Anyone have an answer for me?

Frank b1 · 2025-04-30 20:09:12

I Imagine someone wi
I Imagine someone will know the name of this mathematical sequence and how it progresses the further along the sequence you take it. I have no idea about either but I would like to know.

I'll use an example of a piano keyboard to hopefully describe the sequence as best I can.

If you imagine starting on the extreme left of the keyboard and play, left to right, keys 1 and 2 it will play a short tune. Do this again and will repeat the same tune. Now play keys 1, 2 and 3 to give another (three note) tune. Now we have to go backwards to play keys 2, 1 and 2 followed by 3, 2 and 1 followed by 2, 3 and 2 followed by 1, 2 and 3 at which point you've just repeated the initial tune. So it's a continuous, rolling, back and forth sequence which finally brings you back to the tune you first played.

You will have noticed that to repeat a three note tune you had to go back and forth five times to get you back to the initial tune.

If you then apply the same continuous, rolling, back and forth method to playing a four note tune it would go like this..... 1, 2, 3, 4 then 3, 2, 1, 2 then 3, 4, 3, 2 and finally back to 1, 2, 3, 4 at which point you've just repeated the initial four note tune. But this time you only had to go back and forth four times to repeat the initial four note tune. So fewer times than you did for the three note tune.

Expanding these sequences to a five note tune will see you going back and forth nine times but for a six note tune only six times. For a seven note tune it will be fifteen times but for an eight note tune only eight times.

So, so far, it appears that for tunes with an even amount of notes you will go back and forth the same amount of times as the number of notes in the initial tune. Four times for four notes, six times for six notes, eight times for eight notes etc, etc. For a tune with an uneven amount of notes there seems to be a formula you can use to calculate how many times you will have to go back and forth to get to the point where you are back to the original tune. At these shortish tunes at least, if you multiply the amount of notes in the original tune by 2 then subtract that total by 1 you get your answer. So for a three note tune, multiply by 2 gives you 6, minus 1 gives you 5. Which is the amount of times you had to go back and forth. For a five note tune multiply by 2 gives you 10 minus 1 gives you 9 which is the amount of times you have to go back and forth to get back to the original nine note tune. I Imagine this formula might continue ad infinitum for both even and uneven numbers but I might be wrong. That would seem too simple. If, for example, if you played a forty note tune, would you only have to go back and forth forty times but for a forty-one note tune have to do so eighty-one times (41 x 2 = 82 - 1 = 81)?

If I've explained this all clearly enough and you understand my query, is there a mathematical name or formula for this and how does the formula evolve ad infinitum?

Thanks in advance for any responses.

Frank

Bob · 2025-04-30 21:32:48

hi Frank b1

Welcome to the forum.

I'd better start by saying I cannot play a piano. But we have one in the house and others can play it.

123, 212, 321, 232, 123

I'm not following how each pattern progresses to the next. Why not 312 or even 333 ?

Please would you explain how you are going from one to the next and why 212 follows from 123 for example.

Bob

Frank b1 · 2025-04-30 21:55:02

Hi Bob and thanks fo
Hi Bob and thanks for your reply.

Yes, I can understand that it might be difficult to follow the sequence initially but if you do try it out on your piano a few times I'm sure you'll see what I mean. Especially so if you listen to the first three notes and wait for them to be repeated. 212 follows 123 because you've just gone from left to right (123) then continued but by going immediately right to left and then back again (212). From there all you're doing is "rolling over" the sequence until you're back to the original three note tune. I can only suggest that you try it a few times until it makes sense. Using a piano isn't the only way to investigate this but it does have the advantage of giving you the musical notes to bring it together.

Bob · 2025-05-01 04:51:18

Ok thanks. I've got it now. And thanks for the problem. I went to the gym this afternoon; not because I enjoy it; but to keep fit. So it was great to have something to think about whilst rowing and cycling.

I'll start with the even cases; let's look at n = 8

The notes are 12345678 | 76543212 | 34567876 | and so on.

I'll call these | the bar lines. They occur every 8 notes. I noticed something important if you leave the bar lines out.

12345678765432 || 12345678765432 || 12345678765432 || 12345678765432 || .....
There's a sequence that repeats every 14 notes. I've put in lines to split up the sequences. I'll call these lines sem lines for sequence markers.

What you want is for a bar line to coincide with a sem line. (note: 14/2 = 7 and 8 are co-prime ie have only 1 as a common factor )

8 doesn't divide 14, nor 28, nor 42 but it does divide 56. So after 56 notes the bar and sem lines coincide and the bar count is 7.

In general if n is an even number the repeating sequence is 2n - 2 long. n and n-1 are coprime so the bar and sem lines coincide after n(n-1) notes. At that point n-1 bars have occurred so the pattern starts again at the nth bar. This is what you have discovered experimentally.

I haven't fully developed the odd case yet so I'll do some work on that and post again.

Bob

Bob · 2025-05-01 07:09:31

Odd cases.

I used MS WORD and lots of copy and paste to construct some evidence:

N = 3
123 bar 2 sem12 bar 32 sem1 bar 232 bar sem1232

repeats after 4 bars

N = 5
12345 bar 432 sem12 bar 34543 bar 2 sem1234 bar 5432 sem1 bar 23454 bar 32 sem 123 bar 45432 bar sem12345432

repeats after 8 bars

N = 7
1234567 bar 65432 sem12 bar 3456765 bar 432 sem1234 bar 5676543 bar 2 sem123456 bar 765432 sem1 bar 2345676 bar 5432 sem123 bar 4567654 bar 32 sem12345 bar 6765432 bar sem123456765432

repeats after 12 bars

N = 9
123456789 bar 8765432 sem12 bar 345678987 bar 65432 sem1234 bar 567898765 bar 432 sem123456 bar 789876543 bar 2 sem12345678 bar 98765432 sem1 bar 234567898 bar 765432 sem123 bar 456789876 bar 5432 sem12345 bar 678987654 bar 32 sem1234567 bar 898765432 bar sem1234567898765432

repeats after 16 bars

So the repeat starts after the point where bar and sem occur next to each other .

