Note that when you use a sequence (of consecutive numbers like here, or consecutive 6n\pm1, or consecutive roots) ordered by ascending a_i, \prod\limits_{i=0}^{k-1} \left( 1 + \frac{1}{3a_i}\right)<\prod\limits_{i=0}^{k-1} \left( 1 + \frac{1}{3a_0+i}\right)<\prod\limits_{i=0}^{k-1} \left( 1 + \frac{1}{3a_0}\right)=\left( 1 + \frac{1}{3a_0}\right)^k
It is better to use the first product since it gives a larger margin before reaching 2
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