Thanks for putting "Pisano Period" in the description; that brings up a Wikipedia article having more I wanted to know about them. _[edit: This led me into something I've been exploring but didn't realize had a name: "carryless arithmetic" ... really really appreciate this video!]_
Thank you again for allowing me to join your class . will you be able to show videos on all of the fibonacci sequences. I cant stop watching your videos. I dont watch many , but your videos i can replay over and over again.❤❤❤❤❤❤❤❤❤❤❤😊😊😊😊😊😊😊😊😊😊😊😊😊
For real. I studied engineering in school because I liked math, but I was more a physics and chemistry kinda guy. But now, I'm legitimately thinking, "maybe I should start teaching myself more math"
Here's a suggestion for a future video: One over 24th Prime number equals the Fibonacci Sequence overlayed with itself with each concecutive digit being offset by one; as in continuously subtracting the next fibonacci number from 1/89 while moving over one digit ends up zeroing out the result to infinite digits.
This can be extended out beyond the actual Finonacci numbers to explore how any pair of start numbers lead to a pattern of the last digits. In base ten you actually get 6 different chains. The first is the 60 digit chain in the video, 011235831459437077415617853819099875279651673033695493257291. Then there is a 20 digit chain of all even numbers, 02246066280886404482. There's also a 12 digit chain, 134718976392, a 4 digit chain, 2684, and a 3 digit chain, 055. The final chain is the null chain, which could be represented by a single zero, but takes in a single addition of last digits (0+0) and therefore maybe should be represented as 00, but then the Fibonacci process means that we need the first two digits to add to the third so maybe it needs to be represent as 000, except that we wrap the last addition of digits around from the end to the start in all other examples so maybe 00 or even 0 is an ok representation.
This is the first time in decades I've ever seen anyone else mention these. I remember "discovering" them in my tweens and then never knowing how to look them up / what they were. Thanks for noting this 😊
@@oddlyspecificmath I feel like I recognise your user name, as if I've maybe seen videos by you or something, but your channel has no content, so maybe I'm mixing you up with someone with a similar username. As for these chains I think I have seen them covered somewhere else in a youtube video or something, but again I could be mistaken.
And, unsurpringly, if we add the periods of these 6 chains 60+20+12+4+3+1, we get the magic number 100. Wonderful! (A good reason to call the 000 chain as having one zero.😉)
@@landsgevaer Every 2-digit number is represented in the chains above; my (too young to really understand) method was to tally + cross out all the pairs in case I'd missed a pattern.
@@oddlyspecificmath Yes, excellent. 👍 The periods of all cycles of the last n digits in base b must add to b^(2n). Which implies no individual cycle can be bigger than that, as explained in the video.
There was an episode of MathNet that had tiles in a mosaic encoded with mod 5 (maybe 6?) Fibonacci numbers. It has caused me to play with this concept ever since. Glad to see I'm not the only one poking at it!
Just when I thought I knew everything about Fibonacci….. brilliant as always. And you’ve just written my Friday lesson for my top set maths class. Thanks Domotro! 😃👍
This class is much nicer than the one I had in school because of the cool enthusiastic teacher and the lack of homework. Although, I wouldn't mind solving something for a change.
The video proves that any starting numbers must eventually end in a cycle. It is also true though that those starting numbers must themselves be part of that cycle. (The reason being that the Fibonacci sequence is invertible, i.e. you can uniquely calculate it backwards too, as opposed to for instance the Collatz sequence.)
When Fib(n) / Fib(n-1) approaches Φ ... and the reappearance of the Fib() sequence in the tail later on _also_ approaches Φ (at least that far down) ... does this somehow mean that Φ is composed of overlapping versions of itself?
