ROSECODE 537

Egyptian fractions

C_K_Yang · Math · 2019-01-05

Let A(n) be the number of Egyptian fractions for which $\frac{1}{x_{1}} + \frac{1}{x_{2}} + . . . + \frac{1}{x_{k}} = 3$ (for any k>0) with $0 < x_{1} < x_{2} < . . . < x_{k} \leq n$

For example, A(24)=1, there is only one Egyptian fraction which is equal to 3.

$\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \frac{1}{10} + \frac{1}{15} + \frac{1}{18} + \frac{1}{20} + \frac{1}{24} = 3$

You are given A(30)=25

Find A(200)