Suppose you have an alphabet with countably many letters. Every letter has a particular weight (for instance, as in the game of Scrabble). There are a total of $n^2$ letters that have weight $n$.


Given any word in this alphabet, let the weight of that word be the sum of the weights of its letters (again, as in Scrabble). It follows that there are roughly exponentially many words of weight $W$. 


I am sampling words of weight $W$, uniformly at random. My somewhat vague question is: what does a ``generic'' word look like, for large $W$? This can be made precise in a few ways:


1. What is the expected value of the number of letters comprising a word of weight $W$? How many of these letters are expected to be distinct?
2. Does a generic word of weight $W$ have a letter that appears only once?
3. What is the expectation for the number of letters that appear only once in the word?


This is quite far from my field of expertise, so even simple pointers to references are much appreciated.



Generic words of given weight
Here is a back of envelope computation. There will be no rigor whatsoever, just a cookbook approach that may be acceptable to a physicist but which every self-respecting mathematician should frown upon. It can give a plausible (but not guaranteed) answer to some of your questions if we just want the general order of magnitude without too high precision but if you want more than that, we'll need to count in honest. 


If what I write below looks like gibberish, it's OK. As I said, I just did the calculus, not the actual analysis. If your model corresponds to some physical reality and you know the values of some constants involved, it will be funny to compare them with what follows from my predictions. If you want more, like the distribution laws for the number of occurrences of individual letters (or, God forbid, joint distribution for several letters), you need to consult a true expert, not an amateur like myself. In any case, have fun reading and feel free to ask questions. Also watch for idiotic mistakes in algebra: it is quite late here now... :-)


The main local generating function for counting words of weight $W$ assuming typical length about $L$ and decent length concentration (the point is to kill factorials in the denominator; we will still be be off by the factor $e^L$, but we need a typical structure, not the counts).


$F(z,w)=\exp(Lz(w+4w^2+9w^3+16w^4+\dots))=\exp\left(Lz\frac{w+w^2}{(1-w)^3}\right)$


Everything is nice and positive, so the mountain pass in the circle method is a sure bet. We need to find the controlling radius $r$ for the weight $W$, so, for $z=1$, we need to minimize 
$Lz\frac{r+r^2}{(1-r)^3}-W\log r$.
Alas, we are certain to end up with $r$ noticeably less than $1$, so we have to differentiate in honest:
$$
r\frac{4+r-2r^2}{(1-r)^4}=\frac WL
$$
We also need the consistency equation, which says that we, indeed, get $L$ as a typical length with $z=1$, i.e., $\frac\partial{\partial z}(Lz\frac{r+r^2}{(1-r)^3}-L\log z)|\_{z=1}=0$, i.e., $r+r^2=(1-r)^3$ so $r=0.284774761\dots$ and 
$W/L=4.486414\dots$.


This is a bit counterintuitive because it predicts, in particular, that the typical
word consists mainly of the letters of low weights and the option to diversify, which should lead to more possibilities, is really pretty useless. Let's run the sanity check. Suppose that we are looking at words using a few low weight letters. Then they are just $A^W$ with some fixed $A>1$. Let us add a letter of weight $n$ in proportion $p$. We get the new number about $A^{(1-pn)W}\left(\frac ep\right)^{pW}$, so to get the largest count, we have to solve $Cn=-\log p$ whence the portion of letters of length $n$ is decaying exponentially in $n$. Thus, the cost of diversification is prohibitively high here and the answer we got makes sense.


Now, how many different letters typically? The generating function to consider now is
$$
G(s)=\prod\_{k\ge 1}[1+s(e^{Lr^k}-1)]^{k^2}\,.
$$
The typical number $D$ of distinct letters would correspond to the zero derivative of $\log G(s)-D\log s$ with respect to $s$ at $1$. Thus, we expect something like
$$
\sum\_k k^2(1-e^{-Lr^k})\approx\int\_0^\infty t^2(1-e^{-Le^{-\rho t}})\,dt
$$ 
where $\rho=-\log r$. The second factor is just a sharp cutoff at $t=\frac{\log L}{\rho}\approx\frac{\log W}{\rho}$, so, let's say 
$$
D\approx \frac 1{3\rho^{3}}\log^3 W= 0.168209\log^3 W.
$$


What's next? Ah, the typical number $U$ of letters appearing once! Now we need to place $s$ only on one term in the exponent, so we go to 
$$
H(s)=\prod\_{k\ge 1}[e^{Lr^k}-Lr^k(1-s)]^{k^2}\,.
$$
Same routine with setting the derivative of $\log H(s)-U\log s$ to $0$, we get 
$$
H\approx\sum\_k k^2 Lr^k e^{-Lr^k}\,.
$$ 
This one is harder because we run through the bump very quickly, i.e., in constant time. It also suggests an "oscillating asymptotics", i.e., that the arithmetic nature of $W$ introduces an effect that does not decay with size. However, we can still play our usual game around $\log L/\rho$ and get the quasi-asymptotics
$$
0.7927\log^2 W \le U\le 0.7996 \log^2 W\,.
$$
Alas, as I said, the quasi-asymptotics is, probably, all we can hope for here.


