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#1 |
"NT"
May 2022
U.S.
19 Posts |
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I recently proposed the following sequence to the OEIS (it’s still being reviewed). I wonder if any of you have any ideas/insights into it. It goes like this: a(1)=1, a(2)=0; for n > 2, a(n) is the number of times a(n - 1 - a(n-1)) has appeared in the sequence. In other words, you look at n, which tells you how many terms to go back by, then you count how many terms there are in the sequence of the value you land on and that is your next term. So beginning with 1,0 as given, here are the first 60 or so terms: 1, 0, 1, 1, 3, 1, 1, 5, 5, 5, 1, 3, 3, 3, 6, 3, 5, 5, 5, 5, 1, 7, 1, 1, 9, 5, 9, 8, 8, 9, 9, 1, 4, 2, 10, 4, 10, 4, 1, 3, 2, 11, 4, 11, 4, 2, 2, 5, 5, 2, 10
I would conjecture that every positive integer will eventually appear. I have no idea what happens with large numbers. I’m curious about the size of the gaps between numbers, also if there’s any constant which the ratio of consecutive terms approaches to the limit. Also what happens for the same sequence starting on different numbers (like 2,0 or 3,0). Lots of questions:) Thanks in advance! (I hope this is the right forum to post to, I wasn’t sure where else it would fit better) By the way, in case someone uses MatLab here’s code someone on the OEIS wrote: (MATLAB) function a = A354971(max_n) a = [1 0]; for n = 3:max_n a = [a length(find(a == a(n-1-a(n-1))))]; end end % Thomas Scheuerle, Jun 15 2022 |
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#2 |
"Matthew Anderson"
Dec 2010
Oregon, USA
3×389 Posts |
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Wow!
That is impressive, oreotheory! I too have a comment pending at OEIS. See oeis.org/A272176/ smile :--) Mine has to do with pairs of prime numbers. best of luck with yours (I haven't read yours in detail yet.) Have a nice day. Regards, MCA Last fiddled with by MattcAnderson on 2022-06-17 at 02:52 Reason: didn't add clickable link yet. |
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#3 |
"NT"
May 2022
U.S.
100112 Posts |
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Hi Matt, Thanks for the encouragement, am curious to know your comment about the p+44 series.
It would be awesome if we had a subforum for number series. And maybe another one for visualizations of number series (since they can be beautiful and illuminating in their own right). A separate topic for cellular automata would be cool too… Best, Neal (or oreotheory) PS My series is fairly simple; the formal definition just seems intimidating, but the steps as described are really simple |
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#4 |
Jan 2021
California
1101110002 Posts |
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I don't like this series because it grows very slowly, and computation of any element requires looking at every previous element. There's no elegant way of working on part of the series or extending the series without having the entire series up to the point that it's being worked on.
Last fiddled with by slandrum on 2022-06-17 at 08:05 |
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#5 | |
"NT"
May 2022
U.S.
19 Posts |
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#6 |
"NT"
May 2022
U.S.
19 Posts |
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I admittedly posted this in the lounge and didn't get too much engagement, so I thought I'd try here. I recently proposed the following sequence to the OEIS (it’s still being reviewed). I wonder if any of you have any ideas/insights into it. It goes like this: a(1)=1, a(2)=0; for n > 2, a(n) is the number of times a(n - 1 - a(n-1)) has appeared in the sequence. In other words, you look at n, which tells you how many terms to go back by, then you count how many terms there are in the sequence of the value you land on and that is your next term. So beginning with 1,0 as given, here are the first 60 or so terms: 1, 0, 1, 1, 3, 1, 1, 5, 5, 5, 1, 3, 3, 3, 6, 3, 5, 5, 5, 5, 1, 7, 1, 1, 9, 5, 9, 8, 8, 9, 9, 1, 4, 2, 10, 4, 10, 4, 1, 3, 2, 11, 4, 11, 4, 2, 2, 5, 5, 2, 10
I would conjecture that every positive integer will eventually appear. I’m curious about the size of the gaps between numbers, also if there’s any constant by which it increases to the limit. Also what happens for the same sequence starting on different numbers (like 2,0 or 3,0). Lots of questions:) Really I'm interested in anything you might notice or come up with. Thanks in advance! |
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#7 |
"NT"
May 2022
U.S.
19 Posts |
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P.S. Here's code people at the OEIS wrote in a few different languages:
(MATLAB) function a = A354971( max_n ) a = [1 0]; c = zeros(1, max_n); c(1:2) = [1 1]; for n = 3:max_n m = a(n-1-a(n-1))+1; a = [a c(m)]; c(c(m)+1) = c(c(m)+1)+1; end end % Thomas Scheuerle, Jun 15 2022 (PARI) { nb = [0]; for (n=1, #a=vector(81), print1 (a[n] = if (n==1, 1, n==2, 0, nb[1+a[n-1-a[n-1]]])", "); if (#nb < 1+a[n], nb = concat(nb, vector(#nb))); nb[1+a[n]]++) } \\ Rémy Sigrist, Jun 18 2022 (Python) from collections import Counter from itertools import count, islice def agen(): a = [None, 1, 0]; inventory = Counter(a); yield from a[1:] for n in count(3): c = inventory[a[n-1-a[n-1]]] a.append(c); inventory.update([c]); yield c print(list(islice(agen(), 81))) # Michael S. Branicky, Jun 18 2022 |
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#8 | |
6809 > 6502
"""""""""""""""""""
Aug 2003
101×103 Posts
299416 Posts |
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BTW, I merged the threads. Last fiddled with by Uncwilly on 2022-06-25 at 04:54 |
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#9 |
"NT"
May 2022
U.S.
1316 Posts |
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Gotcha, thanks. I assume mods are the only ones who can move a thread to another forum.
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#10 |
"NT"
May 2022
U.S.
1316 Posts |
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Here's the sequence: https://oeis.org/A354971
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