Yeaaah, many thanks Kris! What about all the sequences starting with an odd a(1)? We would then get a sequence of « terms coming back after k steps » under this procedure. The seq T would be: 3,6,9,18,19,28,37 — a miss for 36 but what about a possible hit with 369? Or 36918? The seq U would be: 5,10,11,12,13,16,17,... we look for 510, etc. Bravo again for yr first result, Kris! Best, É.
Le 24 nov. 2020 à 15:19, Kris Katterjohn <katterjohn@gmail.com> a écrit :
Hello again,
I think I understand your procedure correctly because I can reproduce the terms 1, 2, ..., 124.
If I did this correctly then 12369 appears as the 2679th term:
1, 2, 3, 6, 9, ..., 12366, 12369, 12378, ...
Kris Katterjohn
On Tue, Nov 24, 2020 at 12:14:42PM +0100, Éric Angelini wrote: Hello Math-Fun Say we add the last odd digit of S to a(n): S=1,2,3,6,9,18,19,28,37,44,51,52,57,64,71,72, 79,88,97,104,105,110,111,112,113,116,117,124,... Say we have a hit when the concatenation of two or more initial terms of S reappear in S; will we ever find a hit here (the hit « 123 » was missed by a single unit with « 124 », in the above example; the next possibilities would be with 1236, 12369, 1236918,... but the margin of this post, etc. Best, É.
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