RE is this worth submitting?

[Tautócrona]

Andrew Weimholt:
>>If x^0 is considered a power of x, then the sequence becomes
>>1, 2, 6, 12, 25, 85, 194, 590, 1695, 4879, 19077, 83994, 167988, 1041780

Jud Macranie:
>I'm working on it.  It appears that each term is at least twice as
>large as the previous one.  Has that been proved/conjectured?

Let S={a1,a2,a3,...,ak} be the first k numbers of the seq, and N the first number not 
generated by S. Then the numbers 1...N-1 can be generated by S.

So, if we include now a_(k+1)=N to form S', we can take (a_(k+1))^1 = N and form the 
numbers N+{1...N-1}, taking the same combinations for the powers of a1...ak that we took 
in S for the numbers 1...N-1, hence covering the range [N+1,2N-1].

Therefore, yes! The minimum a_(k+1) possible is 2·ak!

Regards. Jose Brox