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.
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