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A075421
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Trajectory of n under the Reverse and Add! operation carried out in base 4 (presumably) does not reach a palindrome and (presumably) does not join the trajectory of any term m < n.
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17
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290, 318, 719, 795, 799, 1210, 3903, 4199, 4207, 4219, 4236, 4278, 4279, 4294, 4326, 4333, 4334, 4338, 4402, 4598, 4662, 4726, 5046, 5357, 6157, 6174, 7246, 7247, 7295, 7407, 7549, 8063, 8191, 9211, 12319, 12431, 12463, 12539, 15487, 16519, 16587
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OFFSET
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1,1
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COMMENTS
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For 318 (cf. A075153), 266718 (cf. A075466) and 270798 (cf. A075467) one can prove that the base 4 trajectory does not contain a palindrome. A proof for 290 (cf. A075299) has not been found up to now. 4398859679359 is another known candidate (obtained from a remark of David J. Seal, cf. Links) for a term whose trajectory is provably palindrome-free, but is not secured that it does not join the trajectory of some term m < n. - If the trajectory of an integer k joins the trajectory of a smaller integer which is a term of the present sequence, then this occurs after very few Reverse and Add! steps (at most 28 for k < 20000). On the other hand, the trajectories of the terms listed above do not join the trajectory of any smaller term within at least 1000 steps.
David J. Seal (see LINKS) observed a cyclic pattern (length 6) in the trajectories that can be represented by an extended right regular grammar with production rules:
S -> S_a | S_b | S_c | S_d | S_e | S_f,
S_a -> 1033202000232 T_a, T_a -> 222 T_a | 2302333113230
S_b -> 2022321332331 T_b, T_b -> 111 T_b | 1223001203131
S_c -> 10002003002212 T_c, T_c -> 222 T_c | 3221333101333
S_d -> 103312202321111 T_d, T_d -> 111 T_d | 1102023122000
S_e -> 110200123122222 T_e, T_e -> 222 T_e | 2231232001301
S_f -> 213301021321111 T_f, T_f -> 111 T_f | 1113213003312
Within the first 471 terms of this sequence we observed three trajectories with a cyclic pattern (length 6) that can be represented by a context-free grammar with production rules:
S -> S_a | S_b | S_c | S_d | S_e | S_f,
S_a -> 10 T_a 00, T_a -> 3 T_a 0 | T_a0,
S_b -> 11 T_b 01, T_b -> 0 T_b 3 | T_b0,
S_c -> 22 T_c 12, T_c -> 0 T_c 3 | T_c0,
S_d -> 10 T_d 000, T_d -> 3 T_d 0 | T_d0,
S_e -> 11 T_e 301, T_e -> 0 T_e 3 | T_e0,
S_f -> 22 T_f 312, T_f -> 0 T_f 3 | T_f0.
The terminating strings in these context-free grammars are given by:
n 2 359 371
a(n) 318 266718 270798
T_a0 33230 33230000001033230 3323001033230
T_b0 03123 03123010001103123 0312302103123
T_c0 01313 01313120002201313 0131320201313
T_d0 33323 33323000001033323 3332300103323
T_e0 03222 03222301001103222 0322201113222
T_f0 02111 02111312002202111 0211112222111
From the fact that both, right regular grammars and context-free grammars occur, we wonder if other trajectories can be represented by context-sensitive grammars as well, by which other trajectories can be proven never to end up in a palindromic string? (End)
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LINKS
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EXAMPLE
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719 is a term since the trajectory of 719 (presumably) does not lead to an integer which occurs in the trajectory of 290 or of 318.
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MATHEMATICA
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limit = 10^3; (* Assumes that there is no palindrome if none is found before "limit" iterations *)
utraj = {};
Select[Range[0, 17000], (x = NestWhileList[# + IntegerReverse[#, 4] &, #, # !=IntegerReverse[#, 4] & , 1, limit];
If[Length[x] >= limit && Intersection[x, utraj] == {},
utraj = Union[utraj, x]; True,
utraj = Union[utraj, x]]) &] (* Robert Price, Oct 16 2019 *)
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CROSSREFS
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KEYWORD
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base,nonn
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AUTHOR
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STATUS
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approved
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