A common use of continuations is in the implementation of control structures for: returning from a function, breaking from a loop, exception handling etc. Most languages (like Java, C++ etc) provide these features as part of the core language. Some languages don't (e.g: Scheme). Instead, these languages expose continuatiions as first class objects and let the programmer define new control structures. Thus Scheme should be looked upon as a programming language toolkit, not a complete language in itself.

In Clojure, we almost never need to use continuations directly, because almost all the control structures are provided by the language/VM combination. Still, first class continuations can be a powerful tool in the hands of the competent programmer. Especially in Scheme, continuations are better than the equivalent counterparts in other languages (like the setjmp/longjmp pair in C). This article has more details on this.

BTW, it will be interesting to know how Rich Hickey justifies his opinion about continuations. Any links for that?

Continuations in Clojure
Reading a paper regarding Bernoulli numbers, and I stumbled onto a definition.  First let
$$\frac{x}{e^x-1}=\sum_{k=0}^{\infty}B_k\frac{x^k}{k!}$$
The author then goes on to define new terms.
Let
$$\beta_n=(-1)^n\frac{B_n}{n}, n\ge 1$$
noting that $\beta_n=B_n/n$ except for $\beta_1=-B_1=1/2$ since $B_n=0$ when $n$ is odd and greater than 1.  Now he defines
$$\beta_n^{(j)}=\frac1{j!}\sum_{i_1+i_2+...+i_j=n}\binom{n}{i_1,i_2,...,i_j}\beta_{i_1}...\beta_{i_1}$$
I'm not even sure where to start with this, but, I suppose I'm confused about the role of the bound $i_1+i_2+...+i_j=n.$  I would have to say that the sum of the $i_k$'s are integer partitions of a number $n$.  Should I take this as unique integers?  Or is this simply all possible integer partitions of $n$? So for $n=4$, so i take
$$\{4, 3+1, 2+2, 2+1+1, 1+1+1+1\}$$
or 
$$\{4, 3+1, 1+3, 2+2, 2+1+1, 1+2+1, 1+1+2, 1+1+1+1\}$$
And what of the $j$?  So this tells me, for example if $j=2$ that my summation is 
$$\beta_4^{(2)}=\frac{1}{2!}\left[\binom{4}{2,2}\frac{B_2}{2}\frac{B_2}{2}+\binom{4}{3,1}\frac{B_3}{3}\frac{B_1}{1}\right]$$
Or do i have to include the case of $\binom{4}{1,3}$.  Or am i completely off to begin with?
