Would you count it manually? Part 2

Count the number of subsets with no consecutive integers:

For $k\le 40$ , choosing $k$ elements, none of which are consecutive, is equivalent to placing $k$ bars in distinct places between $40-k$ balls. Therefore, there are ${41-k \choose k}$ possibilities.

Therefore, the answer is $\displaystyle2^{40}-\sum_{k=0}^{40}{41-k \choose k}=2^{40}-\sum_{k=0}^{20}{41-k \choose k}=2^{40}-F(42)$ , where $F(42)$ is the $42^{\text{nd}}$ Fibonacci number.

With a calculator, we evaluate and get $\boxed{1099243713480}$ .