I've previously thought about the mod 10 digit cycles, and discovered some pretty interesting things: As you mentioned, the last digits of Fibonacci numbers cycle with a period of 60. Let's start with the numbers 0 and 1, and write the last digits of every number in the cycle: 011235831459437 077415617853819 099875279651673 033695493257291 Notice the way I wrote it: four lines, each of which consist of 15 numbers. These four lines are kind of "sub-cycles" in a way, because they all show a similar pattern: they all start with a zero, and they feature a five in the sixth and eleventh positions. For the other numbers, it's almost like they've been put through a cipher for every sub-cycle, where every digit except 0 and 5 transforms into something else. 1 becomes 7, 2 becomes 4, 3 becomes 1, 4 becomes 8, 6 becomes 2, 7 becomes 9, 8 becomes 6, and 9 becomes 3. Generally, each digit becomes 7 multiplied by itself, modulo 10, which makes sense considering how the "sub-cycles" started: whenever there gets to be a 0 in the cycle, whatever number precedes it becomes almost a "seed" of sorts, starting a new cycle by appearing twice. However, this cycle does not tell the full story: let's look at every 2-digit number that appears in this cycle, allowing them to start with a 0. For instance, the first few would be 01, 11, 12, 23, etc. Note that numbers like 70 and 10 are included in this list: 70 just spans across two lines (which are just visual aids to emphasize the sub-cycles), and 10 spans from the end of one cycle to the beginning of the next. How many of these numbers are there? Well, every number in this cycle only depends on the two numbers before it, so if a two-digit number repeats, we have necessarily completed the cycle. Thus, because the cycle has length 60, there are exactly 60 two-digit numbers in the cycle. That means that there are 40 that we don't have! For instance, we will never see the digit string "22", corresponding to two consecutive Fibonacci numbers ending with a 2, and this should make sense: you showed in the video that the parities of the Fibonacci numbers repeat in an "even-odd-odd" pattern, making two consecutive even Fibonacci numbers impossible. But what if we start a chain at 22 and go from there? Or actually, let's start a chain at 02, to match our 01 chain from earlier. We'll get 22 as our next two-digit number anyway. When we do this, we get: 02246 06628 08864 04482 Here, we see a cycle of 20, made of four sub-cycles of length 5. Because we started with two even numbers, all of the numbers in this cycle are even, so I like to call this one "the even cycle". (I call the first one we discovered "the normal cycle"). We again see a sort of "cipher" pattern: to get from one "sub-cycle" to the next, turn all of the 2s into 6s, the 4s into 2s, the 6s into 8s, and the 8s into 4s. This "cipher" corresponds to a multiplication by 3. This cycle actually has the same structure as the cycle you get by taking the last digits of the Fibonacci numbers in base 5: you can think of it as multiplying each digit in the original chain by 2, and multiplying by 2 in mod 10 collapses numbers that are congruent modulo 5 into the name number. Instead of multiplying by 2, we can multiply the normal chain by other numbers, but multiplying by 3, 7, or 9 just results in a phase-shifted version of the normal chain, and multiplying by 4, 6, or 8 just results in a phase-shifted version of the even chain. The only other numbers mod 10 that provide unique chains are 5 and 0, but their chains are considerably more boring: the chain for 5 is 055 (I call it "the five chain".) and the chain for 0 is 0 (I call it "the zero chain".) Much like how the even chain corresponded to taking the Fibonacci numbers mod 5, the five chain corresponds to taking the Fibonacci numbers mod 2, and here it's easier to see: the Fibonacci numbers' parities cycle in an "even-odd-odd" cycle, where the "even" corresponds to the 0 in the five chain and the two "odd"s correspond to the two 5s. I guess, by analogy, the zero chain would correspond to the Fibonacci numbers modulo 1? Multiplying by 0 is the same as multiplying by 10, and every integer modulo 1 is 0, so it makes enough sense. However, all of these chains we've talked about so far have a combined length of 84, meaning there are still 16 two-digit numbers we can construct that haven't appeared in any chain so far! To find these chains, let's look back at the normal chain for a second. 01123 58314 59437 07741 56178 53819 09987 52796 51673 03369 54932 57291 I've added spaces every five numbers to better show some of the patterns. Let's have a look back at the 0s and 5s in this chain. We recognized earlier that every subchain started with a 0, and had a 5 as its sixth and eleventh digits, so looking at only every fifth digit (starting from the first), we get "055055055055..." This is the five chain again! So the five chain is sort of contained within the normal chain, because if you take every fifth digit from the normal chain, you get the five chain! This begs the question: What if we take every fifth digit starting from a different point in the chain? Let's try it out starting from the first 1: 189 763 921 347 Surprisingly, this sequence of numbers follows the rules of Fibonacci: every digit is the sum of the two previous ones in the cycle modulo 10. More than that, it's a completely new chain, this time of length 12. I like to call this one "the Lucas chain", because it shows up when dealing with Lucas numbers (a sequence like the Fibonacci numbers but the first two numbers are 2 and 1, and the normal Fibonacci rule applies from there, so the first few Lucas numbers are 2, 1, 3, 4, 7, 11, 18, ...). The "21" in the middle of the chain corresponds to the first two Lucas numbers. We obtained this chain through this "take every fifth number" process, which I like to call "compression". Let's see what else we can do with compression. If we compress the even chain, we get: 2684 I call this one "the even Lucas chain" because it's sort of a combination of the even chain and the Lucas chain. You can obtain it as I just did by taking the even chain and compressing it, or you can take the Lucas chain and multiply it by 2. It's a short chain, consisting of just four elements. And, with that, we have successfully found all of the chains in base 10. The combined lengths of these chains is 100, meaning that we have accounted for every two-digit combination, and every one of them will fall into either the normal chain, the even chain, the five chain, the zero chain, the Lucas chain, or the even Lucas chain. Any multiplications or compressions on any of these chains will just yield chains we've talked about previously: for instance, multiplying the Lucas chain by five just yields the normal five chain, and compressing the Lucas chain just yields a phase-shifted version of the Lucas chain. I haven't analyzed what would happen if you tried to do this procedure in other bases, but there would almost certainly be different structures, but they would probably be similar to the ones we saw in base 10. (Edit: The fives don't occur at the fifth and tenth positions, they occur at the sixth and eleventh positions.)
