Question

How to produce a sequential count in octal, starting at zero in Python

Answer 1

# Fixed‑width octal (e.g., always 3 digits)

for i in range(65):
    print(f"{i:3d} = {i:03o}")



'''
run:

  0 = 000
  1 = 001
  2 = 002
  3 = 003
  4 = 004
  5 = 005
  6 = 006
  7 = 007
  8 = 010
  9 = 011
 10 = 012
 11 = 013
 12 = 014
 13 = 015
 14 = 016
 15 = 017
 16 = 020
 17 = 021
 18 = 022
 19 = 023
 20 = 024
 21 = 025
 22 = 026
 23 = 027
 24 = 030
 25 = 031
 26 = 032
 27 = 033
 28 = 034
 29 = 035
 30 = 036
 31 = 037
 32 = 040
 33 = 041
 34 = 042
 35 = 043
 36 = 044
 37 = 045
 38 = 046
 39 = 047
 40 = 050
 41 = 051
 42 = 052
 43 = 053
 44 = 054
 45 = 055
 46 = 056
 47 = 057
 48 = 060
 49 = 061
 50 = 062
 51 = 063
 52 = 064
 53 = 065
 54 = 066
 55 = 067
 56 = 070
 57 = 071
 58 = 072
 59 = 073
 60 = 074
 61 = 075
 62 = 076
 63 = 077
 64 = 100

'''

 

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Answer 2

# With conversion function

def to_octal(n):
    return format(n, "o")

for i in range(65):
    print(i, "=", to_octal(i))



'''
run:

0 = 0
1 = 1
2 = 2
3 = 3
4 = 4
5 = 5
6 = 6
7 = 7
8 = 10
9 = 11
10 = 12
11 = 13
12 = 14
13 = 15
14 = 16
15 = 17
16 = 20
17 = 21
18 = 22
19 = 23
20 = 24
21 = 25
22 = 26
23 = 27
24 = 30
25 = 31
26 = 32
27 = 33
28 = 34
29 = 35
30 = 36
31 = 37
32 = 40
33 = 41
34 = 42
35 = 43
36 = 44
37 = 45
38 = 46
39 = 47
40 = 50
41 = 51
42 = 52
43 = 53
44 = 54
45 = 55
46 = 56
47 = 57
48 = 60
49 = 61
50 = 62
51 = 63
52 = 64
53 = 65
54 = 66
55 = 67
56 = 70
57 = 71
58 = 72
59 = 73
60 = 74
61 = 75
62 = 76
63 = 77
64 = 100

'''

 

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