Nice ! Now looking for the N-bonacci numbers :) Hum I just found something interesting : - The N-bonacci polynomial [x^n - (1+x+x²+...+x^(n-1))] can also be written as (2x^n - x^(n+1) - 1)/(1-x) - Because we look for its roots (other than 1), the N-bonacci constant (the golden ratio for N=2 for example) is also the root of this numerator (2x^n - x^(n+1) - 1) - When n gets big, x^n and x^(n+1) overshadow the -1, hence we have 0 ≈ 2x^n - x^(n+1) which gives x ≈ 2. the ratio between two consecutive numbers of the N-bonacci sequence tends towards 2 when N grows to infinity - Furthermore, by setting x ≈ 2-ε, with ε very small, we can show that x goes like 2 - 1/2^n
The 1, x, x^2 etc sequence was introduced with no explanation, and the generation of the initial shapes was also passed over in a heartbeat. I generally love Numberphile videos and learn something from them, but here they were so eager to get to the cool shape demos that they didn’t give me any sense of where it came from.
If you already know that the ratio between consecutive terms converges, then the 1, x², x³, ... thing comes from the following: suppose a, b, c, d are consecutive terms of the tribonnaci sequence, then d/c = (a+b+c)/c = a/c + b/c + 1 = (a/b)(b/c) + (b/c) + 1 If x is the limit of the ratio between a term and the previous term, then taking the limit of the expression before, we have x = (1/x)(1/x) + 1/x + 1 = 1/x² + 1/x + 1 In other words, x³ = x² + x + 1
Or, if you prefer, if you start with a sequence a_1, a_2, a_3, a_4, ... with non zero terms, consider the ratios R_1 = 1 and R_n = a_n/a_{n-1} Notice that, for example, a_5/a_2 = (a_5/a_4)*(a_4/a_3)*(a_3/a_2) = R_5*R_4*R_3 Knowing this, suppose the original sequence is recursive, so a_{n+k+1} = F(a_n, ... , a_{n+k}) with F a continuous function in k+1 variables such that c*F(x_0, ... , x_k) = F(c*x_0, ... , c*x_k) for any scalar c. Of course, for c different of zero, F(x_0, ... , x_k)/c = F(x_0/c, ... , x_k/c) Then R_{n+k+1} R_{n+k} ... R_{n+1} = a_{n+k+1}/a_n = F(a_n, ... , a_{n+k})}/a_n = F(a_n/a_n , ... , a_{n+k}/a_n) = F(1, R_{n+1}, ... , R_{n+k}*R_{n+k-1}*...*R_{n+1}) So, if R_n converges to x, taking limit when n goes to infinity of the last expression, we obtain x^{k+1} = F(1, x, ... , x^k) Cheers! If F is the sum of the coordinates of a N-dimensional vector F(x_1, ... , x_n) = x1 + ... x_n we have the N-bonacci numbers. So, the N-bonacci constant satisfies the polynomial x^n = 1+x+ ... + x^{n-1}
I'm loving the animations, especially in the last couple of videos. It's super helpful to see the transitions between states and how things fit together!
Quite often the Tribonacci sequence starts 0, 0, 1 (you get the same constant). If you do this then the sequence goes 0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, 10609, ... . Expressed in this form, 1, 4, 81, 3136, 10609 are the only known square Tribonacci numbers (Sloane A128911). If there is a larger, it is enormous (I checked up to a few hundred million digits using GnuMP).
If you need more detail than what was provided, take a course in some odd numbers course at your local ... friggin insane asylum. This fellow got the point across quite well, compared it to the Fibonacci version, gave the formula, gave animated and physical representations, and gave the only place it is seen outside of theory in the object he called a "snub cube" ... reallydude, were you even paying attention?
TwoPizzas if you mean (2^n)+1, then it’s cuz we are looking at too few terms. Even Fibonacci 1 1 2 3 5 looks like 2^n + 1, but it breaks at F(5)=8. And tribonacci 1 1 1 3 5 9 17 also, but breaks at T(7)=31. Now, when does Nonabonacci break?
@@pocarski Well, not exactly. If you add even more terms, they eventually fall behind 2^n, because they don't ever double the numbers at the previous stage. They get close, but not quite.
The golden ratio is important in the construction of some aperiodic tilings, particularly the Penrose tiling. The spiral self-similar tiling reproduced here must be aperiodic, so this is an example of an aperiodic tiling which uses the tribonacci constant instead.