The 60 mod 10 digits in the Fibonacci cycle sum to 280. I thought I'd look at other near relation sequences. In the Jacobsthal sequence (OEIS A001045), with the recursion An = An-1 + 2(An-2) differing from the Fibonacci An = An-1 + 1(An-2) , the cycle is only 4 digits long, with 1 1 3 5 summing to 10. In the sequence An = An-1 + 3(An-2), OEIS A006130, the cycle is 24 digits long, splitting up into 4 sub-cycles similar to the ones you saw in Fibonacci, and sums to exactly 100. For An = An-1 + 4(An-2), OEIS A006131, the cycle is short again at 1 1 5 9 9 5 summing to 30. The only pattern I can see so far is that these sums are multiples of 10, and that the last three sums themselves sum to 140, half the Fibonacci 280.
So the last digits of the Fibonacci sequence of numbers themselves follow a sequence of 60 before repeating, at least in base ten. I just looked at the so-called Jacobstahl sequence, OEIS A001045, which only has such a cycle of 4, namely 1 1 3 5 before repeating. How do the differences in the recursions account for this? They are An = An-1 + 2(An-2) for the Jacobsthal, while that of the Fibonacci is An = An-1 + (1)An-2.
I wonder what cool properties these Fibonacci-like sequences have addition (standard) -> n = (n-1) + (n-2) subtraction -> n = (n-1) - (n-2) complex -> n = (n-1) + (n-2)i Eisenstein -> n = (n-1) + (n-2)w
I looked into subtraction, n = (n-1) - (n-2). Starting with 0 1 it goes 0 1 1 0 -1 -1 0 1 1 0 -1 -1 0 1 1 0 ... which doesn't seem to get us anywhere interesting. But if we move on to the Pell sequence, n = 𝟮(n-1) - (n-2), it goes 0 1 2 3 4 5 6 ... In general I think subtractive sequences are cooler than their familiar additive versions. What have you found?
@@chrisg3030 finding the ratio between values for the eisenstein int, there may be a bug in my code.. it seems to stay pretty small, but sometimes randomly is very large
Thank you, Combo Class. You make education fun, and you also make jokes from time to time, unlike some teachers who just go, "And then blah is equal to b+a+h-g³/√π÷5.4" . Keep doing rhe great work, and you will be one of the top Education Channels! Remember me when you get to the top.
i thought this would come down to the fibonacci sequence as a whole re-repeating in its last digtis at some point... (is there some known sequence of that type?)
I had made a previous episode about base phi so didn’t want to overlap on too much of the same material here. But yeah that’s an interesting pattern I might show in a bonus video sometime
@@ComboClass Yeah, I was doing the conversion for phinary, and it also occurred to me: If the “ones” place represents the radix^0, hence the *center* of the logarithmic scale, why does the radix point come *after* the “ones” place? So I’ve started using diacritics under the “ones” place in my notes. This is what I have for the first 7: 0, 1, 1, 10̥01, 100̥01, 1000̥1001, 10001̥0001. I got those digits from a list online though, so I’m not 100% on the accuracy. I gotta just bite the bullet and script something in Python to generate the digits.
I noted on a title card later in the episode that it’s been proven that 6n is also a limit for the length any mod n can have it’s cycle. So, checking the ones smaller than that limit, no bases have a period of exactly b^2
Smaller bases have fewer options in a given number of digits, so there is almost certainly some correlation. Haven't done the math to prove it, though.
There are so many patterns in this, that I don’t even know where to start… I’m playing with a bunch of Python code checking out cycle lengths, starting numbers, etc
This is interesting because if because of the birthday paradox, if the fibonacci sequence was random, it would repeat after only 14.142 number... yet it goes up to 60. It's like the Fibonacci sequence was anti-random. It's even more amazing when you think that some sequences cannot happen. like 00.
Couldn't you say 0 0 does generate a sequence if you use the Fibonacci rule of adding each pair of numbers to get a third. It's just in this case you always get a 0, so the cycle repeats after just 1 term instead of the 60 you get when starting 0 1.
I've only gotten up to base 24 so far, but I noticed that there are only certain bases where the last digit representation in said base has a periodic repeating cycle which begins and ends with zero *and has no other zeros in between*. So far the only bases 2-24 with this property are 2, 4, 11, and 22. Whyyyyyy? Aka what causes some bases to have single periods between last digit zeros and others to have multiple cycles with zeros in between before going back to 1, 1..
