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A054646
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Smallest number to give 2^(2n) in a hailstone (or 3x + 1) sequence.
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6
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1, 3, 21, 75, 151, 1365, 5461, 14563, 87381, 184111, 932067, 5592405, 13256071, 26512143, 357913941, 1431655765, 3817748707, 22906492245, 91625968981, 244335917283, 1466015503701, 5212499568715, 10424999137431
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OFFSET
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1,2
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COMMENTS
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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
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REFERENCES
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J. Heleen, Final Peak Sequences for Hailstone Numbers, 1993, preprint. [Apparently unpublished as of June 2017]
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LINKS
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FORMULA
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EXAMPLE
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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
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PROG
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(Haskell)
a054646 1 = 1
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CROSSREFS
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KEYWORD
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easy,nice,nonn
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AUTHOR
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STATUS
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approved
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