User blog comment:WaxPlanck/Busy Beavers with Adaptive Lower Bounds/@comment-80.98.179.160-20171230195903

Sigma(5)>=11111111111111111111111...11111111111111111 with 23024 1s, thus it's

$$\sum_{n=0}^{23024}\lfloor\frac{10^n-1}9\rfloor$$