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The number of 3x+1 steps in the Collatz iteration of A033491(n), the least number requiring n iterations.
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%I #5 Jul 29 2013 00:50:36

%S 0,0,0,0,1,1,2,2,2,2,2,3,3,4,4,5,5,5,6,6,6,6,7,7,7,8,8,8,9,9,9,10,10,

%T 11,11,11,11,12,12,12,13,13,13,14,14,15,15,15,16,16,16,17,17,18,18,18,

%U 18,19,19,20,20,20,21,21,22,22,23,23,24,24,24,24,25,25,25,26,26,27,27,28

%N The number of 3x+1 steps in the Collatz iteration of A033491(n), the least number requiring n iterations.

%C It appears that a(n) is the maximum number of 3x+1 steps for all numbers requiring n Collatz iterations, which is row n of A127824.

%F For large n, a(n) ~ n * log(2)/log(6).

%e 24 is the smallest number that requires 10 Collatz iterations. The iteration uses two 3x+1 steps to produce 24, 12, 6, 3, 10, 5, 16, 8, 4, 2, 1. Hence a(10)=2.

%t col[n_] := NestWhileList[If[EvenQ[#], #/2, 3 # + 1] &, n, # > 1 &]; Table[k = 1; While[Length[(y = col[k])] - 1 != n, k++]; Count[y, _?OddQ] - 1, {n, 80}] (* _Jayanta Basu_, Jul 27 2013 *)

%K nonn

%O 1,7

%A _T. D. Noe_, Apr 30 2010