In a dictionary with more than a few words, some "created corruptions" in the lookup table/index collide with each other. (EDIT: e.g. for "hand" and "and" the overlap of the deletion-corrupted sets is substantial.) You only know which/how many by running the algo against a concrete dictionary. So, these expressions are at best a rough guide.
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This actually has a closed form!
numRecs = -1 + 2^n + (n nCr e+1) * 2F1(1, e-n+1; e+2; -1)
n nCr k combinatorial choice, 2F1 is the hypergeometric function
In a dictionary with more than a few words, some "created corruptions" in the lookup table/index collide with each other. (EDIT: e.g. for "hand" and "and" the overlap of the deletion-corrupted sets is substantial.) You only know which/how many by running the algo against a concrete dictionary. So, these expressions are at best a rough guide.