Another really cool thing is that the short cycle period between last digit zeros (which is also the whole cycle sometimes) is always equal to the index number of the first Fibonacci which is evenly divisible by the base n. Examples: In Base 2, every 3rd Fibonacci number ends in 0, and F(3)=2 is the first F number divisible by the Base number, 2. In Base 10, every 15th F number ends in a zero, and F(15) = 610 is the first one which is evenly divisible by 10.
Following similarly, the long-cycle period n (where it hits zero and then starts over) is always a multiple of the short periods length (given that there are a short and long period, rather than a single repeating string between zeros)
So in Base 10 for example again.. the short period between zeros is 15, and F(15) is the first one divisible by 10, and the long period between zeros is 60, and F(60) is the 4th F number divisible by 10.
But why on Earth do some bases only have single periods, and others have short and long ones? And why are the first four bases with single period strings base 2, 4, 11, and 22? And why do some bases with both short and long periods have a long period which is double the short period, while others have a long period which is a greater than 2 multiple of the short period ? All these questions and more. Hahha.
@@stickfiftyfive Have you considered looking at what happens in each base with all possible numbers rather than just those within the usual Fibonacci sequence. For example in base ten there are 6 separate sequences of lengths 60, 20, 12, 4, 3 and 1 that between them cover all 100 possible additions of two last digits. Of those the 60 digits sequence has four zeros in it overall, or three between the start of each cycle, and the 20 digit sequence also has four zeros, or three between the start of each cycle. The 12 digit and 4 digit sequences don't have any zeros at all, not even at the start/end of each cycle, and the 3 digit sequence has only the one zero, or no zeros between the start of each sequence. Finally the one digit sequence is literally just the digit zero. Maybe there are further patterns in this if it is extrapolated accross other bases?
@@ComboClass I keep hearing this about bases 10 and 12, how 12 is better because of its divisors. Maybe maybe, but 10 has a cool decomposition: 1+2+3+4 = 10. This allows you to represent quantities using a stone or whatever on a cross-hair grid ( big + sign). Top right is 1, bottom right 2, bottom left 3, top left 4. Keep it there and put another stone in the top right, that's 5. Keep going like that and when all four quadrants are occupied by one stone each, that's 10. Mind you 12 has a cool decomposition: 1+2+3+3+2+1. This is what allows a 12 hour clock to log rise and fall in tidal speed as well as level.
I don’t think I’m mixing them up. I used mod 12 as an example of what the last digits of base 12 do, since (like I’ve shown in previous episodes) the patterns are the same
But that's only half of the story. The cycle does truly repeat beginning at F(300). You can verify that by comparing F(151) mod 100 with F(301) mod 100.
Hey Domotro, I'm a mathematician & I would like to share some of my research regarding different sizes of of infinity. I am not joking when I say I have discovered a 3rd size of infinity but my research goes far beyond that. At some point I am able to prove that the current definition of a real number is wrong and that imaginary numbers must be a subset to real numbers. I really want to share more with you about this but I have a feeling your the right person to discuss this with.
I learned the Fibonacci sequence very early in my development; but it took a very long time before I discovered more than just the basic concept. It's always nice to find interesting Fibonacci patterns that are instructive for young people.
At this point in time and history I think we got to ask: what are YOUR opinions on the industrial revolution and its consequences for the human race? xD
Thanks for watching! Check the description for more cool links and stuff :)
Thanks for putting "Pisano Period" in the description; that brings up a Wikipedia article having more I wanted to know about them. _[edit: This led me into something I've been exploring but didn't realize had a name: "carryless arithmetic" ... really really appreciate this video!]_
Hi, I would like you to review my mathematical research; I've made some big discoveries.
Youre amazing when areyou going post your video on triangles with the fibonacci sequence anxious and excited.
Thank you again for allowing me to join your class . will you be able to show videos on all of the fibonacci sequences. I cant stop watching your videos. I dont watch many , but your videos i can replay over and over again.❤❤❤❤❤❤❤❤❤❤❤😊😊😊😊😊😊😊😊😊😊😊😊😊
Nothing is more interesting than somebody who is has tons of energy to talk about mathematics in realms beyond what they teach in school.
For real. I studied engineering in school because I liked math, but I was more a physics and chemistry kinda guy. But now, I'm legitimately thinking, "maybe I should start teaching myself more math"
look at the 25th and 26 th number it begins than see my awkward Work With Numbers Letters Shapes Dots and Dashes
Here's a suggestion for a future video: One over 24th Prime number equals the Fibonacci Sequence overlayed with itself with each concecutive digit being offset by one; as in continuously subtracting the next fibonacci number from 1/89 while moving over one digit ends up zeroing out the result to infinite digits.
