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A128333
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a(0) = 0; for n > 0, a(n) = a(n-1)/2 if that number is an integer and not already in the sequence, otherwise a(n) = 3*a(n-1) + 1.
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2
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0, 1, 4, 2, 7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 25, 76, 38, 19, 58, 29, 88, 44, 133, 400, 200, 100, 50, 151, 454, 227, 682, 341, 1024, 512, 256, 128, 64, 32, 97, 292, 146, 73, 220, 110, 55, 166, 83, 250, 125, 376, 188, 94, 47, 142, 71, 214, 107, 322, 161
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,3
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COMMENTS
| Other than a(0) = 0, the sequence misses all multiples of 3. Does it eventually hit all positive non-multiples of 3?
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LINKS
| Nick Hobson, Table of n, a(n) for n = 0..10000
Nick Hobson, Python program
Index entries for sequences related to 3x+1 (or Collatz) problem
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EXAMPLE
| Consider n = 3. We have a(3) = 2 and try to divide by 2. The result, 1, is certainly an integer, but we cannot use it because 1 is already in the sequence. So we must multiply by 3 and add 1 instead, getting a(4) = 3*2 + 1 = 7.
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CROSSREFS
| Cf. A126038, A005132.
Sequence in context: A151890 A108167 A050105 * A201414 A199814 A195347
Adjacent sequences: A128330 A128331 A128332 * A128334 A128335 A128336
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KEYWORD
| easy,nonn
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AUTHOR
| Nick Hobson (nickh(AT)qbyte.org), Feb 27 2007
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