A330591 - OEIS

%I #34 Feb 23 2020 16:57:21

%S 16,21,26,101,83,83,145,145,220,158,145,207,114,114,450,114,357,357,

%T 282,419,419,494,494,494,494,494,494,494,543,494,543,799,799,543,543,

%U 799,543,543,799,799,791,791,791,791,861,861,861,861,998,998,998,861,861,861

%N Number of Collatz steps to reach 1 starting from 6^n + 1.

%C The Collatz transform maps any positive integer k to k/2 if k is even or 3*k+1 if k is odd. There is a famous unsolved problem which says that, starting with any positive integer k, repeated application of the Collatz transform will eventually reach 1 or equivalently enter the cycle (4,2,1).

%C This sequence is related to A179118 and A212653, which look at the stopping times of numbers of the form 2^n+1 and 3^n+1 respectively. We note that there exist several sequences of arithmetic progressions with common difference 1 in the former and with common difference -1 in the latter. This sequence looks at stopping times of numbers of the form 2^n*3^n+1 where we see that there exist arithmetic progressions with common difference 1+(-1)=0. This is an interesting result that requires further investigation.

%F a(n) = A006577(6^n+1).

%e a(2)=21 because the Collatz trajectory of 6^2 + 1 = 37 is

%e   37 -> 112 -> 56 -> 28 -> 14 ->  7 -> 22 ->

%e   11 ->  34 -> 17 -> 52 -> 26 -> 13 -> 40 ->

%e   20 ->  10 ->  5 -> 16 ->  8 ->  4 ->  2 ->  1, which is 21 steps.

%t n=200;

%t For [j=1, j<n,j++,

%t   i=6^j+1;

%t   count=0;

%t   While[i!=1,

%t     count=count+1;i=If[Mod[i,2]==0,i/2,3*i+1]];Print[count]]

%o (Python) from decimal import *

%o n=1000

%o for j in range(2,n):

%o   i=6**j+1

%o   count=0

%o   while(i!=1):

%o     if(i%2==0):

%o       i=i//2

%o     else:

%o       i=3*i+1

%o     count=count+1

%o   print(count)

%o (PARI) nbsteps(n) = if(n<0, 0, my(s=n, c=0); while(s>1, s=if(s%2, 3*s+1, s/2); c++); c); \\ A006577

%o a(n) = nbsteps(6^n+1); \\ _Michel Marcus_, Dec 21 2019

%Y Cf. A006577, A179118, A212653.

%K nonn,easy

%O 1,1

%A _Aranya Kumar Bal_, Dec 18 2019