4:40 I love hearing the huge clock in the background here!
your feelings are irrational
This can be extended out beyond the actual Finonacci numbers to explore how any pair of start numbers lead to a pattern of the last digits. In base ten you actually get 6 different chains. The first is the 60 digit chain in the video, 011235831459437077415617853819099875279651673033695493257291. Then there is a 20 digit chain of all even numbers, 02246066280886404482. There's also a 12 digit chain, 134718976392, a 4 digit chain, 2684, and a 3 digit chain, 055. The final chain is the null chain, which could be represented by a single zero, but takes in a single addition of last digits (0+0) and therefore maybe should be represented as 00, but then the Fibonacci process means that we need the first two digits to add to the third so maybe it needs to be represent as 000, except that we wrap the last addition of digits around from the end to the start in all other examples so maybe 00 or even 0 is an ok representation.
This is the first time in decades I've ever seen anyone else mention these. I remember "discovering" them in my tweens and then never knowing how to look them up / what they were. Thanks for noting this 😊
@@oddlyspecificmath I feel like I recognise your user name, as if I've maybe seen videos by you or something, but your channel has no content, so maybe I'm mixing you up with someone with a similar username.
As for these chains I think I have seen them covered somewhere else in a youtube video or something, but again I could be mistaken.
And, unsurpringly, if we add the periods of these 6 chains 60+20+12+4+3+1, we get the magic number 100.
Wonderful!
(A good reason to call the 000 chain as having one zero.😉)
@@landsgevaer Every 2-digit number is represented in the chains above; my (too young to really understand) method was to tally + cross out all the pairs in case I'd missed a pattern.
@@oddlyspecificmath Yes, excellent. 👍
The periods of all cycles of the last n digits in base b must add to b^(2n). Which implies no individual cycle can be bigger than that, as explained in the video.
There was an episode of MathNet that had tiles in a mosaic encoded with mod 5 (maybe 6?) Fibonacci numbers. It has caused me to play with this concept ever since. Glad to see I'm not the only one poking at it!
Just when I thought I knew everything about Fibonacci….. brilliant as always. And you’ve just written my Friday lesson for my top set maths class. Thanks Domotro! 😃👍
Hated math in school. You make this stuff make sense. I love it. Keep it going!
your feelings are irrational
This class is much nicer than the one I had in school because of the cool enthusiastic teacher and the lack of homework. Although, I wouldn't mind solving something for a change.
Did you see the final exam episode from last season?
The video proves that any starting numbers must eventually end in a cycle.
It is also true though that those starting numbers must themselves be part of that cycle.
(The reason being that the Fibonacci sequence is invertible, i.e. you can uniquely calculate it backwards too, as opposed to for instance the Collatz sequence.)
look at the 25th and 26 th number it begins than see my awkward Work With Numbers Letters Shapes Dots and Dashes
When Fib(n) / Fib(n-1) approaches Φ ... and the reappearance of the Fib() sequence in the tail later on _also_ approaches Φ (at least that far down) ... does this somehow mean that Φ is composed of overlapping versions of itself?
I've previously thought about the mod 10 digit cycles, and discovered some pretty interesting things:
As you mentioned, the last digits of Fibonacci numbers cycle with a period of 60. Let's start with the numbers 0 and 1, and write the last digits of every number in the cycle:
011235831459437
077415617853819
099875279651673
033695493257291
Notice the way I wrote it: four lines, each of which consist of 15 numbers. These four lines are kind of "sub-cycles" in a way, because they all show a similar pattern: they all start with a zero, and they feature a five in the sixth and eleventh positions. For the other numbers, it's almost like they've been put through a cipher for every sub-cycle, where every digit except 0 and 5 transforms into something else. 1 becomes 7, 2 becomes 4, 3 becomes 1, 4 becomes 8, 6 becomes 2, 7 becomes 9, 8 becomes 6, and 9 becomes 3. Generally, each digit becomes 7 multiplied by itself, modulo 10, which makes sense considering how the "sub-cycles" started: whenever there gets to be a 0 in the cycle, whatever number precedes it becomes almost a "seed" of sorts, starting a new cycle by appearing twice.
However, this cycle does not tell the full story: let's look at every 2-digit number that appears in this cycle, allowing them to start with a 0. For instance, the first few would be 01, 11, 12, 23, etc. Note that numbers like 70 and 10 are included in this list: 70 just spans across two lines (which are just visual aids to emphasize the sub-cycles), and 10 spans from the end of one cycle to the beginning of the next. How many of these numbers are there? Well, every number in this cycle only depends on the two numbers before it, so if a two-digit number repeats, we have necessarily completed the cycle. Thus, because the cycle has length 60, there are exactly 60 two-digit numbers in the cycle. That means that there are 40 that we don't have! For instance, we will never see the digit string "22", corresponding to two consecutive Fibonacci numbers ending with a 2, and this should make sense: you showed in the video that the parities of the Fibonacci numbers repeat in an "even-odd-odd" pattern, making two consecutive even Fibonacci numbers impossible. But what if we start a chain at 22 and go from there? Or actually, let's start a chain at 02, to match our 01 chain from earlier. We'll get 22 as our next two-digit number anyway. When we do this, we get:
02246
06628
08864
04482
Here, we see a cycle of 20, made of four sub-cycles of length 5. Because we started with two even numbers, all of the numbers in this cycle are even, so I like to call this one "the even cycle". (I call the first one we discovered "the normal cycle"). We again see a sort of "cipher" pattern: to get from one "sub-cycle" to the next, turn all of the 2s into 6s, the 4s into 2s, the 6s into 8s, and the 8s into 4s. This "cipher" corresponds to a multiplication by 3. This cycle actually has the same structure as the cycle you get by taking the last digits of the Fibonacci numbers in base 5: you can think of it as multiplying each digit in the original chain by 2, and multiplying by 2 in mod 10 collapses numbers that are congruent modulo 5 into the name number.