The respective numbers of large tiles you need go 0, 0, 1, 1, 2, 4, 7, 13, which is a sequence generated by the same recurrence but with a different starting condition. Each size needs as many small tiles as the previous size needed large tiles (because enlargement turns a large tile into one tile of each size, and small tiles are not created in any different way). The respective numbers of medium tiles you need go 0, 1, 0, 1, 2, 3, 6, 11, which is a sequence generated by the same recurrence but with yet another starting condition.
Man gotta love numberphile but that explanation was sooo poor... I’m talking about that x^3=x^2+x+1. I mean I got it but I’m thinking about someone not knowing what’s that about and they get no clue for what that is
Arokace the tribonacci constant is the limiting ratio between successive entries in the sequence, is the solution of x^3-x^2-x-1=0, and is the dimension of the fractal puzzle pieces
I was referring to everything. Was it subtraction between the powers, I thought it was addition. And what was x suppose to equal in each example he gave (1 + x + x^2 = x^3)
And I just realized I was looking at the power part wrong but I still have a question about it, how did he get to using x^position to find the constant.
@@arokace x is the ratio between neighboring numbers of the sequence, in the limit. If you just pick 4 terms in a row and find ratios of neighbors, they will be different ratios, but as you look farther and farther along, the ratios get closer to a constant. It's similar to the way 5/3, 8/5, 13/8, 21/13 ratios get closer to phi, the golden ratio. Thus to solve for golden ratio we think about an ideal set of neighbors and they can be expressed as 1, x, x^2 and we write 1 + x = x^2 to express the recursive rule. Solve that for Fibonacci ratio (phi). For the Tribonacci ratio, you solve 1 + x + x^2 = x^3.
he fails to really explain where the usage of the X variable comes from near the beginning. Its cool that those polynomials relate back to the discussed constants but by failing to explain that part correctly it really is confusing.
Sloppy video indeed. As for the Fibonacci quadratic, this is how you can get it: Suppose the quotient F(n+1)/F(n) has a limit i.e. reaches some value L when n gets "big enough" (yeah, yeah, I know...). For such n it must then be true that F(n+1)/F(n) = F(n+2)/F(n+1) = L. Now substitute F(n+2)=F(n+1)+F(n) gives: F(n+1)/F(n) = (F(n+1)+F(n))/F(n+1). Rewrite as F(n+1)/F(n) = 1 + F(n)/F(n+1) Replace F(n+1)/F(n) by L: L = 1 + 1/L Multiply by L: L^2 = L + 1. And there you have it: x^2 = x + 1 Same trick for Tribonacci and up. Same idea, bit more involved... Note that this only shows that IF there is a limit, THEN this limit is the solution to x^2 - x - 1 = 0. It does not prove there actually _is_ a limit.
Why don't we talk about Fibonacci series extended backwards? Like ...5,-3,2,-1,1,0,1,1,2,3,5... Tribonacci backwards is ...9,-7,1,3,-3,1,1,-1,1,1,1,3,5... It looks cool
Going Fibonacci style but in threes should really be 0 0 1 -> 1 0 1 1 -> 2 1 1 2 -> 4 1 2 4 -> 7 2 4 7 -> 13 4 7 13 -> 24 7 13 24 -> ... since Fibonacci is about a sequence that grows from a single 1 dropped into an endless number line of zeroes. So 1,1,2,4,7,13,24... Adding 3 numbers together to give the next cannot give 3 1s in a row unless you _specifically_ start from 1 1 1.
What's remarkable is that if you look up the Tribonacci constant's closed form expression, it's surprisingly complicated for a number so easy to generate.
Numbers are very hard for me as I am a visual and tactical minded type. This was a great little clip that made something complex easy to understand. Thanks for taking the time to make it! Also, I very much want to play with your wooden tiles, they look like fun. 😁
Wait... what?! Where does a fribonacci series being represented by 1+x+x²+... come from?! Did I miss something? It is kind of represented as so obvious it needs no explanation.
We know that the ratio of successive fibonacci numbers converges to some value x, and the fibonaccis are defined as F(0) = F(1) = 1 and F(n) = F(n-1)+F(n-2). The fibonaccis don't grow exactly at x, but they do approximately: F(0) = 1 F(1) = 1 ~ 1*x F(2) = 2 ~ 1*x*x etc. In other words, we're approximating the fibonacci sequence as a geometric series with common ratio x. When we apply the recurrence relation: then F(2) = F(1) + F(0) => x^2 = x + 1. Solving this gives that x is the golden ratio, aka the fibonacci constant. Therefore the fibonacci series grows at approximately the golden ratio, and in the limit as n->infinity, this approximation gets better and better.
