%I
%S 1,3,21,75,151,1365,5461,14563,87381,184111,932067,5592405,13256071,
%T 26512143,357913941,1431655765,3817748707,22906492245,91625968981,
%U 244335917283,1466015503701,5212499568715,10424999137431
%N Smallest number to give 2^(2n) in a hailstone (or 3x + 1) sequence.
%C In hailstone sequences, only even powers of 2 are obtained as a final peak before descending to 1. [I assume this should really say: "These are numbers whose 3x+1 trajectory has the property that the final peak before descending to 1 is an even power of 2."  _N. J. A. Sloane_, Jul 22 2020]
%C For n>1, this a bisection of A010120. For n=3,6,7,9,12,15,16,18,19,21, we have a(n)=(4^n1)/3, the largest possible value because one 3x+1 step produces 2^(2n).  _T. D. Noe_, Feb 19 2010
%D J. Heleen, Final Peak Sequences for Hailstone Numbers, 1993, preprint. [Apparently unpublished as of June 2017]
%F For n > 1: a(n) = A070167(A000302(n)).  _Reinhard Zumkeller_, Jan 02 2013
%e The "3x+1" sequence starting at 21 is 21, 64, 32, 16, 8, 4, 2, 1, ..., and is the smallest start which contains 64 = 2^(2*3). So a(3) = 21.  _N. J. A. Sloane_, Jul 22 2020
%o (Haskell)
%o a054646 1 = 1
%o a054646 n = a070167 $ a000302 n  _Reinhard Zumkeller_, Jan 02 2013
%K easy,nice,nonn
%O 1,2
%A _Jeff Heleen_, Apr 16 2000
