How many factors a number has I've been wondering how  just adding 1 to a number affects the number of factors.  12 has 6 factors (quite a lot)  but 13 is prime ( I think of 12 as being a sort of factor instability).  On the other hand 14 'resists' change because it has the same number of factors as 15.  23 is prime but 24 has lots of factors as if there is something about 23 that allows the number of factors to dramatically increase.  Clearly the number of factors can increase  an increasing instability or decrease a decreasing stability or resisting stability.  I'd like to know what the proportion of each type of number is. Are there equal numbers of stable and unstable or not? What's the most consecutive stable numbers it is possible to have? It's just interest nothing more. Karl
 A: Not much has been proved about the connection between the numbers of divisors of consecutive integers. There can be dramatic drops and rises, because of the primes, but not much else is known.
One partial exception has to do with numbers that, in your terminology, resist change.
Let $d(n)$ be the number of divisors of $n$. Heath-Brown proved that there are infinitely many integers $n$ such that $d(n)=d(n+1)$.  Thus there are infinitely many $n$ that resist change.  Later, Pinner proved that for any positive integer $B$, there are infinitely many integers $n$ such that $d(n)=d(n+B)$.  Heath-Brown gives an asymptotic estimate that shows that the phenomenon is in fact not very rare.
Please follow this link for a survey of properties of the divisor function $d(n)$.  You can find the paper by Pinner here.  It is now known that for any positive integer $a$, there are infinitely many $n$ such that $d(n)=d(n+1)=24a$.
There has been work, mainly computational, on sequences of more than two consecutive integers that all have the same number of divisors. Edit: Please see the comments of @Gerry Myerson below. 
It is still an open question whether there are infinitely many $n$ such that $d(n)=d(n+1)=d(n+2)$.
The proofs of the results of Heath-Brown and Pinner are difficult.  There is a University of Waterloo Master's thesis by Pechenick that gives a less condensed proof of the result of Heath-Brown, and mentions and/or proves a number of other related results, for example about the number of prime divisors of consecutive integers.
A: It's extremely erratic.
When you add 1 to a number, you replace the set of its prime factors with a different, non-overlapping set of prime factors; the two numbers cannot have any prime factors in common.
For example:
$901 = 17 \times 53$
$902 = 2\times 11\times 41$.
So 901 has four factors: $1,\ \ 17,\ \ 53,\ \ 17\times53$.
902 has eight factors $2,\ \ 11,\ \ 41,\ \ 2\times 11,\ \ 2\times 41,\ \ 11\times 41,\ \ 2\times11\times 41$.
The fact that the two sets of prime factors are always completely disjoint from each other makes the problem non-trivial.
(The fact that the two sets of prime factors are always completely disjoint from each other is also basic to proving that there actually are infinitely many prime numbers.)
The Online Encyclopedia of Integer Sequences has this entry: http://oeis.org/A000005
A: You might find reading something on "highly composite numbers" such as here or here.  Perhaps also of interest, it's not all too hard to find a sequence of composite numbers as long as you like say of length n-1.  Just pick a number n, and consider n!.  Then your sequence goes n!-2, n!-3, ..., n!-n.