If you take the limiting case that the ratio F(n+1) / F(n) tends to some constant g. Then with a bit of handwaving F(n+1) = g F(n), F(n+2) = g F(n+1) = g^2 F(n). Now use the recurence formula F(n+2)=F(n)+F(n+1), and divide by F(n) you get g^2 = 1 + g. Solving that gives the golden ratio. For sufficiently big n the ratio of sucessive fibonanchi numbers will be 1: g : g^2 ....
@@ea-nasir420 I think the handwaving part is just that lim (n->infinity) F(n+1) / F(n) = g is true but you instead use that F(n+1) / F(n) = g for very large n. But in principle by being more careful it should be something like F(n+1) / F(n) = g + O(1/n) or something like that.
He left out a step: Ax³ = Ax² + Ax + A. Pick *A* from far into the series. If you assume the ratios between terms _eventually_ converge to *x,* then the next terms must be very close to A, Ax, Ax², Ax³, ... The original rule means Ax³ = Ax² + Ax + A (getting closer the bigger A is). From that: x³ = x² + x + 1
The numbers in the thumbnail are the sequence used in the fractal pattern shown in the video. The palest one has area 1; the medium shade has area 2 (1+1); the darkest has area 4 (1+1+2). These are combined to give the next, of area 7 (1+2+4), and so on for the rest, giving the areas as the sequence: 1, 1, 2, 4, 7, 13, 24, 44, 81, ...
6:09 I noticed it has property of high double-number frequency in the Tribonacci constant. I can see 55, 11, 55, 6600, 22, 77 and already in this short part of the sequence.
If this video left you anything but confused, you know either so much that you don't need math class or so little that you didn't even notice how they rushed through half a dozen fields of mathematics without even trying to explain anything.
Number of different sized shapes in each shape: Small: 1 0 0 1 1 2 4 7 ... Medium: 0 1 0 1 2 3 6 11 ... Large: 0 0 1 1 2 4 7 13 ... The Small and Large seem to follow the same sequence (but at a shift), but the Medium is different. And all of them are a sort of Tribonnaci sequence but with a different start than 1 1 1...
this is the type of the video that you can understand only if you knew all this before. x2 x 1 and x3 x2 x 1 - where they come from for example? yeah, I can google the connection, but without it the video looks like set of random facts and coincidences
i wonder if you flip a tile on its side (standing up) and you would measure the circumference (1d) how would it relate to the tiles getting larger and also how would the circumference, measured going away and/or perpendicular hold up vesus the the distance in the circumference going towards you
I find it interesting how most of the things talked about in this video are kind of summations of things I find on other channels, such as 3blue1brown, Vsauce, etc. The 6 degrees theory idea thing continues to govern my life.
Oren Rosenman it postulates that everything is connected in some way through a somewhat significant link. It doesn’t exclusively pertain to only 6 connections. If you look it up on RUclips, you can find many, many videos talking about the subject. It’s quite fascinating.
Let's broaden out this frequently raised question of what permutation of three 0s and/or 1s the tribonacci should start with. Let's consider all eight of them: 000, 001, 010, 011, 100, 101, 110, 111, and see what and how many distinct number sequences using the tribonacci recurrence relation generates. I find there are 4 (or 5 if you count 0000000...) but please check anyone. Fibonacci seems to be a special case in that whatever permutation of two 0s and/or 1s you start with, 00, 01, 10, 11, you still only ever get the one we're all familiar with (except for the 00 case again). What's the formula relating the number of summands (2 for Fib, 3 for Trib, etc) to the number of distinct possible number sequences?
The reason why the Ancient Greeks didn't consider powers higher than 3 was because they considered numbers as ratios compared to a unit length hence why π, τ and φ were defined the way they were. It was also that reason why they avoided sqrt(2) like it was an elephant in the room. Square numbers and cubic numbers were visualised as unit squares arranged to form a bigger square and cubes that assembled a solid cube respectively-higher powers would have involved hypercubes.
With the Tribonacci however it doesn't work out the same. I tried starting with all the 8 possible permutations of zeros and ones in groups of three: 000, 001, 010, 011, 100, 101, 110, 111 and think I got at least 4 distinct sequences (5 if you include 000 going with just 0s forever of course). But please check. If n is the number of 0 and 1 summands (2 for Fib and 3 for Trib) what's the general formula for n that gives the number of distinct sequences?
@@encounteringjack5699 It pretty much has to be 0, 0, 0, 0, ... You start with n seeds of 1 or some value - so no seeds in this case since n is 0. Then for every term you add the last n terms - so no terms here.
@@encounteringjack5699 The recurrence relation for the tribonacci is A(n) = A(n-1) + A(n-2) + A(n-3); for the fibonacci it's A(n) = A(n-1) + A(n-2); for the monobonacci it's A(n) = A(n-1); which is yes, a sequence of repeated 1s or whatever number you start with, including 0. So respectively 3 terms, 2 terms, and 1 term on the right. As for the nonbonacci it has to be A(n) = ; which has 0 terms on the right, a blank. So your sequence of commas with nothing between them, not even 0, is just about correct.
