[math-fun] Hits when concatenating?
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.
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, Ãric 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, Ã.
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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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Oh, I notice that only the sequences starting with a(1) = 2 or 4 or 6 or 8 or 20 or 22,... will never produce a distinct other term. a(1) = 10 is even and will start a seq, though. Best, É.
Le 24 nov. 2020 à 15:34, Éric Angelini <eric.angelini@skynet.be> a écrit :
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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Le mar. 24 nov. 2020 à 07:15, Éric Angelini <eric.angelini@skynet.be> a écrit :
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;
As it is worded, you should count the appearance of concat(1,2) in 112 and later in 124... But I understand the concatenation must be equal to a term and not just appear in it. For this case I confirm Kris' result of a(2679) = 12369, and no other match below concat(1, 2, 3, 6, 9, 18, 19, 28, 37) = 1'236'918'192'837. (To see that it's enough to pass a(n) = 12'369'181'928, takes < 1 min with PARI code below.) The ratio a(n)/n seems to tend to some value between 4.6 and 4.7. So one would have to go up to ~ n = 2.5e13 to pass a(n) = 123'691'819'283'744 ~ 1.23...e14 and exclude a hit below a(n) = 1.23...e16. -Maximilian (PARI) lastodd(a)=until(!a\=10,bittest(a,0)&&return(L=a%10));L L=!a=0;V=vector(100,i,a+=lastodd(a)) j=c=L=!a=0;for(i=1,oo,c>(a+=lastodd(a))&&next; c==a&&print1("*** ",[i,a]" *** ");print1([c=eval(Str(c,V[j++]))])) the next possibilities would be
with 1236, 12369, 1236918,... but the margin of this post, etc. Best, É.
participants (3)
-
Kris Katterjohn -
M F Hasler -
Éric Angelini