A212653 -id:A212653 - cp's OEIS frontend

A213214 Number of steps to reach 1 in the Collatz (3x+1) problem starting with 3^n - 1.

1, 3, 10, 9, 96, 95, 32, 31, 43, 42, 134, 133, 132, 131, 99, 98, 190, 189, 139, 138, 261, 260, 427, 426, 394, 393, 330, 329, 390, 389, 388, 387, 461, 460, 459, 458, 457, 456, 455, 454, 453, 452, 500, 499, 498, 497, 496, 495, 494, 493, 492, 491, 746, 745, 488

Offset: 1

Michel Lagneau, Mar 02 2013

nonn

It is interesting to note that the quantity 3^n - 1 appears in the Collatz trajectory of 2^n - 1 after n iterations (see the formula).

			a(8) = 31 because A193688(8)=47, and 47 - 2*8 = 31.

Michel Lagneau, Table of n, a(n) for n = 1..1000

Cf. A212653, A075487, A193688.

f[n_]:=Module[{a=3^n-1, k=0}, While[a>1, k++; If[EvenQ[a], a=a/2, a=a*3+1]]; k]; Table[f[n], {n,100}]
Table[Length[NestWhileList[If[EvenQ[#],#/2,3#+1]&,3^n-1,#>1&]]-1,{n,100}] (* Harvey P. Dale, Sep 06 2015 *)

a(n) = A193688(n) - 2*n for n > 1.

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

Original entry on oeis.org

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

Aranya Kumar Bal, Dec 18 2019

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.

			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.

Cf. A006577, A179118, A212653.

Mathematica
```
n=200;
For [j=1, j
				
```

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

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)

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

cp's OEIS Frontend

A213214 Number of steps to reach 1 in the Collatz (3x+1) problem starting with 3^n - 1.

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