Some armchair speculation: Tribonacci doesn't occur in nature because the simplest form of division is into halves, not thirds. For example mitosis produces two identical cells, not three.
@@EyeOnTheTV Let's say you're right and mitosis produces two cells. How can that result in a Fibonacci reproductive rate? 1,2,4,8, ... isn't a Fibonacci sequence. But imagine 1 cell dividing into 2, but only one of those daughter cells subsequently divides into 2, while the other remains quiescent so now we have 3. Then in the next generation the previously quiescent cell divides into 2, while of the 2 daughter cells only 1 divides into 2 while the other remains quiescent, then we have 5. Keep going like that and a Fibonacci develops. The path of least resistance here looks like a compromise between not dividing at all (or remaining quiescent as I put it) and submitting to the need to reproduce. Does such a pattern exist in nature?
I'm not sure how to word this, but here we go. I wonder what the ratio between n-nacci ratios approaches. I wish I had the math ability to work that out. I could call it the Seiden Number.
He says at 1:43 that the next term is the sum of all the previous terms, 1+x+x^2 = x^3. But that's not how summing works, this would just be 1+x+x^2. Can anybody explain what he's on about with this x notation?
What's the difference between 1 1 1 3 5 9 17... and 0 0 1 1 2 4 7 13... ? If you do the same thing (i.e. "seed" the fibonacci sequence with 0 1...) you end up with the same series, but obviously there's two different "tribonacci" sequences here. Maybe interestingly, starting 0[0] 0[1] 1[2] or 0[0] 1[1] 1[2] where [n] is the nth term in the series doesn't affect null initialed seeds of the 3bonacci sequence...
What if you did Fibonacci, except you left a gap between the numbers you add together? (1)1(1) Then add the numbers in parentheses 1(1)1(2) 11123 11123469... Would that have any interesting properties? Apparently the ratio is the root of x^3-x^2-1=0. Weird.
It's known as Narayana's Cows (OEIS A000930), an allusion to Fibonacci's rabbits no doubt. I've been studying it on and off for a few years. One interesting property is -((Cn)^2) + (Cn+1)^2 + (Cn+3)^2 = C(2(n+2)). I'm not sure if I've expressed that generalization correctly, but examples are -(9^2) + 13^2 + 28^2 = 872 and -(13)^2 + 19^2 + 41^2 =1873. It's also a kind of Tribonacci, except when you add three consecutive terms the sum is the next but one along, for example 2+3+4=9. I call the ratio the Bovine ratio symbolized by the Geek letter M (pronounced Moo) and is the root of x + 1/x = x^2, that is Moo + Baby Moo = Moo squared. I wonder if Narayana was thinking of that.
For Fibonacci: 1 is the first term, x is the second term and x^2 is the third term. Thus, 1 + x = x^2. Solve for x and you get (1 + sqrt(5))/2 which is the golden ratio.
@@HighTech636 The thing that kinda confuses me about this explanation is it's not really clear to me why we're allowed to assume the three terms are a geometric progression. In fact, even just looking at the first three terms shows that it isn't. I get that the idea is consecutive terms in the sequence will approach a geometric progression the further you go down but no one ever explains how we can be certain that will happen :/
@@snillie x is the golden ratio. The golden ratio squared is 1 + the golden ratio. The golden ratio cubed is the golden ratio + the golden ratio squared. The golden ratio to the 4th power is the golden ratio squared plus the golden ratio cubed etc. That can be represented as 1, x, x^2, x^3, ...., x^(n-1).
@@snillie I think the best way to frame this is in terms of linear algebra. The Fibonacci recurrence f(n+2) = f(n+1) + f(n) is equivalent to the matrix equation f(n+1) = [[1 1] [1 0]] * f(n), where f is vector-valued, implying that f(n) = [[1 1] [1 0]]^n * f(0). For the Tribonacci numbers, the recurrence is f(n+3) = f(n+2) + f(n+1) + f(n), equivalent to f(n+1) = [[1 1 1] [1 0 0] [0 1 0]] * f(n).
Fibonacci is a name, which was derived from Figlio Bonaccio and means son of Bonaccio. "Tribonacci" would be the grandson of Bonaccio. In Italian this is Nipote Bonaccio or short Nibonacci :-).
