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

1, 3, 21, 75, 151, 1365, 5461, 14563, 87381, 184111, 932067, 5592405, 13256071, 26512143, 357913941, 1431655765, 3817748707, 22906492245, 91625968981, 244335917283, 1466015503701, 5212499568715, 10424999137431

Offset: 1

Author

Jeff Heleen, Apr 16 2000

Keywords

• easy
• nice
• nonn

Comments

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]

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

Examples

			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
		

References

• J. Heleen, Final Peak Sequences for Hailstone Numbers, 1993, preprint. [Apparently unpublished as of June 2017]

Links

• Index entries for sequences related to 3x+1 (or Collatz) problem

Programs

Haskell

a054646 1 = 1
a054646 n = a070167 $ a000302 n  -- Reinhard Zumkeller, Jan 02 2013

Formula

For n > 1: a(n) = A070167(A000302(n)). - Reinhard Zumkeller, Jan 02 2013