Answer: $2^{n+1}-1$ Example: If there is one thief at every station $m=1,2,3,\cdots,2^{n+1}-1$, then Azer can not make it through. Every time he stops, the fuel capacity is halved, so after going $2^n+2^{n-1}+\cdots+2^1+2^0=2^{n+1}-1$ units, the fuel capacity drops below 1 and he can not continue anymore. Now, assume we have less than $2^{n+1}-1$ thieves and we will show there is a way for Azer to pass all thieves with at least 1 fuel. The friend of Azer, Ilham, helps Azer by going through all stations to create a strategy for him. Ilham has $2^i$ hats with color $c_i$ for $i=0,1,2,\cdots,n$. He passes this number line $n+1$ times, for $i=n,n-1,n-2,\cdots,0$, every time starting from 0, and goes as long as he still has hats with color $c_i$, that is until he distributes all of his hats with color $c_i$, or until there is no more thieves along the way to give hats. And then he goes back to 0 to start the next round. In every round, whenever he sees a station with at least one thief who does not wear a hat, he gives one of them the hat with color $c_i$ to wear. After these rounds, since the number of hats ($2^{n+1}-1$) is greater than the number of thieves, Ilham has at least one hat left. If a hat with color $c_n$ is left: That means, in the first round, Ilham sees less than $2^n$ stations with at least one thief. So, Azer finds a safe spot in any consecutive $2^n$ stations and since he starts with $2^n$ fuel, he can refill at those safe spot and pass without even seeing a single thief. If a hat with color other than $c_n$ is left: In this case, we will show that there is a route such that Azer sees at most one thief with color $c_i$ for every $i$, and also there is at least one color such that Azer never sees a thief with hat of that color. This proves the capacity is halved at most $n$ times and with a capacity of at least $2^0=1$, Azer can continue indefinitely. Ilham, tries to find such a route to help Azer, but accidentally, he does not take into account the thieves with hat of color $c_n$. Since there is no hat of color $c_n$ left, the total number of thieves with other colors is less than $2^n-1$, so by induction, Ilham finds a route such that Azer sees at most one thief with color $c_i$ for every $i=0,1,2,\cdots, n-1$, and also there is at least one color such that Azer never sees a thief with hat of that color. However, he may encounter multiple thieves with hat of color $c_n$ in this strategy since Ilham accidentally ignored them. Now, we will fix that. Since $c_n$ hats are distributed in the first round, there is number $s$ such that the first $s$ station either is empty or includes a thief with color $c_n$. And after $s$ stations, there is no thieves with hat color $c_n$. To fix his suggestion, Ilham removes the stations with a thief of color $c_n$ in it except for the last one. By induction, we know the old suggestion works if we started with $2^{n-1}$ fuel and ignored $c_n$. Now, we will show that Azer can reach this station with a $c_n$ in it in the new suggestion and we are done. Since we have $2^n$ stations with $c_n$ in it, until reaching the last station with $c_n$, Ilham only observes stations either empty or $c_n$ in it. Even if there is no empty station to refill, Azer can reach the last station with $c_n$ located at $m=2^n$ since the starting fuel is sufficient for this.

