OFFSET
1,1
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.
FORMULA
a(n) = A006577(6^n+1).
EXAMPLE
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.
MATHEMATICA
n=200;
For [j=1, j<n, j++,
i=6^j+1;
count=0;
While[i!=1,
count=count+1; i=If[Mod[i, 2]==0, i/2, 3*i+1]]; Print[count]]
PROG
(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)
(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
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Aranya Kumar Bal, Dec 18 2019
STATUS
approved