A330591 Number of Collatz steps to reach 1 starting from 6^n + 1.

16, 21, 26, 101, 83, 83, 145, 145, 220, 158, 145, 207, 114, 114, 450, 114, 357, 357, 282, 419, 419, 494, 494, 494, 494, 494, 494, 494, 543, 494, 543, 799, 799, 543, 543, 799, 543, 543, 799, 799, 791, 791, 791, 791, 861, 861, 861, 861, 998, 998, 998, 861, 861, 861

Offset: 1

Author

Aranya Kumar Bal, Dec 18 2019

Keywords

• nonn
• easy

Comments

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).

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.

Examples

			a(2)=21 because the Collatz trajectory of 6^2 + 1 = 37 is
  37 -> 112 -> 56 -> 28 -> 14 ->  7 -> 22 ->
  11 ->  34 -> 17 -> 52 -> 26 -> 13 -> 40 ->
  20 ->  10 ->  5 -> 16 ->  8 ->  4 ->  2 ->  1, which is 21 steps.
		

Crossrefs

Cf. A006577, A179118, A212653.

Programs

Mathematica

n=200;
For [j=1, j
				

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
a(n) = nbsteps(6^n+1); \\ Michel Marcus, Dec 21 2019

Python

from decimal import *
n=1000
for j in range(2,n):
  i=6**j+1
  count=0
  while(i!=1):
    if(i%2==0):
      i=i//2
    else:
      i=3*i+1
    count=count+1
  print(count)

Formula

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