Appetizing Apricots

$log_{2} ((2-1)(2+1)(2^2+1)....(2^{32}+1)+1)$

$log_{2} ((2^{2}-1)(2^2+1)....(2^{32}+1)+1)$

$log_{2} ((2^{4}-1)(2^4+1)....(2^{32}+1)+1)$

$log_{2} ((2^{8}-1)(2^8+1)....(2^{32}+1)+1)$

and so on till

$log_{2} (2^{64}-1+1)$

$\boxed{64}$

Well looking it through a binary perspective,it,s seen tht since the powers double from 1 to 2 to 4 and so on to 32,we will get all possible combinations of powers of 2 below 64 in the range repeated only once.For instance consider this: 1(2+1)(2^2+1)(2^4+1) for now on ..We get 2^0 i.e 1 frm seq and by observing we see tht 2power 1 ,2,3,4,5 and 6 are all possible and tht they occur 1 time only. we get this seq 2^0+2^1+2^2+2^3+2^4+2^5+2^6+2^7 as this answer..This is only possible however if consecutive powers increase by TWICE (increasing order) and hence for for abv seq we get 2^0+....2^63 +1(this is at the end) now using geometric formula we have(2^64-1)/2-1 +1=2^64-1+1=2^64 and voila 64,s the answer