Task
a(1) = 1; for n > 1, a(n) = largest proper divisor of n, where n < 101 .


F lpd(n)
   L(i) (n - 1 .< 0).step(-1)
      I n % i == 0
         R i
   R 1
 
L(i) 1..100
   print(‘#3’.format(lpd(i)), end' I i % 10 == 0 {"\n"} E ‘’)