Shannon's Information Theory: how is N(t) I have problems understanding the initial parts of Shannon's "A Mathematical Theory of Communication", namely the definition of "the number sequences of duration t" N(t) on page 3. It is defined as follows:

N(t)=N(t-t1)+N(t-t2)+..+N(t-tn)

where

all sequences of the symbols S1,...Sn are allowed and these symbols
  have durations t1,...tn

It is also stated that

The total number is equal to the sum of the number of sequences ending
  in S1, S2, .. , Sn and these are N(t-t1), N(t-t2), .. , N(t-tn),
  respectively.

My problem is if I have a single possible sequence s1 of duration t1 = 5 and am in point of time t=6, then N(6) would be N(1), which would be 0 although I have a sequence of duration 5 which should be counting. I suppose my way of thinking is completely wrong. Could someone help me me?
 A: The formula $N(t)=N(t-t_1)+N(t-t_2)+\cdots$ is implicitly assuming that the total time slot $t$ is fully occupied by a strict concatenation (no empty spaces) of symbols. 
It's true that this assumption has some obvious "border" problems: if we assume that we have two symbols (I prefer not to consider the "degenerate" case of just one symbol) of lenghts $(3,9)$ then we'd get $N(10)=0$, in spite of having the possibility of fitting one long symbol (or three short ones); even more, $N(t)=0$ for any $t$ that is not multiple of $3$. This problem is rather unimportant, because we usually are assuming a discretized "time" that correspond to the greatest common divisor of $t_1, t_2 \cdots $, and we are interested in the long time asymptotics. 
You should study the standard solution of the constant recursive sequence  and the consequence (used in the paper) that asymptotically $s(n)=x_0^n$ where $x_0$ is the largest root of the characteristic equation. Now, you might ask: wait a moment, what if the recursion has only "even" terms, e.g. $s(n)= s(n-2) + s(n-4)$? In this case, the asymptotic is invalid for odd times, say $s(1001)=0$. Of course, that's true, but that's also unimportant (in terms, of the solution as sum of exponentials, we get a cancellation at odd times).
