A128333 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.
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
Offset: 0
Examples
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.
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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