Shouldn't it start 1, 1, 2? Or 1, 0, 1? The two 1s at the start of the regular Fibonacci are not arbitrary. You start with 1 and add the number before it. As there isn't one it's just 1. For tribonacci, ignoring negatives it's 0 + 0 + 1 = 1, then 0 +1 + 1 = 2, then 1 + 1 + 2 = 4 1,1,2,4,7,12,23 etc. Including negatives it's different still: -1 +0 +1 = 0 (1,0) 0 + 1 + 0 = 1 (1,0,1) ...= 2 (1,0,1,2) etc. and the sequence becomes 1,0,1,2,3,6,9,18 Starting with three 1s just seems plucked out of thin air, or am I missing something?
Mmmm-kay. I'm familiar with the derivation of x^2+x+1=0 for the Fibonacci series, but nothing about that makes it obvious to me that your Tribonacci polynomial should work. I expect you are right, but I'd like to know why. I'd also love to know where the squinkly shape comes from, because surely there's some treasure buried there.
Rod Spade and I would like to know where the x^2 and x^3 came from. It's not explained. It seems contradictory since each number is the sum of previous numbers in the series. In the Fn series, he writes x^2 but that corresponds to '2' which is NOT a square. What's going on here?
Fibonacci can be started with 0,1 0,1,1,2,3,5,etc Tribonacci started with 0,1 looks like this: 0,1,1,2,4,7,13,24 It seems there's some disagreement as to what is the real 'tribonacci' sequence.
Benjamin Wang is right. If you just start with 0,1 then the Tribonacci recurrence doesn't tell you what the next number should be. My preference is to start with 0,0,1
This is an interesting topic, but without reading the comments I wouldn't have had a clue about why you were raising things to powers, the significance of the shapes' jagged edges, or how the two related. I can normally follow your videos well enough but this one seemed very underexplained.
Nice ! Now looking for the N-bonacci numbers :)
Hum I just found something interesting :
- The N-bonacci polynomial [x^n - (1+x+x²+...+x^(n-1))] can also be written as (2x^n - x^(n+1) - 1)/(1-x)
- Because we look for its roots (other than 1), the N-bonacci constant (the golden ratio for N=2 for example) is also the root of this numerator (2x^n - x^(n+1) - 1)
- When n gets big, x^n and x^(n+1) overshadow the -1, hence we have 0 ≈ 2x^n - x^(n+1) which gives x ≈ 2. the ratio between two consecutive numbers of the N-bonacci sequence tends towards 2 when N grows to infinity
- Furthermore, by setting x ≈ 2-ε, with ε very small, we can show that x goes like 2 - 1/2^n
ScienceClic actually, more interesting ideas are Fibonacci but with different starting numbers, and Lucas numbers.
It's the solution to x^n = (x^n - 1)/(x - 1)
The solution to this goes to x = 2 as n approaches infinity
-1/12 no doubt
a recreational mathematician in the making, gold star to you
Ok, now I have to replace the flooring in my house with n-bonacci tiles.
You're gonna need a bigger house.
You need a bigger earth or if there is not enough space move your house to the sun and somehow make the sun and your house bigger
Not n-bonacci, but Tribonacci.
Actually, Rauzy tiles. Where can I buy them? **grin**
@@Amechaniaa you would also need a flat earth ...
"quite an attractive little polyhedron"
I'm going to use this as a compliment
I wish someone would call me an attractive polyhedron. Or rather, infinitely intricate self-similar figure since my name is Fractal lol
There is a small mistake at 5:09. The number of tiles in the last shape is 17 instead of 13.
The 1, x, x^2 etc sequence was introduced with no explanation, and the generation of the initial shapes was also passed over in a heartbeat. I generally love Numberphile videos and learn something from them, but here they were so eager to get to the cool shape demos that they didn’t give me any sense of where it came from.
The shapes are not generated from the sequence. That's just something Rauzy found (don't ask me how).
If you already know that the ratio between consecutive terms converges, then the 1, x², x³, ... thing comes from the following:
suppose a, b, c, d are consecutive terms of the tribonnaci sequence, then
d/c = (a+b+c)/c
= a/c + b/c + 1
= (a/b)(b/c) + (b/c) + 1
If x is the limit of the ratio between a term and the previous term, then taking the limit of the expression before, we have
x = (1/x)(1/x) + 1/x + 1
= 1/x² + 1/x + 1
In other words,
x³ = x² + x + 1
Or, if you prefer, if you start with a sequence
a_1, a_2, a_3, a_4, ...