Also, I think we can find all the possible distributions of $2^{n+1}-1$ thieves such that Azer can not pass through. I feel like for every $i>j$, the last thief with color $c_j$ is not behind the last thief with color $c_i$. For example, when $n=2$, we find the situations below where blue is $c_2$, green is $c_1$ and red is $c_0$. [asy][asy] size(10cm); real a = 1.0, b = 1.0; int m = 7, n = 11; real w = a*m, h = b*n; path out = (-1, -1) -- (-1, h+1) -- (w+1, h+1) -- (w+1, -1) -- cycle; pen blue = RGB(80, 80, 255), green = RGB(80, 255, 80), red = RGB(255, 80, 80); void fill_cell(int row, int column, pen color) { fill((row*a, column*b) -- (row*a+a, column*b) -- (row*a+a, column*b+b) -- (row*a, column*b+b) -- cycle, color); draw((row*a, column*b) -- (row*a+a, column*b) -- (row*a+a, column*b+b) -- (row*a, column*b+b) -- cycle, linewidth(1.0pt)); } void fill_cell_label(int row, int column, pen color, Label entry) { fill((row*a, column*b) -- (row*a+a, column*b) -- (row*a+a, column*b+b) -- (row*a, column*b+b) -- cycle, color); label(entry, (row*a+a/2, (n-column)*b+b/2), p = fontsize(20pt)); } void cell_label(int row, int column, Label entry) { label(entry, (row*a+a/2, (n-column)*b+b/2), p = fontsize(20pt)); } cell_label(0, 0, "1"); cell_label(1, 0, "2"); cell_label(2, 0, "3"); cell_label(3, 0, "4"); cell_label(4, 0, "5"); cell_label(5, 0, "6"); cell_label(6, 0, "7"); fill_cell(0, n - 1, blue); fill_cell(1, n - 1, blue); fill_cell(2, n - 1, blue); fill_cell(3, n - 1, blue); fill_cell(4, n - 1, green); fill_cell(5, n - 1, green); fill_cell(6, n - 1, red); fill_cell(0, n - 3, blue); fill_cell(1, n - 3, blue); fill_cell(2, n - 3, blue); fill_cell(3, n - 3, blue); fill_cell(4, n - 3, green); fill_cell(5, n - 3, green); fill_cell(5, n - 4, red); fill_cell(0, n - 6, blue); fill_cell(1, n - 6, blue); fill_cell(2, n - 6, blue); fill_cell(3, n - 6, blue); fill_cell(3, n - 7, green); fill_cell(4, n - 7, green); fill_cell(5, n - 7, red); fill_cell(0, n - 9, blue); fill_cell(1, n - 9, blue); fill_cell(2, n - 9, blue); fill_cell(3, n - 9, blue); fill_cell(3, n - 10, green); fill_cell(4, n - 10, green); fill_cell(4, n - 11, red); [/asy][/asy] [asy][asy] size(6cm); real a = 1.0, b = 1.0; int m = 7, n = 11; real w = a*m, h = b*n; path out = (-1, -1) -- (-1, h+1) -- (w+1, h+1) -- (w+1, -1) -- cycle; pen blue = RGB(80, 80, 255), green = RGB(80, 255, 80), red = RGB(255, 80, 80); void fill_cell(int row, int column, pen color) { fill((row*a, column*b) -- (row*a+a, column*b) -- (row*a+a, column*b+b) -- (row*a, column*b+b) -- cycle, color); draw((row*a, column*b) -- (row*a+a, column*b) -- (row*a+a, column*b+b) -- (row*a, column*b+b) -- cycle, linewidth(1.0pt)); } void fill_cell_label(int row, int column, pen color, Label entry) { fill((row*a, column*b) -- (row*a+a, column*b) -- (row*a+a, column*b+b) -- (row*a, column*b+b) -- cycle, color); label(entry, (row*a+a/2, (n-column)*b+b/2), p = fontsize(20pt)); } void cell_label(int row, int column, Label entry) { label(entry, (row*a+a/2, (n-column)*b+b/2), p = fontsize(20pt)); } cell_label(0, 0, "1"); cell_label(1, 0, "2"); cell_label(2, 0, "3"); cell_label(3, 0, "4"); cell_label(4, 0, "5"); cell_label(5, 0, "6"); cell_label(6, 0, "7"); fill_cell(0, n - 1, blue); fill_cell(1, n - 1, blue); fill_cell(2, n - 1, blue); fill_cell(3, n - 1, blue); fill_cell(2, n - 2, green); fill_cell(3, n - 2, green); fill_cell(4, n - 2, red); fill_cell(0, n - 4, blue); fill_cell(1, n - 4, blue); fill_cell(2, n - 4, blue); fill_cell(3, n - 4, blue); fill_cell(2, n - 5, green); fill_cell(3, n - 5, green); fill_cell(3, n - 6, red); [/asy][/asy]

Not a hard problem but really interesting. First observation is, if previous gas stations $k<2^n$ have no thief, then the thieves before $k$ are all useless since Azer can skip them to $k.$ Now WLOG all the $k\le N$-th gas stations have at least one thief. Here we get the construction quickly: take $N=2^{n+1}-1.$ Each gas station $k\le N$ have one thief. Then obviously Azer needs to cross at least $n+1$ times before reaching $2^{n+1},$ and he cannot keep moving since $2^n/2^{n+1}=1/2<1.$ Now we only need to prove that if there are at most $2^{n+1}-2$ thieves, Azer can find a method to run forever. WLOG $1\sim 2^n$ have at least one thief, otherwise Azer first run to the station with no thief. Now there are at most $2^{n+1}-2-2^n=2^n-2$ more thieves, so simple idea is induction. Base case is trivial. Assume property holds for $n-1,$ consider $n.$ Azer first run to the $2^n$-th gas station. Then his new fuel capacity is $2^{n-1}$ and there are at most $2^{n-1}-2$ thieves after $2^n$. Use induction hypothesis and we are done. $\Box$

EthanWYX2009 wrote: Azer first run to the $2^n$-th gas station. Then his new fuel capacity is $2^{n-1}$ and there are at most $2^{n-1}-2$ thieves after $2^n$. Use induction hypothesis and we are done. $\Box$ Nice solution! But what if the there are multiple thieves at $2^{n}$ along the number line?

