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%I #23 Mar 04 2020 08:59:13
%S 0,0,7,8,16,19,9,17,20,20,7,10,23,111,18,26,21,21,34,8,29,16,11,24,
%T 112,112,32,19,107,27,14,22,115,14,35,35,22,9,30,17,17,12,118,25,25,
%U 38,113,113,69,33,33,20,20,46,108,46,121,28,28,41,15,15,23,116,116
%N Number of halving and tripling steps to reach 1 from the n-th pure hailstone number in the '3x+1' problem.
%C Previous name was: A006577(k), k = pure hailstone numbers from A061641.
%C Impure hailstone numbers are those which occur in the trajectories of smaller numbers. Thus 5 is impure since it occurs in the trajectory of 3.
%H Amiram Eldar, <a href="/A127933/b127933.txt">Table of n, a(n) for n = 1..10000</a>
%H Douglas J. Shaw, <a href="http://www.fq.math.ca/Papers1/44-3/quartshaw03_2006.pdf">The Pure Numbers Generated by the Collatz Sequence</a>, The Fibonacci Quarterly, Vol. 44, Number 3, August 2006, p. 194.
%F a(n) = A006577(A061641(n)).
%e 3 is the 3rd term of A061641, the trajectory of 3 has 7 terms: (10, 5, 16, 8, 4, 2, 1), so a(3) = 7.
%Y Cf. A006577, A061641.
%K nonn
%O 1,3
%A _Gary W. Adamson_, Feb 08 2007
%E Offset corrected and more terms added by _Amiram Eldar_, Feb 28 2020
%E Name edited by _Michel Marcus_, Feb 28 2020
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