Counting in exclusion and inclusion range

Counting the items in a range has a famous trap: the off-by-one error.
It happens because you must decide whether the two endpoints are included in the count or not.
Simply subtracting the endpoints often gives the wrong answer.

When both endpoints are included, the count is $b - a + 1$.
For example, the years from $2012$ to $2019$ inclusive number $2019 - 2012 + 1 = 8$, not $7$.
You add the extra $1$ to count the starting year itself.

This shows up in sequences, where the terms are numbered $n = 1, 2, 3, \dots$.
If a sequence starts in a given year at $n = 1$, then the value at term $n$ is $\text{start} + (n - 1)$.
The $(n - 1)$ is the inclusive-counting idea again: the first term adds nothing.

A related question asks how many steps or events happen across a span.
There you divide the length of the span by the size of each step.
For instance, if something doubles every $5000$ cycles from cycle $10{,}000$ to cycle $25{,}000$, it doubles $\frac{25{,}000 - 10{,}000}{5000} = 3$ times.

How many integers are there from $3$ to $9$, inclusive?
Since both endpoints are included, use $b - a + 1$.
So it is $9 - 3 + 1 = 7$ integers.

A sequence starts in $2015$, which is term $n = 1$. What year is term $n = 11$?
The year at term $n$ is $\text{start} + (n - 1)$.
So it is $2015 + (11 - 1) = 2025$.

A quantity doubles every $5000$ cycles, from cycle $10{,}000$ to cycle $25{,}000$. How many times does it double?
Divide the span by the step size: $\frac{25{,}000 - 10{,}000}{5000}$.
So it doubles $\frac{15{,}000}{5000} = 3$ times.