你只需要计算最小公倍数，然后除以步长，得到循环排列的个数。

import math

num_cycles = math.lcm(elements,stride)/stride

这里是完整的程序，如果它是有用的任何人。
解决方案，感谢@michael-cao。

"""
    Program to create a square array of a fixed-length list, in a sequential cyclic pattern until
    square is filled from first to last.

    Example using [*, * 1,2,3,4, X, X, X]  in a 3-column list
    2 leadin, 3 leadout, 4 values = 8 elements.

      |------------|  stride = 3  (amount per row)
      |----- first element
        *;   *;0001;1
        0002;0003;   X;2
           X;   X;   *;3
           *;0001;0002;4
        0003;   X;   X;5
           X;   *;   *;6
        0001;0002;0003;7
           X;   X;   X;8
                    |---- last element

      The purpose is to print each row on a paper, put these papers in sequence.
      After last row, the first row is printed again. For this reason, the row must be filled.

    To find amount of lines the following formula is used:

              least common multiple (elementcount, stride)
               ---------------------------------------
                        stride

             example:   lcm(8,3) = 24         24 /3  = 8 rows

    to output the list, the itertools module is used
          * cycle
          * islice

"""
import math
from itertools import cycle, islice

# ----------------------------------------------------
# -- configuration
#
leadin = '*'              # which string for leadin
leadincount =  2          # amount of lead in
leadout = 'X'             # which string for leadout
leadoutcount = 3          # amount of leadout
startvalue = 1            # first number in sequence
endvalue = 3              # last number in sequence
startindex  = 0           # zero-based start index for the list..

# ---- options for output
stride=3                  # amount of columns
numberformat='04d'        # formatting number series
textformat  = '>4s'       # formatting string (lead in/out)
listseparator = ';'       # what separates output list...
outputIndex= True         # print index number next to data
#
# ----------------------------------------------------

dataelements = [leadin]*leadincount + list(range(startvalue, endvalue+1)) + [leadout]*leadoutcount

repeats = int(math.lcm(len(dataelements), stride)/stride)

"""
   format the elements according to needs, text or number
"""
def outgroup(d):
    a = []
    for i in d:
        l = str(i)
        if l.isnumeric():
            a.append(f"{i:{numberformat}}")
        else:
            a.append(f"{l:{textformat}}")
    return a
    

for index, *group in zip(
             range(repeats), 
             *[islice(cycle(dataelements), startindex, None)] * stride
             ):
    dataline = listseparator.join(outgroup(group))
    if outputIndex:
        print(f"{dataline}{listseparator}{index+1}")
    else:
        print(f"{dataline}{listseparator}")