with non zero terms, consider the ratios
R_1 = 1 and R_n = a_n/a_{n-1}
Notice that, for example,
a_5/a_2
= (a_5/a_4)*(a_4/a_3)*(a_3/a_2)
= R_5*R_4*R_3
Knowing this, suppose the original sequence is recursive, so
a_{n+k+1} = F(a_n, ... , a_{n+k})
with F a continuous function in k+1 variables such that
c*F(x_0, ... , x_k) = F(c*x_0, ... , c*x_k)
for any scalar c. Of course, for c different of zero,
F(x_0, ... , x_k)/c = F(x_0/c, ... , x_k/c)
Then
R_{n+k+1} R_{n+k} ... R_{n+1}
= a_{n+k+1}/a_n
= F(a_n, ... , a_{n+k})}/a_n
= F(a_n/a_n , ... , a_{n+k}/a_n)
= F(1, R_{n+1}, ... , R_{n+k}*R_{n+k-1}*...*R_{n+1})
So, if R_n converges to x, taking limit when n goes to infinity of the last expression, we obtain
x^{k+1} = F(1, x, ... , x^k)
Cheers! If F is the sum of the coordinates of a N-dimensional vector
F(x_1, ... , x_n) = x1 + ... x_n
we have the N-bonacci numbers. So, the N-bonacci constant satisfies the polynomial
x^n = 1+x+ ... + x^{n-1}
I'm loving the animations, especially in the last couple of videos. It's super helpful to see the transitions between states and how things fit together!
Your animator is awesome
you're
@@mattgsm your*
@@10KCannabis there*
@@mattgsmTheir*
@@brian7168342 whomst'd've'ly'yaint'nt'ed'ies's'y'es*
Quite often the Tribonacci sequence starts 0, 0, 1 (you get the same constant). If you do this then the sequence goes 0, 0, 1, 1, 2, 4, 7, 13, 24, 44, 81, 149, 274, 504, 927, 1705, 3136, 5768, 10609, ... . Expressed in this form, 1, 4, 81, 3136, 10609 are the only known square Tribonacci numbers (Sloane A128911). If there is a larger, it is enormous (I checked up to a few hundred million digits using GnuMP).
Make a video of the carpenter and the laser wood cutter....
"you want me to make what!?!?!?!"
Thats why I do my own cutting.
@@GeladaMaths Well said... :-)
If it was Clifford Stoll, he'd do it himself...
Just print the pattern, glue it to the wood and cut. Could even saw and sandpaper the piece.
@@disnecessaurorex4908 ever used a coping saw? One piece would take a while to do well enough to work.
Interesting topic but I feel this could have been explained in much greater detail.
Every topic could be explained in much greater detail.
Some recursive xxx detected?
*should have
If you need more detail than what was provided, take a course in some odd numbers course at your local ... friggin insane asylum. This fellow got the point across quite well, compared it to the Fibonacci version, gave the formula, gave animated and physical representations, and gave the only place it is seen outside of theory in the object he called a "snub cube" ... reallydude, were you even paying attention?
Ted Jh LOL
Check out Mathologer's video on tribonacci numbers. Much better.
Nonabonacci just rolls off the tongue, so musical. Can't wait until we get to that one!
1, 1, 1, 1, 1, 1, 1, 1, 1, 9, 17, 33, 65, 129, 257, 513, 1025, 2049... (A127193 on the OEIS).
so it just turns into 2n - 1, that's interesting
@@pocarski Fascinating.
TwoPizzas if you mean (2^n)+1, then it’s cuz we are looking at too few terms. Even Fibonacci 1 1 2 3 5 looks like 2^n + 1, but it breaks at F(5)=8. And tribonacci 1 1 1 3 5 9 17 also, but breaks at T(7)=31. Now, when does Nonabonacci break?
@@pocarski Well, not exactly. If you add even more terms, they eventually fall behind 2^n, because they don't ever double the numbers at the previous stage. They get close, but not quite.
The golden ratio is important in the construction of some aperiodic tilings, particularly the Penrose tiling. The spiral self-similar tiling reproduced here must be aperiodic, so this is an example of an aperiodic tiling which uses the tribonacci constant instead.
The respective numbers of large tiles you need go 0, 0, 1, 1, 2, 4, 7, 13, which is a sequence generated by the same recurrence but with a different starting condition. Each size needs as many small tiles as the previous size needed large tiles (because enlargement turns a large tile into one tile of each size, and small tiles are not created in any different way). The respective numbers of medium tiles you need go 0, 1, 0, 1, 2, 3, 6, 11, which is a sequence generated by the same recurrence but with yet another starting condition.
The Numberphile Podcast is great! Mathematicians talking about their lives is very interesting.
I've only seen Grant's video.
MATH Genius
You mean you can’t find the others?
@@dedwarmo No, I was interested only in that one.
