
COMMENTS

A permutation of the natural numbers. Proof: Let k be the smallest number that 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 fixedpoints 49, 57, 65, 81, 85, 93, 121, 133, 153, 169, 185, 201, 209, 217, 225, 253, 261, 289, 297, ...  John W. Layman, Nov 09 2005
The trajectory of 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, Nov 09 2005
Conjecture: For all odd primes p, a(p1) = p. Equivalently, it appears that if an initial 0 is appended (the smallest divisor of 0, the zeroth triangular number), then the fixed points in this include the odd primes.  Enrique Navarrete, Jul 24 2019 [Wording of the equivalent property corrected by Peter Munn, Jul 27 2019]
From Peter Munn, Jul 27 2019: (Start)
The above conjecture is true.
For odd k, k appears by term k. Proof: choose m such that k1 <= m <= k and T(m) is odd. k is a divisor of T(m) and (by induction) all smaller odd divisors have occurred earlier, so a(m) = k if k has not occurred earlier.
For even k, k appears by term 2k1, as k divides T(2k1) and by induction all smaller divisors have occurred earlier.
For odd prime p, the first triangular number p divides is T(p1) = p*(p1)/2. But (p1)/2 and any smaller divisors have occurred by term (p1)1, so a(p1) = p.
(End)
For a generalization of the construction, see A309200.  N. J. A. Sloane, Jul 25 2019
Regarding iteration cycles, for length 2 there are many additional ones after the mentioned (2,3): (50, 75), (122, 183), (174, 203), (194, 291), (338, 507), etc.; for length 3: (1734, 4335, 2312), (4804, 6005, 8407), (7494, 18735, 9992), (8994, 10493, 13491), (12548, 18822, 21959), etc.; for length 4: (84326, 126489, 149487, 91992), (94138, 98417, 135761, 141207), (255206, 382809, 638015, 364580), (345928, 487444, 609305, 680063), (384350, 422785, 499655, 399724), etc. The trajectories of 10 and other families (14, 40, 60, 72, 78, 88, 96, etc.) are best thought of as being continuations of sequences arriving from infinity: ..., 451160, 300774, 300775, 186140, 124094, 124095, 62048, 31304, 25044, 20870, 20871, 13914, 13915, 10934, 10935, 7290, 7291, 14582, 14583, 9722, 9723, 6482, 6483, 4322, 4323, 2882, 4061, 12183, 9138, 9139, 11882, 17823, 8912, 6684, 5570, 5571, 2786, 4179, 2090, 2091, 1394, 1395, 698, 1047, 524, 350, 351, 176, 132, 114, 115, 145, 365, 915, 458, 459, 414, 415, 208, 152, 102, 103, 52, 53, 159, 80, 54, 55, 44, 45, 69, 105, 265, 371, 186, 341, 589, 1121, 1947, 1298, 1299, 866, 867, 578, 579, 290, 435, 218, 219, 146, 147, 74, 111, 56, 38, 39, 26, 27, 18, 19, 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, 49847, 28484, 35605, 89015, 74180, 74181, 101041, 210061, 8297449, ...  Hans Havermann, Jul 26 2019
