A290101 a(n) = dropping time for the modified Collatz problem, where x -> 3x+1 if x is odd, and x -> either x/2 or 3x+1 if x is even (minimal number of any such steps to reach a lower number than the starting value n); a(1) = 0 by convention.

0, 1, 6, 1, 3, 1, 11, 1, 3, 1, 8, 1, 3, 1, 11, 1, 3, 1, 6, 1, 3, 1, 8, 1, 3, 1, 16, 1, 3, 1, 19, 1, 3, 1, 6, 1, 3, 1, 11, 1, 3, 1, 8, 1, 3, 1, 16, 1, 3, 1, 6, 1, 3, 1, 8, 1, 3, 1, 11, 1, 3, 1, 19, 1, 3, 1, 6, 1, 3, 1, 13, 1, 3, 1, 8, 1, 3, 1, 13, 1, 3, 1, 6, 1, 3, 1, 8, 1, 3, 1, 11, 1, 3, 1, 13, 1, 3, 1, 6, 1, 3, 1, 16, 1, 3, 1, 8, 1, 3

Offset: 1

Author

Antti Karttunen, Aug 20 2017

Keywords

• nonn

Comments

In contrast to the "3x+1" problem (see A006577, A102419), here you are free to choose either step if x is even. The sequence counts the minimum number of optimally chosen steps which leads to a value smaller than the value we started from.

Examples

			Starting from n = 27, the following is a shortest path leading to a value smaller than 27: 27 -> 82 -> 41 -> 124 -> 373 -> 1120 -> 560 -> 280 -> 140 -> 70 -> 35 -> 106 -> 53 -> 160 -> 80 -> 40 -> 20. It has 16 steps, thus a(27) = 16. Note the 3x+1 step from 124 to 373 which is not allowed in the ordinary Collatz problem.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10000
• Index entries for sequences related to 3x+1 (or Collatz) problem

Crossrefs

Cf. A127885, A290100, A290102, A000012 (even bisection).

Differs from A102419 for the first time at n=27, where a(27) = 16, while A102419(27) = 96.

Programs

PARI

A290101(n) = { if(1==n,return(0)); my(S, k); S=[n]; k=0; while( S[1]>=n, k++; S=vecsort( concat(apply(x->3*x+1, S), apply(x->x\2, select(x->x%2==0, S) )), , 8);  ); k } \\ After Max Alekseyev's code for A127885

Python

from sympy import flatten
def ok(n, L):
    return any(i < n for i in L)
def a(n):
    if n==1: return 0
    L=[n]
    i = 0
    while not ok(n, L):
        L=set(flatten([[3*k + 1, k//2] if k%2==0 else 3*k + 1 for k in L]))
        i+=1
    return i
print([a(n) for n in range(1, 121)]) # Indranil Ghosh, Aug 31 2017

Formula

a(n) <= A102419(n).

a(n) <= A127885(n) [apart from any hypothetical -1's in A127885].