I think Terry Tao made a comment about difficulties in factoring some number that was off by a factor of 2. However, I haven't been able to find a reference, (I went through several years of his blog). Maybe some one could point me in the right direction.
Edit My motivation is the discovery of the need for a factor of two in my function to get the greatest prime divisor from a square-free number (without having to factor the entire thing.) I just want to assure myself that this wasn't his issue.
Edit--Greatest prime divisor of any square-free number >2 $$ \begin{aligned} GPD = \left \{ \begin{aligned} &\gcd(sfn, 2\ sfn) && \text{if }sfn \text{ is odd}\\ &\gcd(sfn, sfn/2) && \text{if }sfn \text{ is even} \end{aligned} \right. \end{aligned} $$ Edit 16 March 2024 I cannot get the greatest prime from a square-free without factoring the whole thing.
But while I was searching for the formula, I discovered a variation to Mertens conjecture that has the power of two critical to the code. I show the formulas below:
{A019565 Square-frees---ordered lexicographically} $$ \begin{array}{|c|cc|cc|cccc|cccccccc|}\hline p_0&1&&&&&&&&&&&&&&&\\ p_1&1&2&\phi&&&&&&&&&&&&&\\ p_2&1&2& 3& 6&\phi&&&&&&&&&&&\\ p_3&1&2& 3& 6& 5& 10& 15&30&\phi&&&&&&&\\ p_4&1&2& 3& 6& 5& 10& 15&30& 7& 14& 21& 42& 35& 70& 105& 210\\ \mu&1& -1& -1& 1& -1& 1& 1& -1& -1& 1& 1& -1& 1& -1& -1& 1\\ \hline &1&2& 3& 5& 6& 7& 10& 14& 15& 21& 30& 35& 42& 70& 105& 210\\ \hline \end{array},$$ where bottom row is the normal divisors.
{A 4-element cycle repeats null product}$$1=\prod_{m=0} 1\leftarrow \prod_{n=4m+1}^{4m+4}\text{A019565}(n)^{\mu(\text{A019565}(n))}$$ {Alternate 2-elements per 2-cycle repeats null product}$$1=\prod_{m=0} 1\leftarrow \prod_{n=2m+1}^{2m+2}\text{A332382}(n)^{\mu(\text{A332382}(n))}(\text{A332382}(n)2)^{\mu(\text{A332382}(n)2)}$$ {A 2-element cycle repeats null sum}$$0=\sum_{m=0} 0\leftarrow \sum_{n=2m+1}^{2m+2}{\mu(\text{A019565}(n))}$$ {Alternate 2-elements per cycle repeats null sum}$$0=\sum_{n=1}{\mu(\text{A332382}(n))+\mu(\text{A332382}(n)2)}$$ {Indexes needed from line $p_4$ above} Single step indexing is needed for \text{A019565} and \text{A332382}-- using odds from $p_4$. 25 Mar 2024 I've found a function that steps through these two sequences. It is based on the index to the prime instead the value of the prime. It is resource intensive, so not practical. However, I think it might be provable.
I'd still like a reference to Terry Tao's comment.