N = 3 at bar 5
N = 5 at bar 9
N = 7 at bar 13
N = 9 at bar 17

formula 2N - 1 which is what you have stated.

Complete proof:

If N is the bar length then new bars start every N notes.

The 1 to N and back down sequence is always 2N - 2 that's every note twice less once for each end point.

So the bars will start again at every common multiple of N and 2(N-1). But is that the first? ie Is it the lowest common multiple?

When N is even, no it isn't because N and N-1 have no common factor (except 1) and N is even so it does have a common factor with 2(N-1) namely 2. So the LCM is N(N-1) leading to a repeat bar at N-1 plus 1 ie at N.

When N is odd, N and 2(N-1) have no common factors (except 1) so the reset is at N.2(N-1) and so is after 2(N-1) bars. So the repear bar occurs at 2(N -1) plus 1 ie at 2N -1.

Bob

Frank b1 · 2025-05-02 20:56:03

Thanks Bob for all that effort. I'm glad you understood what I was trying say. The piano analogy was the best way I could think of describing the query although, of course, there were many other options. Did having a piano at your side actually help? Was that just a little bit of luck?

I normally have very little interest in anything mathematical. Number theory, equations, mathematical formulas are things I left behind me in my school days and they were a long time ago. This puzzle just came to me out of the blue when I was laying in bed and staring at the slats of the window shutters and began counting them up and down til I noticed what was happening. I needed to know a little more about it.

So, your responses seem to confirm what I was seeing but you did that using mathematics to provide you with what you, I think, call Complete Proof. A process which, I imagine, is essential to uphold the theory or the working out of the problem.

I think I still have a couple of unanswered questions though. The first I think I probably already know the answer to. And that is, as I mentioned in my first post, do these patterns continue ad infinitum? The example I mentioned was if you played a forty note tune, would you only have to go back and forth forty times but for a forty-one note tune have to do so eighty-one times (41 x 2 = 82 - 1 = 81)?

I imagine that's correct which, to me, is a simple but quite amazing fact. To know with certainty beforehand that if you played a 3,027 note tune, to get to the point of repeating the initial tune, you'd need to play the sequence through 6,053 times (3,027 x 2 = 6,054 - 1 = 6,053), is sort of "cool" as they say these days. But, I suppose, you'd need complete proof of that too to be absolutely confident that it is actually correct. So, has that ever been done?

Which brings me to my second unanswered question. Has all this been recognized before? The answer must be yes because I'm 1000 % certain I'm not the first person to have queried this over the past several thousand years of mathematical problem solving. So, if the two "phenomena" (the odd one and the even one) have been recognized before, someone must have given them names, or whatever the equivalent would be in mathematical terms. I can't imagine, after all this time, the only way to ask another person about this would be in the long-winded, analogy-finding way I just did with you and for you to then have to go through all the effort you went through to research and prove the logistics of it. Someone must have done all that a very long time ago and put it on the record books as a well identified problem with an actual formulaic solution and gave it a moniker. We two can't be the first. If so, we'd give the whole process a name so that anyone else after us would instantly know what we're talking about.

So, do you know, or are you able to find out, if there is an established term or name or written descriptive of any sort for what we've just "discovered"?

I hope you don't mind me asking and I do thank you for time. I imagine you probably found it quite interesting to work on.

Bob · 2025-05-02 22:16:59

You're welcome. I enjoyed working it out so you did me a favour by setting the problem.

My 'complete proof' is meant to cover all cases, so if I'm right it does carry on doing this as infinitum. The problem with proves is, sometimes, they have flaws in them. Andrew Wiles surprised the mathematical world by coming up with a proof of Fermat's last theorem. His proof was taken away by other mathematicians for close scrutiny. They did find some points where he had jumped ahead too fast and he went back to the drawing board to clean up the criticisms (which he did!).

The common multiple bit is fairly robust. Then I had to show the lowest common multiple. I think it's ok too but maybe others will spot a weakness. When I was at university we spent a lot of time showing even the most obvious results are generally true so I've had practice.

You have to be careful assuming results just because they work for a few easy numbers. Have a look here:

http://www.mathisfunforum.com/viewtopic.php?id=30690

Another example is the statement N^2 + N + 41 is a prime number generator. That is if you substitute N = 1, 2, 3 etc
each time you get a prime number.

If you try some N values it look promising. But the statement is FALSE. Can you find a value of N for which it fails?

Has someone already come up with your theorem? Not as far as I know. I tried an internet search and came up with nothing. So I suggest you claim it as yours and give it a fancy name. The history of maths is full of disputed claims (eg. Leibnitz and Newton over calculus) so you'll be in with a select crowd.

But now, can you pull this out of the sphere of 'pure maths' and find a useful application?

Best wishes,

Bob

Math Is Fun Forum

#1 2025-04-30 20:09:12

Anyone have an answer for me?

#2 2025-04-30 21:32:48

Re: Anyone have an answer for me?

#3 2025-04-30 21:55:02

Re: Anyone have an answer for me?

#4 2025-05-01 04:51:18

Re: Anyone have an answer for me?

#5 2025-05-01 07:09:31

Re: Anyone have an answer for me?

#6 2025-05-02 20:56:03

Re: Anyone have an answer for me?

#7 2025-05-02 22:16:59

Re: Anyone have an answer for me?

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