Approximation formula for the lower bounds of nxnxn cube in slice turn metric
I did not take into account, that k>=n/2 successive slice moves on the same axis might be equivalent to <=n/2 different slice moves on this axis. So the number of sequences of length k in this model will be greater than the number Tom derives with a full syllable analysis. But for large n the relative difference will get negligible because from a probabilistic point of view there are "almost no" sequences with this property. The advantage of this simplified model is, that it can be handled by methods of Linear Algebra. Especially the number of sequences with length k can be derived from the k-th power of some nxn matrix A.
I got the following results:1. There are 9N sequences of length 1 and 9n(7.5n-1.5) sequences of length 2.
2. The greatest eigenvalue of A and thus the asymptotic branching factor is bf[n]=3/(1.5^(1/n)-1)
From this we get the approximation formulacountApprox[n] = 9n(7.5n-1.5) bf[n]^(k-2)
for the number of sequences of length k. For the number of sequences of length <= k we then getcountSumApprox[n] = 9n(7.5n-1.5) (bf[n]^(k-1)-1)/(bf[n]-1)
I implemented Chris Hardwicks formula for the number of positions of an nxnxn cube aspos[n_]:=(24*2^10*12!)^(Mod[n,2])*7!*3^6*24!^Quotient[n^2-2n,4]/4!^(6 Quotient[(n-2)^2,4])
Searching the first n that countSumApprox[n]>=pos[n] gives for 2<=n<=30{6,15,32,48,72,95,124,153,189,224,265,306,353,400,452,504,562,620,682,746,814,882,956,1029,1108,1187,1270,1354,1443}
The consistency with Tom's h-s result for n<= 20 is perfect for k>=6. If we only take even n the formula for pos[n] gets simpler.By solving countSumApprox[n]=pos[n] for k and doing further simplification under the assumption of large n we finally get for an approximation of the lower bound:
approxLowerBoundEven[n_]:=Ceiling[(Log[7!*3^6]+(n^2-2n)/4*Log[24!]-3/2(n-2)^2*Log[4!])/Log[bf[n]]]
For the even 2<=n<=30 we get {6,32,72,124,189,265,353,452,562,683,814,956,1108,1270,1443} which still gives remarkably good values for small n.For n=100 we get 13394 also in perfect accordance with Tom's result. For n=1000, 10000 and 100000 we get 1001335, 79633836 and 6607110479. Maybe Tom can supply his values occasionally.
Because there is no strict estimation for the error approxLowerBoundEven will not give a true lower bound for the number of moves. But still it is interesting that such a simple formula gives a very good approximation. Even if you plug odd n into the formula the results are quite accurate.