CareerCup

Just a correction:

2^(n/2)*3-1

Recursive relation is:

T(n)  = 2 T((n-1)/2) + 1

Total operations required will be : 2^((n-1)/2) -1

T(n) = 2.T(n-2) + 3

T(n,r) = 2t(k,r)+t(n-k,r-1)
Where n is number of disks and r is number of pegs.

I don't know if it's optimal, but I managed to improve the minimum steps needed for a 3 peg problem, in a 4 peg problem.
let T(n) be the amount of steps needed to move "n" discsto another peg

T(n) = 2*T(n-2)+3

explanation: you move all but 2 discs to a helper,
then move the n-1 disc to helper2,
move the last disc to the destination,
then from helper2 to destination,
and lastly the n-2 discs to detination.

ultimately if sums up to

2^(n/2)-1

I used the 3 and 4 discs as a test and it seems to work.