Instead of multiplying by 2, we can multiply the normal chain by other numbers, but multiplying by 3, 7, or 9 just results in a phase-shifted version of the normal chain, and multiplying by 4, 6, or 8 just results in a phase-shifted version of the even chain. The only other numbers mod 10 that provide unique chains are 5 and 0, but their chains are considerably more boring: the chain for 5 is
055 (I call it "the five chain".)
and the chain for 0 is
0 (I call it "the zero chain".)
Much like how the even chain corresponded to taking the Fibonacci numbers mod 5, the five chain corresponds to taking the Fibonacci numbers mod 2, and here it's easier to see: the Fibonacci numbers' parities cycle in an "even-odd-odd" cycle, where the "even" corresponds to the 0 in the five chain and the two "odd"s correspond to the two 5s. I guess, by analogy, the zero chain would correspond to the Fibonacci numbers modulo 1? Multiplying by 0 is the same as multiplying by 10, and every integer modulo 1 is 0, so it makes enough sense.
However, all of these chains we've talked about so far have a combined length of 84, meaning there are still 16 two-digit numbers we can construct that haven't appeared in any chain so far! To find these chains, let's look back at the normal chain for a second.
01123 58314 59437
07741 56178 53819
09987 52796 51673
03369 54932 57291
I've added spaces every five numbers to better show some of the patterns. Let's have a look back at the 0s and 5s in this chain. We recognized earlier that every subchain started with a 0, and had a 5 as its sixth and eleventh digits, so looking at only every fifth digit (starting from the first), we get "055055055055..." This is the five chain again! So the five chain is sort of contained within the normal chain, because if you take every fifth digit from the normal chain, you get the five chain! This begs the question: What if we take every fifth digit starting from a different point in the chain? Let's try it out starting from the first 1:
189
763
921
347
Surprisingly, this sequence of numbers follows the rules of Fibonacci: every digit is the sum of the two previous ones in the cycle modulo 10. More than that, it's a completely new chain, this time of length 12. I like to call this one "the Lucas chain", because it shows up when dealing with Lucas numbers (a sequence like the Fibonacci numbers but the first two numbers are 2 and 1, and the normal Fibonacci rule applies from there, so the first few Lucas numbers are 2, 1, 3, 4, 7, 11, 18, ...). The "21" in the middle of the chain corresponds to the first two Lucas numbers. We obtained this chain through this "take every fifth number" process, which I like to call "compression".
Let's see what else we can do with compression. If we compress the even chain, we get:
2684
I call this one "the even Lucas chain" because it's sort of a combination of the even chain and the Lucas chain. You can obtain it as I just did by taking the even chain and compressing it, or you can take the Lucas chain and multiply it by 2. It's a short chain, consisting of just four elements.
And, with that, we have successfully found all of the chains in base 10. The combined lengths of these chains is 100, meaning that we have accounted for every two-digit combination, and every one of them will fall into either the normal chain, the even chain, the five chain, the zero chain, the Lucas chain, or the even Lucas chain. Any multiplications or compressions on any of these chains will just yield chains we've talked about previously: for instance, multiplying the Lucas chain by five just yields the normal five chain, and compressing the Lucas chain just yields a phase-shifted version of the Lucas chain. I haven't analyzed what would happen if you tried to do this procedure in other bases, but there would almost certainly be different structures, but they would probably be similar to the ones we saw in base 10.
(Edit: The fives don't occur at the fifth and tenth positions, they occur at the sixth and eleventh positions.)
Thanks for sharing, that seems very cool and I will read through that more thoroughly later!
Also you get the award for longest TH-cam comment in Combo Class history I'm pretty sure
The 60 mod 10 digits in the Fibonacci cycle sum to 280. I thought I'd look at other near relation sequences.
In the Jacobsthal sequence (OEIS A001045), with the recursion An = An-1 + 2(An-2) differing from the Fibonacci An = An-1 + 1(An-2) , the cycle is only 4 digits long, with 1 1 3 5 summing to 10.
In the sequence An = An-1 + 3(An-2), OEIS A006130, the cycle is 24 digits long, splitting up into 4 sub-cycles similar to the ones you saw in Fibonacci, and sums to exactly 100.
