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COMMENTS
| A permutation of the natural numbers. Proof: Let k be smallest number whiuch does not appear. Let n_0 be such that by term n_0 every number < k has appeared. Let m be smallest multiple of k > n_0. Then T(2m) is divisible by k and so a(2m) = k, a contradiction.
Known cycles are: (1), (2, 3), (4, 5, 15, 8, 6, 7), (9), (16, 17, 51, 34, 35, 30, 31), (25) and {28, 29, 87, 58, 59, 118, 119, 68, 46, 47, 94, 95, 48} and the additional fixed-points 49, 57, 65, 81, 85, 93, 121, 133, 153, 169, 185, 201, 209, 217, 225, 253, 261, 289, 297, ... - John W. Layman (layman(AT)math.vt.edu), Nov 09 2005
The cycle containing 10 begins {10, 11, 22, 23, 12, 13, 91, 161, 189, 285, 429, 473, 869, 957, 1437, 2157, 3237, 4857, 7287, 4164, 3470, 4511, 2256, 1464, 1172, 782, 783, 392, 294, 413, 531, 342, 343, 172, 173, 519, 346, 347, 694, 1735, 1388, 926, 927, 464, 248, 166, 167, 84, 70, 71, 36, 37, 703, 352, 353, 1059, 706, 2471, 1412, 1413, 2121, 7427, 6366, 6367, 3184, 1820, 1214, 1215, 608, 336, 337, 4381, 28483, ...) and cannot be further determined without calculating at least the first 28483 terms of {a(n)}. - John W. Layman (layman(AT)math.vt.edu), Nov 09 2005
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