Man gotta love numberphile but that explanation was sooo poor... I’m talking about that x^3=x^2+x+1. I mean I got it but I’m thinking about someone not knowing what’s that about and they get no clue for what that is
I stopped the video at that point, I can confirm it doesn't make any sense.
compassion for the weak is not something among the virtues of mathematicians XD
@@STAR0SS Same.
@@-raven6485 Got that right.
STAR0SS you get it from generating functions.
Fourbonacci: 1, 1, 1, 1, 4, 7, 13, 25, 49, 94, 181, 349, 673, 1297, 2500, 4819...
Constant: 1.927562
I'm so lost, there's no connection or correlation between the sequence, x to powers, and the fractal puzzle pieces.
Arokace the tribonacci constant is the limiting ratio between successive entries in the sequence, is the solution of x^3-x^2-x-1=0, and is the dimension of the fractal puzzle pieces
@@lucas29476 I think he is referring to the shape of the pieces.
I was referring to everything. Was it subtraction between the powers, I thought it was addition. And what was x suppose to equal in each example he gave (1 + x + x^2 = x^3)
And I just realized I was looking at the power part wrong but I still have a question about it, how did he get to using x^position to find the constant.
@@arokace x is the ratio between neighboring numbers of the sequence, in the limit. If you just pick 4 terms in a row and find ratios of neighbors, they will be different ratios, but as you look farther and farther along, the ratios get closer to a constant.
It's similar to the way 5/3, 8/5, 13/8, 21/13 ratios get closer to phi, the golden ratio.
Thus to solve for golden ratio we think about an ideal set of neighbors and they can be expressed as 1, x, x^2 and we write 1 + x = x^2 to express the recursive rule. Solve that for Fibonacci ratio (phi).
For the Tribonacci ratio, you solve 1 + x + x^2 = x^3.
Video: Tribonacci numbers
Subtitles: *Trib inochi* numbers (0:47)
Also Subtitles: *ribon Archie* numbers (0:53)
Subtitles again: *Trib annachi* numbers (1:07)
Me: _What the hell?_
When you enable English (auto-generated) subtitles
And ribbon achi
what would you even need the subtitles for?
@@NinjarioPicmin If you're deaf or in a quiet place...
@@branthebrave true
The first Numberphile video that had me completely lost. Maybe you could do a follow-up?
I agree. Lost and in search of a point, and sadly, not entertained.
he fails to really explain where the usage of the X variable comes from near the beginning. Its cool that those polynomials relate back to the discussed constants but by failing to explain that part correctly it really is confusing.
Sloppy video indeed. As for the Fibonacci quadratic, this is how you can get it:
Suppose the quotient F(n+1)/F(n) has a limit i.e. reaches some value L when n gets "big enough" (yeah, yeah, I know...).
For such n it must then be true that F(n+1)/F(n) = F(n+2)/F(n+1) = L.
Now substitute F(n+2)=F(n+1)+F(n) gives: F(n+1)/F(n) = (F(n+1)+F(n))/F(n+1).
Rewrite as F(n+1)/F(n) = 1 + F(n)/F(n+1)
Replace F(n+1)/F(n) by L: L = 1 + 1/L
Multiply by L: L^2 = L + 1. And there you have it: x^2 = x + 1
Same trick for Tribonacci and up. Same idea, bit more involved...
Note that this only shows that IF there is a limit, THEN this limit is the solution to x^2 - x - 1 = 0. It does not prove there actually _is_ a limit.
@@chaosme1ster One Internet point for you, kind sir.
I can't believe Edmund thought this was trivial enough not to explain it...
Why don't we talk about Fibonacci series extended backwards?
Like ...5,-3,2,-1,1,0,1,1,2,3,5...
Tribonacci backwards is ...9,-7,1,3,-3,1,1,-1,1,1,1,3,5...
It looks cool
It's basically the normal Fibonacci series but the terms alternate their signs.
Tribonacci backwards is ...9,-7,1,3,-3,1,1,-1,1,1,1,3,5...
@@vaibhavagarwal3925 That series looks cooler.
Joseph Mellor And the ratio of consecutive terms approaches the other root -1/φ
No I haven’t found out about the Numberphile podcast, and frankly I don’t care. You’re one of the best channels ever made.
NumberPhile: *uploads this video on June 3rd
My RUclips Notification on June 7 at 9:34 AM: NuMbErPhILe HaS JuSt UpPLoAdEd A nEw ViDeO
👏🏻👏🏻👏🏻👏🏻👏🏻
For those curious, the exact value of the Tribonacci constant is 1/3 (1 + (19 - 3 sqrt(33))^(1/3) + (19 + 3 sqrt(33))^(1/3)).
JorWat25
You are a genius. Thanks.
