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COMMENTS
| On 3 elements ABC, some tower (Halmos, "Naive Set Theory" among many) that begins with the empty set can be written without loss of generality as {0, A, AB, ABC}. But we need to have sets B, C, BC, AC included somewhere too so that the thing is "largest", i.e., includes every subset of {A,B,C}. For ABCD, there are 3^3 ways to include B,C,D into AB,AC,AD,BC,BD,CD and 2^5 ways to include AC,AD,BC,BD,CD into ABC,ABD,ACD,BCD. So a(4) = 3^3*2^5.
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EXAMPLE
| a(2) = 4 because:
(1) 0->A A->AB B->AB C->AB AB->ABC AC->ABC BC->ABC ABC->ABC maximal
(2) 0->A A->AB B->AB C->AC AB->ABC AC->ABC BC->ABC ABC->ABC maximal
(3) 0->A A->AB B->AC C->AB AB->ABC AC->ABC BC->ABC ABC->ABC maximal
(4) 0->A A->AB B->AC C->AC AB->ABC AC->ABC BC->ABC ABC->ABC maximal
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