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A284668 Numbers that have the largest Collatz total stopping time of all numbers below 10^n. The smallest number is chosen in case of ties.

%I A284668 #38 Mar 23 2025 21:54:01
%S A284668 9,97,871,6171,77031,837799,8400511,63728127,670617279,9780657630,
%T A284668 75128138247,989345275647,7887663552367,80867137596217,
%U A284668 942488749153153,7579309213675935,93571393692802302,931386509544713451
%N A284668 Numbers that have the largest Collatz total stopping time of all numbers below 10^n. The smallest number is chosen in case of ties.
%C A284668 Collatz stopping time is defined as the number of steps that a number n takes to converge to 1 using one of the following steps:
%C A284668   0) if n is 1, stop.
%C A284668   1) if n is even, divide n by 2 (n/2).
%C A284668   2) if n is odd, multiply n by 3 and add 1 (3n+1).
%C A284668 Subsequence of A006877. The first tie occurs at a(10) which is tied with 9780657631. - _Jens Kruse Andersen_, Feb 23 2021
%H A284668 Gary T. Leavens and Mike Vermeulen, <a href="https://doi.org/10.1016/0898-1221(92)90034-F">3x+1 Search Programs</a>, Computers & Mathematics with Applications. 24 (11): 79-99 (1992).
%H A284668 Eric Roosendaal, <a href="http://www.ericr.nl/wondrous/delrecs.html">3x+1 delay records</a>
%H A284668 <a href="/index/3#3x1">Index entries for sequences related to 3x+1 (or Collatz) problem</a>
%F A284668 a(n) = max{i} (steps(i) for i in range from 1 to 10^n-1).
%F A284668 max(i) returns the i with the maximum steps(i) value.
%F A284668 where steps(n) is defined as follows
%F A284668 steps(n)= 0 if n=1.
%F A284668           1+steps(n/2) if n is even.
%F A284668           1+steps(3*n+1) if n is odd.
%e A284668 For n=1, steps(1) to steps(9) take the following values: 0, 1, 7, 2, 5, 8, 16, 3, 19; the maximum of all those is 19 which occurs for steps(9) therefore a(1)=9.
%t A284668 Table[Last@Ordering@Array[If[#>1,#0@If[OddQ@#,3#+1,#/2]+1,0]&,10^k],{k,4}] (* _Giorgos Kalogeropoulos_, Apr 01 2021 *)
%o A284668 (Python)
%o A284668 def steps(n):
%o A284668     if n==1:
%o A284668         return 0
%o A284668     else:
%o A284668         if (n%2)==0:
%o A284668             return 1+steps(n//2)
%o A284668         else:
%o A284668             return 1+steps(3*n+1)
%o A284668 def max_steps(i):
%o A284668     a=max([[i, steps(i)] for i in range(1, 10**(i))], key=lambda x:x[1])
%o A284668     return a[0]
%Y A284668 Cf. A006577, A006877.
%K A284668 nonn
%O A284668 1,1
%A A284668 _Rahul Chand_, Apr 01 2017
%E A284668 Clarified and extended by _Jens Kruse Andersen_, Feb 23 2021