, so $100-3=97$ choices.

Hence $|A|=98\cdot 6\cdot 97.$ By symmetry, $|B|=98\cdot 6\cdot 97.$

So $|A|+|B|=2\cdot 98\cdot 6\cdot 97.$

3. Count $|A\cap B|$.

Here both conditions hold:

• $b_1,b_2,b_3$ are consecutive,
• $b_2,b_3,b_4$ are consecutive.

Since all four entries are distinct, the two 3-element sets $\{b_1,b_2,b_3\}, \qquad \{b_2,b_3,b_4\}$ share exactly the two elements $b_2,b_3$.

Two distinct sets of 3 consecutive integers having exactly 2 common elements must be of the form $\{k,k+1,k+2\}, \quad \{k+1,k+2,k+3\}$ or $\{k+1,k+2,k+3\}, \quad \{k,k+1,k+2\}.$ Thus the four numbers involved must be four consecutive integers: $\{k,k+1,k+2,k+3\}, \qquad k=1,2,\dots,97.$ So there are $97$ choices for the set of values.

Now we count valid arrangements $(b_1,b_2,b_3,b_4)$ of these four numbers.

The middle pair $(b_2,b_3)$ must be the common part of two consecutive triples, so it must be either $\{k+1,k+2\}.$ And since order matters, $(b_2,b_3)$ can be: $(k+1,k+2) \quad \text{or} \quad (k+2,k+1),$ so $2$ choices.

Then the remaining two numbers $k$ and $k+3$ must occupy $b_1$ and $b_4$. Once $(b_2,b_3)$ is fixed, there are $2$ ways:

• $b_1=k,\ b_4=k+3$,
• $b_1=k+3,\ b_4=k$.

Each of these indeed makes both triples consecutive.

Hence for each $k$, number of arrangements in $A\cap B$ is $2\times 2=4.$ Therefore $|A\cap B|=97\cdot 4.$

4. Apply inclusion-exclusion.

$|A\cup B|=2\cdot 98\cdot 6\cdot 97 - 97\cdot 4.$ Factor out $97$: $|A\cup B|=97(2\cdot 98\cdot 6 -4).$ Now, $2\cdot 98\cdot 6=1176,$ so $1176-4=1172.$ Thus $|A\cup B|=97\cdot 1172.$ Compute: $1172\cdot 97=1172(100-3)=117200-3516=113684.$

So the required number is $\boxed{113684}.$

5. Compare with stored answer.

Stored correct answer is $18915$, but our derived answer is $113684$. Therefore, the stored answer does not match the correct count.