AnSoLiN wrote: Answer: $2^{n+1}-1$ Example: If there is one thief at every station $m=1,2,3,\cdots,2^{n+1}-1$, then Azer can not make it through. Every time he stops, the fuel capacity is halved, so after going $2^n+2^{n-1}+\cdots+2^1+2^0=2^{n+1}-1$ units, the fuel capacity drops below 1 and he can not continue anymore. Now, assume we have less than $2^{n+1}-1$ thieves and we will show there is a way for Azer to pass all thieves with at least 1 fuel. The friend of Azer, Ilham, helps Azer by going through all stations to create a strategy for him. Ilham has $2^i$ hats with color $c_i$ for $i=0,1,2,\cdots,n$. He passes this number line $n+1$ times, for $i=n,n-1,n-2,\cdots,0$, every time starting from 0, and goes as long as he still has hats with color $c_i$, that is until he distributes all of his hats with color $c_i$, or until there is no more thieves along the way to give hats. And then he goes back to 0 to start the next round. In every round, whenever he sees a station with at least one thief who does not wear a hat, he gives one of them the hat with color $c_i$ to wear. After these rounds, since the number of hats ($2^{n+1}-1$) is greater than the number of thieves, Ilham has at least one hat left. If a hat with color $c_n$ is left: That means, in the first round, Ilham sees less than $2^n$ stations with at least one thief. So, Azer finds a safe spot in any consecutive $2^n$ stations and since he starts with $2^n$ fuel, he can refill at those safe spot and pass without even seeing a single thief. If a hat with color other than $c_n$ is left: In this case, we will show that there is a route such that Azer sees at most one thief with color $c_i$ for every $i$, and also there is at least one color such that Azer never sees a thief with hat of that color. This proves the capacity is halved at most $n$ times and with a capacity of at least $2^0=1$, Azer can continue indefinitely. Ilham, tries to find such a route to help Azer, but accidentally, he does not take into account the thieves with hat of color $c_n$. Since there is no hat of color $c_n$ left, the total number of thieves with other colors is less than $2^n-1$, so by induction, Ilham finds a route such that Azer sees at most one thief with color $c_i$ for every $i=0,1,2,\cdots, n-1$, and also there is at least one color such that Azer never sees a thief with hat of that color. However, he may encounter multiple thieves with hat of color $c_n$ in this strategy since Ilham accidentally ignored them. Now, we will fix that. Since $c_n$ hats are distributed in the first round, there is number $s$ such that the first $s$ station either is empty or includes a thief with color $c_n$. And after $s$ stations, there is no thieves with hat color $c_n$. To fix his suggestion, Ilham removes the stations with a thief of color $c_n$ in it except for the last one. By induction, we know the old suggestion works if we started with $2^{n-1}$ fuel and ignored $c_n$. Now, we will show that Azer can reach this station with a $c_n$ in it in the new suggestion and we are done. Since we have $2^n$ stations with $c_n$ in it, until reaching the last station with $c_n$, Ilham only observes stations either empty or $c_n$ in it. Even if there is no empty station to refill, Azer can reach the last station with $c_n$ located at $m=2^n$ since the starting fuel is sufficient for this. Nice solution! But what if the last station with $c_n$ located at $m=2^n$ has thieves with color $c_i$ for $i=n,n-1,n-2,\cdots,0$ ?