For An = An-1 + 4(An-2), OEIS A006131, the cycle is short again at 1 1 5 9 9 5 summing to 30.
The only pattern I can see so far is that these sums are multiples of 10, and that the last three sums themselves sum to 140, half the Fibonacci 280.
So the last digits of the Fibonacci sequence of numbers themselves follow a sequence of 60 before repeating, at least in base ten. I just looked at the so-called Jacobstahl sequence, OEIS A001045, which only has such a cycle of 4, namely 1 1 3 5 before repeating. How do the differences in the recursions account for this? They are An = An-1 + 2(An-2) for the Jacobsthal, while that of the Fibonacci is An = An-1 + (1)An-2.
I wonder what cool properties these Fibonacci-like sequences have
addition (standard) -> n = (n-1) + (n-2)
subtraction -> n = (n-1) - (n-2)
complex -> n = (n-1) + (n-2)i
Eisenstein -> n = (n-1) + (n-2)w
I looked into subtraction, n = (n-1) - (n-2). Starting with 0 1 it goes 0 1 1 0 -1 -1 0 1 1 0 -1 -1 0 1 1 0 ... which doesn't seem to get us anywhere interesting. But if we move on to the Pell sequence, n = 𝟮(n-1) - (n-2), it goes 0 1 2 3 4 5 6 ...
In general I think subtractive sequences are cooler than their familiar additive versions. What have you found?
The complex like ones spin around and diverge as you might expect, I'm not sure about any patterns in the actually numbers
@@chrisg3030 finding the ratio between values for the eisenstein int, there may be a bug in my code.. it seems to stay pretty small, but sometimes randomly is very large
not enough floating point
Well done sir.
The Fibonacci numbers (golden ratio) and the harmonic series (music), are the fingerprints of God, perfectly designed.
I like your passion keep it alive!
Thank you, Combo Class. You make education fun, and you also make jokes from time to time, unlike some teachers who just go, "And then blah is equal to b+a+h-g³/√π÷5.4" . Keep doing rhe great work, and you will be one of the top Education Channels! Remember me when you get to the top.
thank you for making your shorts account otherwise I dont think id have ever found your full channel
i thought this would come down to the fibonacci sequence as a whole re-repeating in its last digtis at some point... (is there some known sequence of that type?)
Big D slangin' those digits
wow! the pattern is so intereting. it is inspiring me so much. !'d better look at closely fibonacci numbers. thanks😁
Love these videos. Subbed.
wonderful class , many thanks!
wow, very nice! I can't imagine how much time you put into this video. Thank you!
Grandfather Clock feature 4:38
The digit sums of the Fibonacci numbers form a sequence that repeats every 24 Fibonacci numbers.
look at the 25th and 26 th number repeat than see my awkward Work With Numbers Letters Shapes Dots and Dashes
Very interesting, but sorely disappointed you didn’t explore the patterns of Fibonacci numbers in phinary
I had made a previous episode about base phi so didn’t want to overlap on too much of the same material here. But yeah that’s an interesting pattern I might show in a bonus video sometime
@@ComboClass Yeah, I was doing the conversion for phinary, and it also occurred to me: If the “ones” place represents the radix^0, hence the *center* of the logarithmic scale, why does the radix point come *after* the “ones” place? So I’ve started using diacritics under the “ones” place in my notes.
This is what I have for the first 7: 0, 1, 1, 10̥01, 100̥01, 1000̥1001, 10001̥0001. I got those digits from a list online though, so I’m not 100% on the accuracy. I gotta just bite the bullet and script something in Python to generate the digits.
Are there any bases that repeat every b^2?
I noted on a title card later in the episode that it’s been proven that 6n is also a limit for the length any mod n can have it’s cycle. So, checking the ones smaller than that limit, no bases have a period of exactly b^2
I was thinking does it ever catch up to itsel()v -so for for y sequence 1 contains only an earlier number like f(sil)=1111x... ,,,(shrug),,,
does the base correlate to the amount of numbers before the Fibonacci sequence resets?
Smaller bases have fewer options in a given number of digits, so there is almost certainly some correlation. Haven't done the math to prove it, though.
There are so many patterns in this, that I don’t even know where to start… I’m playing with a bunch of Python code checking out cycle lengths, starting numbers, etc
Just realized I got way too few whiteboards in my backyard
This is interesting because if because of the birthday paradox, if the fibonacci sequence was random, it would repeat after only 14.142 number... yet it goes up to 60. It's like the Fibonacci sequence was anti-random. It's even more amazing when you think that some sequences cannot happen. like 00.
Couldn't you say 0 0 does generate a sequence if you use the Fibonacci rule of adding each pair of numbers to get a third. It's just in this case you always get a 0, so the cycle repeats after just 1 term instead of the 60 you get when starting 0 1.
