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A054646 Smallest number to give 2^(2n) in a hailstone (or 3x + 1) sequence. 6

%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^n-1)/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 [Apparently unpublished as of June 2017]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

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Last modified June 14 04:16 EDT 2021. Contains 345018 sequences. (Running on oeis4.)