Task

The objective is to write a function that finds the sum of all positive multiples of 3 or 5 below n.

Show output for n = 1000.

This is is the same as Project Euler problem 1.

Extra credit: do this efficiently for n = 1e20 or higher.




  [ dup 1+ * 2 / ]                  is triangulared ( n --> n )
 
  [ 1 - 
    dup   3 / triangulared  3 *
    over  5 / triangulared  5 * +
    swap 15 / triangulared 15 * - ] is sum-of-3s&5s ( n --> n )
 
      1000 sum-of-3s&5s echo cr 
 
  10 20 ** sum-of-3s&5s echo cr