Wangguanhua wrote: AnSoLiN wrote: Answer: $2^{n+1}-1$ Example: If there is one thief at every station $m=1,2,3,\cdots,2^{n+1}-1$, then Azer can not make it through. Every time he stops, the fuel capacity is halved, so after going $2^n+2^{n-1}+\cdots+2^1+2^0=2^{n+1}-1$ units, the fuel capacity drops below 1 and he can not continue anymore. Now, assume we have less than $2^{n+1}-1$ thieves and we will show there is a way for Azer to pass all thieves with at least 1 fuel. The friend of Azer, Ilham, helps Azer by going through all stations to create a strategy for him. Ilham has $2^i$ hats with color $c_i$ for $i=0,1,2,\cdots,n$. He passes this number line $n+1$ times, for $i=n,n-1,n-2,\cdots,0$, every time starting from 0, and goes as long as he still has hats with color $c_i$, that is until he distributes all of his hats with color $c_i$, or until there is no more thieves along the way to give hats. And then he goes back to 0 to start the next round. In every round, whenever he sees a station with at least one thief who does not wear a hat, he gives one of them the hat with color $c_i$ to wear. After these rounds, since the number of hats ($2^{n+1}-1$) is greater than the number of thieves, Ilham has at least one hat left. If a hat with color $c_n$ is left: That means, in the first round, Ilham sees less than $2^n$ stations with at least one thief. So, Azer finds a safe spot in any consecutive $2^n$ stations and since he starts with $2^n$ fuel, he can refill at those safe spot and pass without even seeing a single thief. If a hat with color other than $c_n$ is left: In this case, we will show that there is a route such that Azer sees at most one thief with color $c_i$ for every $i$, and also there is at least one color such that Azer never sees a thief with hat of that color. This proves the capacity is halved at most $n$ times and with a capacity of at least $2^0=1$, Azer can continue indefinitely. Ilham, tries to find such a route to help Azer, but accidentally, he does not take into account the thieves with hat of color $c_n$. Since there is no hat of color $c_n$ left, the total number of thieves with other colors is less than $2^n-1$, so by induction, Ilham finds a route such that Azer sees at most one thief with color $c_i$ for every $i=0,1,2,\cdots, n-1$, and also there is at least one color such that Azer never sees a thief with hat of that color. However, he may encounter multiple thieves with hat of color $c_n$ in this strategy since Ilham accidentally ignored them. Now, we will fix that. Since $c_n$ hats are distributed in the first round, there is number $s$ such that the first $s$ station either is empty or includes a thief with color $c_n$. And after $s$ stations, there is no thieves with hat color $c_n$. To fix his suggestion, Ilham removes the stations with a thief of color $c_n$ in it except for the last one. By induction, we know the old suggestion works if we started with $2^{n-1}$ fuel and ignored $c_n$. Now, we will show that Azer can reach this station with a $c_n$ in it in the new suggestion and we are done. Since we have $2^n$ stations with $c_n$ in it, until reaching the last station with $c_n$, Ilham only observes stations either empty or $c_n$ in it. Even if there is no empty station to refill, Azer can reach the last station with $c_n$ located at $m=2^n$ since the starting fuel is sufficient for this. Nice solution! But what if the last station with $c_n$ located at $m=2^n$ has thieves with color $c_i$ for $i=n,n-1,n-2,\cdots,0$ ? That is not possible, because we are building the new suggestion on top of the initial suggestion which ignores $c_n$. So, in this station, we have all the hats of color $c_1,c_2,\cdots,c_{n-1}$ (ignoring $c_n$), so it can not be in the set of stations suggested initially, because of the induction hypothesis. We take a working suggestion for the case without $c_n$ and we just remove suggestions (if we need to) and do not add any. In the worst case where we do not remove any suggestions (because we already have a single $c_n$ in the suggested stations), the only difference between these two situations is that we have double starting fuel, and extra one thief with hat $c_n$. These cancel out.