Th: The numbers 1, 2, 3, 4, …, 3n-1, 3n are reachable for every n

Proof by induction:

base case n=2: 1, 2, 3, 4, 5 and 6 are all reachable (1–>1, 1–>4–>2, 1–>4–>13–>6–>3, 1–>4, 1–>4–>13–>40–>20–>10–>5, 1–>4–>13–>6) This enfolds the base case n=1. We need n=2 in the base case too because the step case won’t work with n=1

step case: Let 1, 2, 3, 4, …, 3n-1, 3n be reachable (n>1)

1. n ist reachable and the algorithm reaches 3n+1 with one (3x+1)-step so 3n+1 is reachable

2. 2n+1 ist less than or equal to 3n thus reachable and one (3x+1)-step with this seed gives 6n+4 and one (ceil(x/2))-step on 6n+4 gives 3n+2 therefore 3n+2 is reachable

3. 2n+2 ist less than or equal to 3n (here we need n>1) thus reachable and one (3x+1)-step with this seed gives 6n+7 and one (ceil(x/2))-step on 6n+7 gives ceil(3n+3.5) = 3n+3 thus 3n+3 is reachable

and from the other end when the audience say 1

In all other cases the magician loses against the best strategy of the audience. Assume he has put his card at position 14.

Then the audience always tells 15 so his card always remains at position 14. When 27 cards are left the audience tells 14 and the

magician loses.