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A339614
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Inputs n that yield a record-breaking value of A008908(n)/(log_2(n)+1) for the Collatz conjecture.
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0
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1, 3, 7, 9, 27, 26623, 35655, 52527, 142587, 156159, 230631, 626331, 837799, 1723519, 3542887, 3732423, 5649499, 6649279, 8400511, 63728127, 3743559068799, 100759293214567, 104899295810901231
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OFFSET
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1,2
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COMMENTS
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The metric A008908(n)/(log_2(n)+1) is always equal to 1 for any power of 2 (where 1 is the smallest possible value).
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LINKS
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EXAMPLE
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a(1) = 1, which is trivial, because the first element in any sequence is record setting.
a(5) = 27, because A008908(n)/(log_2(n)+1) yields a maximum value at n=27 among the first 27 elements, and there are 4 record-breaking elements beforehand.
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PROG
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(Python)
import math
oeis_A006877 = [1, 2, 3, 6, 7, 9, 18, 25, 27, 54, 73, 97, 129, 171, 231, 313, 327, 649, 703, 871, 1161, 2223, 2463, 2919, 3711, 6171, 10971, 13255, 17647, 23529, 26623, 34239, 35655, 52527, 77031, 10623, 142587, 156159, 216367, 230631, 410011, 511935, 626331, 837799, 1117065, 1501353, 1723519, 2298025, 3064033, 3542887, 3732423, 5649499, 6649279, 8400511, 11200681, 14934241, 15733191, 31466382, 36791535, 63728127]
def stopping_time(n):
time = 1
while n>1:
n = 3*n + 1 if n & 1 else n//2
time += 1
return time
def stopping_time_metric(n):
time = stopping_time(n)
logarithmic_distance = (math.log(n, 2)+1)
return float(time/logarithmic_distance)
result = []
record_stopping_time_metric = stopping_time_metric(record_input)
result.append(record_input)
for n in range(1, len(oeis_A006877)):
current_stopping_time_metric = stopping_time_metric(current_input)
if current_stopping_time_metric > record_stopping_time_metric:
record_input = current_input
record_stopping_time_metric = current_stopping_time_metric
result.append(record_input)
for n in range(len(result)):
print(result[n], end=", ")
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CROSSREFS
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KEYWORD
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nonn,more
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AUTHOR
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STATUS
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approved
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