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.

9, 97, 871, 6171, 77031, 837799, 8400511, 63728127, 670617279, 9780657630, 75128138247, 989345275647, 7887663552367, 80867137596217, 942488749153153, 7579309213675935, 93571393692802302, 931386509544713451

Offset: 1

Author

Rahul Chand, Apr 01 2017

Keywords

• nonn

Comments

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:

0) if n is 1, stop.

1) if n is even, divide n by 2 (n/2).

2) if n is odd, multiply n by 3 and add 1 (3n+1).

Subsequence of A006877. The first tie occurs at a(10) which is tied with 9780657631. - Jens Kruse Andersen, Feb 23 2021

Examples

			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.
		

Links

• Gary T. Leavens and Mike Vermeulen, 3x+1 Search Programs, Computers & Mathematics with Applications. 24 (11): 79-99 (1992).
• Eric Roosendaal, 3x+1 delay records
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A006577, A006877.

Programs

Mathematica

Table[Last@Ordering@Array[If[#>1,#0@If[OddQ@#,3#+1,#/2]+1,0]&,10^k],{k,4}] (* Giorgos Kalogeropoulos, Apr 01 2021 *)

Python

def steps(n):
    if n==1:
        return 0
    else:
        if (n%2)==0:
            return 1+steps(n//2)
        else:
            return 1+steps(3*n+1)
def max_steps(i):
    a=max([[i, steps(i)] for i in range(1, 10**(i))], key=lambda x:x[1])
    return a[0]

Formula

a(n) = max{i} (steps(i) for i in range from 1 to 10^n-1).

max(i) returns the i with the maximum steps(i) value.

where steps(n) is defined as follows

steps(n)= 0 if n=1.

1+steps(n/2) if n is even.

1+steps(3*n+1) if n is odd.

Extensions

Clarified and extended by Jens Kruse Andersen, Feb 23 2021