Erdős Problem 69 #
Reference: erdosproblems.com/69
Tao observed that erdos_69
is a special case of erdos_257
, since
$$ \sum_{n\geq 2}\frac{\omega(n)}{2^n} = \sum_p \frac{1}{2^p - 1}. $$
Reference: erdosproblems.com/69
Tao observed that erdos_69
is a special case of erdos_257
, since
$$ \sum_{n\geq 2}\frac{\omega(n)}{2^n} = \sum_p \frac{1}{2^p - 1}. $$