A008833.asm

; A008833 o=1: Largest square dividing n.
; Coded manually 2021-02-25 by Antti Karttunen, https://github.com/karttu
; Note that A000188(n) = A000196(A008833(n)), A007913(n) = n / A008833(n).

add $0,1 ; Add one, because A008833 is offset=1 sequence.
mov $1,1 ; Initialize the result-register, will successively contain those squares that divide n
mov $3,1 ; Odd numbers: 1, 3, 5, 7, 9, ...	
mov $4,1 ; Square candidates: 1, 4, 9, 16, 25, ... (partial sums of the above)
mov $2,$0 ; Make a copy of an argument, to be used as
lpb $2 ; a loop-counter
  add $3,2 ; Get the next odd number
  add $4,$3 ; Get the next square
  mov $5,$0 ; Get a work copy of the original n
  mod $5,$4 ; Does our square candidate divide it?
  cmp $5,0 ; now $5 = 1 if it did divide.
  mov $6,$4
  sub $6,$1 ; $6 = (current_square_candidate - the_last_dividing_square)
  mul $6,$5 ; $6 = (current_square_candidate - the_last_dividing_square) if current_square_candidate divided n, otherwise 0
  add $1,$6 ; Update thus the largest square that has divided so far n
  mov $5,$0 ; Get a fresh work copy of n again
  add $5,1
  trn $5,$4 ; Check whether the square candidate has grown over n already?
  cmp $5,0
  cmp $5,0 ; Now $5 = 0 if the square candidate has grown over n, otherwise 1
  sub $2,$5 ; thus, either decrement the loop counter by one, or fall out
lpe
mov $0,$1