A339614 Inputs n that yield a record-breaking value of A008908(n)/(log_2(n)+1) for the Collatz conjecture.

1, 3, 7, 9, 27, 26623, 35655, 52527, 142587, 156159, 230631, 626331, 837799, 1723519, 3542887, 3732423, 5649499, 6649279, 8400511, 63728127, 3743559068799, 100759293214567, 104899295810901231

Offset: 1

Author

Matthew Russell Downey, Dec 10 2020

Keywords

• nonn
• more

Comments

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).

Examples

			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.
		

Links

• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A008908, A006877, A033492.

Programs

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_input = oeis_A006877[0]
record_stopping_time_metric = stopping_time_metric(record_input)
result.append(record_input)
for n in range(1, len(oeis_A006877)):
    current_input = oeis_A006877[n]
    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=", ")