# How many arithmetic progressions with integral terms with first term $a$, last term $b$, and at least $n$ terms?

How many different arithmetic progressions with integral terms can be formed with first term $a$, last term $b$, and at least $n$ terms?

$a \lt b$, and $a,b,n \in \mathbb{N}$ and they are always chosen in a manner which gives a valid answer.

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If an arithmetic progression starts with $a$ and ends with $b$, and has $k$ terms, then we must be able to write $b = a+(k-1)t$ for some $t\gt 0$ (and integer, given the conditions).
That means that $b-a$ must be a multiple of $k-1$. Conversely, if $b-a$ is a multiple of $k-1$, $b-a = (k-1)t$, then the arithmetic progression $$a, a+t, a+2t,\ldots,a+(k-1)t = b$$ starts with $a$, ends with $b$, and has exactly $k$ terms.
So the number of arithmetic progressions that start with $a$ and end with $b$ and have at least $n$ terms is equal to the number of integer divisors of $b-a$ that are greater than $n-1$.
Can we count those more easily instead? In other words, given a number $K$, can we count how many divisors greater than a given $N$ it has?
:Thanks for your answer.I could comment on it (probably due to my network which is currently not allowing the concerned script). I know how to compute the number of divisors of K from the prime factorization, but I don't know how to do that with the additional constraints of " greater than a given $N$ ". Please help. – user18844 Nov 3 '11 at 20:39
The number will depend on $N$ and the prime factors of $K$. Every divisor greater than $N$ is "paired off" with a divisor smaller than $K/N$, which may be easier to work with. It would be simple to do with particular $K$ and $N$, but I doubt you can find some nice closed formula (even in terms of the factorization of $K$, you would need to break it up into cases depending how the prime factors grow and their exponents). – Arturo Magidin Nov 3 '11 at 20:45
I don't know any way besides listing the factors and checking. But once you have the factorization that is not hard. For example, if you have $12=2^23^1$, you know it has $(2+1)(1+1)=6$ factors. To list them, they are of the form something from $\{1,2,4\}$ times something from $\{1,3\}$ – Ross Millikan Nov 3 '11 at 20:49