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rewrite the equation as:

206! - 1 = (2^a - 1)(2^b - 1) + (2^a - 2) + ... + 2
start by factoring out (2^a - 1) from the right-hand side:

206! - 1 = (2^a - 1)[(2^b - 1) + (2^(a-1) - 1) + ... + 1]
(2^b - 1) + (2^(a-1) - 1) + ... + 1 = (2^b + 2^(a-1) + ... + 1) - a

= (2^(a+b) - 1)/(2 - 1) - a

= 2^(a+b) - a - 1
206! - 1 = (2^a - 1)[2^(a+b) - a - 1]I got
3 = 2/(1 - 1/(3^a))