# Simplifying a log log function

This starts with the summation function over $$\omega(n)$$: $$S_\omega(x) = \sum_{n = 1}^{x} \omega(n) = x \log \log x + B_1 x + o(x)$$ I was trying to separate this into congruence classes modulo 6. (If there's a straight-forward way, let me know and I'll ask that question too.)

Now I am focused on the following, and have made progress (note the 6x limit): $$S_\omega(x) = S_\hat{0}(x) + S_\hat{1}(x) \; + \;... + \; S_\hat{5}(x) \\ S_\chi(6x) = \sum_{n = 1}^{6x}{\omega(n)\chi_1} = S_\hat{1}(6x) + S_\hat{5}(6x) \\ (several \; skipped \; steps \; here) \\ S_\chi(6x) = S_\omega(6x) + S_\omega(x) - S_\omega(3x) - S_\omega(2x) - \left\lfloor\frac{5x}{3} \right\rfloor$$ That is as clean as I can get it, so I substitute the approximations in: $$S_\chi(6x) = 6x \log\log 6x + x \log\log x - 3x \log\log 3x - 2x \log\log 2x \\ + B_1 2x - \left\lfloor\frac{5x}{3}\right\rfloor + o(x)$$ Everything I try from here either breaks it, or makes is worse. Can this be simplified?