Seeking the algorithm to generate this table of numbers

I'm looking for a way to generate the following the series of numbers given the version number on the left. For example, given "2", generate [6, 18]. The series of numbers is from here.

 V2  6  18
 V3  6  22
 V4  6  26
 V5  6  30
 V6  6  34
 V7  6  22  38
 V8  6  24  42
 V9  6  26  46
V10  6  28  50
V11  6  30  54
V12  6  32  58
V13  6  34  62
V14  6  26  46  66
V15  6  26  48  70
V16  6  26  50  74
V17  6  30  54  78
V18  6  30  56  82
V19  6  30  58  86
V20  6  34  62  90
V21  6  28  50  72   94
V22  6  26  50  74   98
V23  6  30  54  78  102
V24  6  28  54  80  106
V25  6  32  58  84  110
V26  6  30  58  86  114
V27  6  34  62  90  118
V28  6  26  50  74   98  122
V29  6  30  54  78  102  126
V30  6  26  52  78  104  130
V31  6  30  56  82  108  134
V32  6  34  60  86  112  138
V33  6  30  58  86  114  142
V34  6  34  62  90  118  146
V35  6  30  54  78  102  126  150
V36  6  24  50  76  102  128  154
V37  6  28  54  80  106  132  158
V38  6  32  58  84  110  136  162
V39  6  26  54  82  110  138  166
V40  6  30  58  86  114  142  170

I've come up with the following Python code which generates the correct sequences, but contains an "exceptional" condition for V32, expressed in the if statement.

import math

def pattern(ver):
    w = 21 + ((ver - 1) * 4)       # Width
    h = 6                          # First item
    t = (w - 1) - 6                # Last item
    r = t - h                      # Distance between first and last
    n = math.ceil(r/28)            # Number of items in list - 1

    if (w == 145):
        intervals = 26             # Anomaly
    else:
        intervals = math.ceil(r/n) # Ceiling to nearest multiple of 2
        intervals = intervals + (intervals % 2)

    xs = [t]
    for m in (n-1) * [intervals]: xs.append(xs[-1] - m)
    xs.append(h)
    return list(reversed(xs))

for n in range(2, 41):
    print("Version {:2d}: {}".format(n, pattern(n)))

I'm looking for perhaps an improved algorithm where the conditional is not needed. The algorithm might not exist, but I'm hopeful.