Did you know that you were going to make a lot of videos that involve modular arithmetic when you chose your clocks anesthetic? Lol
I've only gotten up to base 24 so far, but I noticed that there are only certain bases where the last digit representation in said base has a periodic repeating cycle which begins and ends with zero *and has no other zeros in between*.
So far the only bases 2-24 with this property are 2, 4, 11, and 22.
Whyyyyyy?
Aka what causes some bases to have single periods between last digit zeros and others to have multiple cycles with zeros in between before going back to 1, 1..
Another really cool thing is that the short cycle period between last digit zeros (which is also the whole cycle sometimes) is always equal to the index number of the first Fibonacci which is evenly divisible by the base n.
Examples:
In Base 2, every 3rd Fibonacci number ends in 0, and F(3)=2 is the first F number divisible by the Base number, 2.
In Base 10, every 15th F number ends in a zero, and F(15) = 610 is the first one which is evenly divisible by 10.
Following similarly, the long-cycle period n (where it hits zero and then starts over) is always a multiple of the short periods length (given that there are a short and long period, rather than a single repeating string between zeros)
So in Base 10 for example again.. the short period between zeros is 15, and F(15) is the first one divisible by 10, and the long period between zeros is 60, and F(60) is the 4th F number divisible by 10.
But why on Earth do some bases only have single periods, and others have short and long ones?
And why are the first four bases with single period strings base 2, 4, 11, and 22?
And why do some bases with both short and long periods have a long period which is double the short period, while others have a long period which is a greater than 2 multiple of the short period ?
All these questions and more. Hahha.
@@stickfiftyfive Have you considered looking at what happens in each base with all possible numbers rather than just those within the usual Fibonacci sequence. For example in base ten there are 6 separate sequences of lengths 60, 20, 12, 4, 3 and 1 that between them cover all 100 possible additions of two last digits. Of those the 60 digits sequence has four zeros in it overall, or three between the start of each cycle, and the 20 digit sequence also has four zeros, or three between the start of each cycle. The 12 digit and 4 digit sequences don't have any zeros at all, not even at the start/end of each cycle, and the 3 digit sequence has only the one zero, or no zeros between the start of each sequence. Finally the one digit sequence is literally just the digit zero. Maybe there are further patterns in this if it is extrapolated accross other bases?
you're awesome :D I enjoyed this.
F(8) +1 👍
Search for "Fibonacci in binary" and look for the patterns.
The 24 pattern is not base 12 exclusive.
I like this classroom 🙂
Youre amazing love your videos
Most excellent video
the fibonnocia sequence restarts at the 25 number
How?
We as a species screwed up by abandoning base 12 in favor of base 10.
Base Six is actually my favorite overall (although base twelve would also be significantly better than base ten)
number reform where?? we could even make the US use the seximal metric system!
@@ComboClass I keep hearing this about bases 10 and 12, how 12 is better because of its divisors. Maybe maybe, but 10 has a cool decomposition: 1+2+3+4 = 10. This allows you to represent quantities using a stone or whatever on a cross-hair grid ( big + sign). Top right is 1, bottom right 2, bottom left 3, top left 4. Keep it there and put another stone in the top right, that's 5. Keep going like that and when all four quadrants are occupied by one stone each, that's 10.
Mind you 12 has a cool decomposition: 1+2+3+3+2+1. This is what allows a 12 hour clock to log rise and fall in tidal speed as well as level.
You are mixing base 12 and mod 12 which are different things ... but we get the point and the video is cool
The last digit, base 12, *is* the number, mod 12, though.
I don’t think I’m mixing them up. I used mod 12 as an example of what the last digits of base 12 do, since (like I’ve shown in previous episodes) the patterns are the same
@@ComboClass ok, I get it, thx
Actually, the first occurrence of two terminating zeros occurs at F(150).
But that's only half of the story. The cycle does truly repeat beginning at F(300). You can verify that by comparing F(151) mod 100 with F(301) mod 100.
It happens at any point where the previous two were 9 then 1.
For just a single digit, and just in base ten, but that’s just a small part of the topic/episode
Hey Domotro, I'm a mathematician & I would like to share some of my research regarding different sizes of of infinity. I am not joking when I say I have discovered a 3rd size of infinity but my research goes far beyond that. At some point I am able to prove that the current definition of a real number is wrong and that imaginary numbers must be a subset to real numbers. I really want to share more with you about this but I have a feeling your the right person to discuss this with.
this room is giving me anxiety
Love math and love u brother 😊
I learned the Fibonacci sequence very early in my development; but it took a very long time before I discovered more than just the basic concept. It's always nice to find interesting Fibonacci patterns that are instructive for young people.
It works without zero..
At this point in time and history I think we got to ask: what are YOUR opinions on the industrial revolution and its consequences for the human race? xD
Comment for the algorithm
Yes bro algorithm plase
Zero is even?
Yes
not particularly surprising...?
ok the omittance is cool
First! f(0)