A Nurse and 100 Patients A nurse is tending 100 patients, ranked from 1 to 100, where patient 1 is the important and patient 100 is the least important.  A patient's health will go down each day that the nurse doesn't see him/her or if the nurse saw a patient with a lower ranking(a bigger number) since the last time she visited him/her.  Each day, the nurse sees the highest ranked who's health is going down.  For example, the nurse will see patient 1 on the first day, then patient 2 on the second day, then patient 1 on the third day, then patient 3 on the fourth day, then patient 1 on the fifth day, etc.
Describe the set of patients that will wake on the $n$th day feeling worse then the day before.

*Clarification
The nurse will very likely get sick, but let's assume he/she does not.  Whenever the nurse sees a patient, the patient will feel better, until the nurse sees someone that is ranked lower(in this case, a rank with a higher number) then him/her.  The ranking will not change for the patients.  The ranking is just a representation of how important each patient is.  
 A: I will try to think about this problem in terms of an OK list:
On day 0, no one is on the OK list.
When a patient is seen, they go on the OK list.
When patient $n$ is seen, anyone with rank $<n$ goes off the OK list.
The nurse will see the most'important' (i.e. With lowest rank number) patient not on the OK list.
A patient feels worse on day $n$ when on day $n-1$ they were not on the OK list, and they were not seen on day $n-1$.
The nurse visits the patients in this order:
1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,5,1,2,1,3,1,2,1,4,1,2,1,3,1,2,1,6,1,2,1,...
This means that patient 1 will feel worse on days 1,3,5,7,...
Patient 2 will feel worse on days 1,2,5,6,9,10,...
Patient 3 will feel worse on days 1,2,3,4,9,10,11,12,17,18,19,20,...
In general, patient $k$ will feel worse worse on day $n$ iff $\lfloor \frac{n-1}{2^{k-1}} \rfloor$ is even.
To see this function at work, I'll evaluate the function for different $n$ and $k$, and output 1 if the function indicates that the patient feels worse:
\begin{array}{|c|c|c|c|C|}
\hline
n (day) & (Patient) k= 1 & 2&3&4\\
\hline
1&1&1&1&1\\
\hline
2&0&1&1&1\\
\hline
3&1&0&1&1\\
\hline
4&0&0&1&1\\
\hline
5&1&1&0&1\\
\hline
6&0&1&0&1\\
\hline
\end{array}
A: Let's play Hanoi with the nurse.


*

*She has to tends for 100 patients [she has to move 100 disks]

*A patient health goes down if not visited [unmoved disks are wainting to be moved]

*She takes care of the highest patient whose health degraded first [she moves the bigger disk that can move]

*a patient cannot move until a lower patient get treated [a disk can move if smaller disks above it are removed]


So according to this woolly analogy, the sequence follows a Sierpinski pattern.
Disk 1 moves every 2 days
Disk 2 moves every 4 days
Disk 4 moves every 8 days
...
Disk N moves every 2^N days
Moving disk : $1213121412131215121312141213121612131214121312151213121412131217...$
Sequence A001511
I don't know if it is the good answer or if I smoked the carpet, but I find this funny.
A: The examples given for the order in which the nurse visits patients follow the pattern that the nurse visits patient $k$ on day $n$ if $n$ is an odd multiple of $2^{k-1},$
that is, if the rightmost $1$ in the binary representation of $n$ is in the place with place-value $2^{k-1}.$
This suggests (but does not prove) that all visits will follow a pattern based on binary representations of numbers.
The rules for who feels better or worse each day seem a little imprecise, so I'll try to make them precise. 
Let a patient be "declining on day $n$" if they wake up on day $n$ feeling worse than the day before, and "stable on day $n$" if they are not declining on day $n.$
By assumption, all patients are declining on day $1.$
In general, on day $n$ the nurse sees the lowest-numbered patient who is declining on day $n.$
If the nurse sees patient number $k$ on day $n,$
then patient $k$ is stable on day $n+1$; furthermore, patients numbered higher than $k$ are stable on day $n+1$ if and only if they were stable on day $n.$
But all patients are aware of their ranks of "importance" from $1$ to $100$ and all are hypochondriacs with a sense of entitlement over all "less important" (higher-numbered) patients, so on the day after the nurse visits patient $k,$ patients $1$ through $k-1$ (inclusive) all are declining.
(They wake up complaining that they feel worse than the day before.)
To prove a correspondence between the nurses visits and the digits of binary numbers--or better still, to prove a correspondence between the set of who feels worse each morning than they did the day before and the digits of binary numbers, begin by 
letting $b_{99}b_{98}b_{97}\ldots b_3b_2b_1b_0$ be a binary numeral of up to $100$ digits, with all digits to the left of the most significant non-zero digit set to zero. That is, we write the decimal number $8$ as
$000\ldots01000$ instead of just $1000.$
Then patient $k$ is declining on the $n$th day if and only if the binary representation of $n-1$ has a zero in the place for place value $2^{k-1}.$
We can prove this by induction.
As a base case, for $n = 1$ the binary representation of $n-1$ has all zeros. But on the first day (day $n = 1$) all patients are declining, hence the statement is true for $n = 1.$
Now suppose the statement is true for some value of $n \geq 1,$ that is, the patients who are declining on the $n$th day are exactly the ones for which the corresponding digits of the binary representation of $n-1$ are $0.$
On the $n$th day, the nurse will see the most important patient who is declining on that day; that is, a patient whose digit in the binary representation of $n-1$ is $0,$ while the digits corresponding to all lower-numbered patients are all $1.$ Then on the $n+1$st day the patient the nurse saw on the $n$th day will be stable (digit $1$) and all lower-numbered patients will be declining (digit $0$).
All higher-numbered patients will have the same status on the $n+1$st day as on the $n$th day.
But of course when we add $1$ to the binary representation of $n-1,$ the rightmost $0$ in $n-1$ changes to $1,$ all digits to the right of it change from $1$ to $0,$ and all digits to the left if it are unchanged.
Hence the statement is true for $n+1,$ that is, the patients who are declining on the $n+1$st day are exactly the ones for whom the corresponding digits of the binary representation of $n$ are $0.$
Notice that this exactly matches Bram28's answer, since the digit with place value $2^{k-1}$ in the binary representation of $n-1$ is $0$ if and only if
$\left\lfloor \frac{n-1}{2^{k-1}} \right\rfloor$ is even.
Moreover, see what happens if you reverse the columns of the table (so $k=1$ is rightmost) and invert all the digits (change $0$ to $1$ and simultaneously change $1$ to $0,$ since Bram28 uses $1$ to signify declining whereas I use $0$): you get a table in which the $n$th row is the binary representation of $n-1.$
The correspondence with the Towers of Hanoi mentioned in zwim's answer is that the identity of the moving disk in the $n$th move of the Towers of Hanoi relates in the same way to the binary representation of $n$ as does the identity of the patient whom the nurse sees on the $n$th day.
