@talzamir : I have worked on the assumption the drop off points are in multiples of 300 miles, given you cannot deliver fuel more than 300 miles without consuming some of the fuel you have already brought out into the desert - which I believe (intuitively rather than mathematically) will be less efficient. For example, you cannot build a depot at the 2400 mile mark, without first storing fuel at the 2700 mile mark (and using some of it in the process).

@forkedknight : I agree there is a more efficient method than the one I posted above. I have re-visited and I agree you need 7 at point F, but I believe you only need :

~ 12 at point E;

~ 23 at point D;

~ 44 at point C;

~ 87 at point B, and

~ 172 at point A.

This leaves no containers of fuel behind. To get 7 to point F:

1) we use 4 fuel to deliver 2 to F and return to E

2) we use 4 fuel to deliver 2 to F and return to E

3) we use 4 fuel to deliver 2 to F and remain at F with 1 container worth of fuel in the tank

Total fuel at point F = 7

Total fuel at point E = 12

To get 12 to point E:

1) we use 4 fuel to deliver 2 to E and return to D (we do this 5 times)

6) we use 3 fuel to deliver 2 to E and remain at E with 0 fuel in the tank.

Total fuel at point E = 12

Total fuel at point D = 23

.....and so on for each point......

The number required flips from odd to even at each point. Working backwards (i.e. from point F to point A), the fuel required at point n-1 (assuming we want to get fuel to point n) is:

if fuel(n) is even : fuel(n-1) = (fuel(n) * 2) - 1

if fuel(n) is odd : fuel(n-1) = (fuel(n) - 1) * 2

There may be a simpler way to express this, but this is what I came up for the moment.

Andrew

P.S. This assumes a process of stock-piling fuel at 300 mile intervals and only moving on to the next point once you have a enough fuel to complete the entire journey. There may be another method for delivering the minimum fuel necessary to complete a single journey as post by talzamir but I haven't looked at that method yet.