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

%I A330591 #34 Feb 23 2020 16:57:21
%S A330591 16,21,26,101,83,83,145,145,220,158,145,207,114,114,450,114,357,357,
%T A330591 282,419,419,494,494,494,494,494,494,494,543,494,543,799,799,543,543,
%U A330591 799,543,543,799,799,791,791,791,791,861,861,861,861,998,998,998,861,861,861
%N A330591 Number of Collatz steps to reach 1 starting from 6^n + 1.
%C A330591 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 A330591 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 A330591 a(n) = A006577(6^n+1).
%e A330591 a(2)=21 because the Collatz trajectory of 6^2 + 1 = 37 is
%e A330591   37 -> 112 -> 56 -> 28 -> 14 ->  7 -> 22 ->
%e A330591   11 ->  34 -> 17 -> 52 -> 26 -> 13 -> 40 ->
%e A330591   20 ->  10 ->  5 -> 16 ->  8 ->  4 ->  2 ->  1, which is 21 steps.
%t A330591 n=200;
%t A330591 For [j=1, j<n,j++,
%t A330591   i=6^j+1;
%t A330591   count=0;
%t A330591   While[i!=1,
%t A330591     count=count+1;i=If[Mod[i,2]==0,i/2,3*i+1]];Print[count]]
%o A330591 (Python) from decimal import *
%o A330591 n=1000
%o A330591 for j in range(2,n):
%o A330591   i=6**j+1
%o A330591   count=0
%o A330591   while(i!=1):
%o A330591     if(i%2==0):
%o A330591       i=i//2
%o A330591     else:
%o A330591       i=3*i+1
%o A330591     count=count+1
%o A330591   print(count)
%o A330591 (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 A330591 a(n) = nbsteps(6^n+1); \\ _Michel Marcus_, Dec 21 2019
%Y A330591 Cf. A006577, A179118, A212653.
%K A330591 nonn,easy
%O A330591 1,1
%A A330591 _Aranya Kumar Bal_, Dec 18 2019