For anyone interested in the general "n-bonacci" case, the ratio for an n-bonacci sequence seems to approach 2 as n approaches infinity
so how is the shape defined? If I wanted to draw/3D print it how do I go about constructing one of these?
Going Fibonacci style but in threes should really be
0 0 1 -> 1
0 1 1 -> 2
1 1 2 -> 4
1 2 4 -> 7
2 4 7 -> 13
4 7 13 -> 24
7 13 24 -> ...
since Fibonacci is about a sequence that grows from a single 1 dropped into an endless number line of zeroes.
So 1,1,2,4,7,13,24...
Adding 3 numbers together to give the next cannot give 3 1s in a row unless you _specifically_ start from 1 1 1.
What's remarkable is that if you look up the Tribonacci constant's closed form expression, it's surprisingly complicated for a number so easy to generate.
Numbers are very hard for me as I am a visual and tactical minded type. This was a great little clip that made something complex easy to understand. Thanks for taking the time to make it! Also, I very much want to play with your wooden tiles, they look like fun. 😁
Wait... what?!
Where does a fribonacci series being represented by 1+x+x²+... come from?! Did I miss something? It is kind of represented as so obvious it needs no explanation.
We know that the ratio of successive fibonacci numbers converges to some value x, and the fibonaccis are defined as F(0) = F(1) = 1 and F(n) = F(n-1)+F(n-2). The fibonaccis don't grow exactly at x, but they do approximately:
F(0) = 1
F(1) = 1 ~ 1*x
F(2) = 2 ~ 1*x*x
etc.
In other words, we're approximating the fibonacci sequence as a geometric series with common ratio x. When we apply the recurrence relation: then
F(2) = F(1) + F(0) => x^2 = x + 1.
Solving this gives that x is the golden ratio, aka the fibonacci constant. Therefore the fibonacci series grows at approximately the golden ratio, and in the limit as n->infinity, this approximation gets better and better.
If you take the limiting case that the ratio F(n+1) / F(n) tends to some constant g. Then with a bit of handwaving F(n+1) = g F(n), F(n+2) = g F(n+1) = g^2 F(n). Now use the recurence formula F(n+2)=F(n)+F(n+1), and divide by F(n) you get g^2 = 1 + g. Solving that gives the golden ratio. For sufficiently big n the ratio of sucessive fibonanchi numbers will be 1: g : g^2 ....
@@SalixAlba256 Thanks. I was also baffled by this, but your explanation makes perfect sense.
@@SalixAlba256 Can you explain the handwaving or will I need to delve into some mathematical article? :)
@@ea-nasir420 I think the handwaving part is just that lim (n->infinity) F(n+1) / F(n) = g is true but you instead use that F(n+1) / F(n) = g for very large n. But in principle by being more careful it should be something like F(n+1) / F(n) = g + O(1/n) or something like that.
He left out a step: Ax³ = Ax² + Ax + A.
Pick *A* from far into the series. If you assume the ratios between terms _eventually_ converge to *x,* then the next terms must be very close to A, Ax, Ax², Ax³, ... The original rule means Ax³ = Ax² + Ax + A (getting closer the bigger A is). From that: x³ = x² + x + 1
Bob Stein, Yes, exactly. And that's *not* an obvious step to just skip it!
The numbers on the thumbnail actually are different.
They Start with only two ones instead of 3 as in the video.
The numbers in the thumbnail are the sequence used in the fractal pattern shown in the video. The palest one has area 1; the medium shade has area 2 (1+1); the darkest has area 4 (1+1+2). These are combined to give the next, of area 7 (1+2+4), and so on for the rest, giving the areas as the sequence: 1, 1, 2, 4, 7, 13, 24, 44, 81, ...
a snub cube which is quite an attractive little polyhedron - My new favorite sentence of all time.
A great pickup line ;)
What is 1,1,2,4,7, etc from. It is the sequence on the preview for this video
rbm10101 that’s the true tribonacci sequence, it shouldn’t be 1,1,1,3,5,9…
It should be 1,1,2,4,7,13,24…
I remember trying to solve the n-bonacci problem on my own a few years ago, and it took me almost a week to finally figured out the solution
The sequence should start with “1 1 2”, to follow the pattern of adding the prior 3. Numbers before the sequence are considered 0
That’s exactly what I was thinking!
Despite how I've watched all the fractal and Fibonacci videos here you still lost me lol... but I am equally intrigued!
You need to make a series of three videos on this topic.
Then you can combine them together to make a fourth video...
I felt this was very poorly explained compared to the usual excellent Numberphile videos :/
Steve Mason damm
It is really as simple as showing the Tribonacci constant and then showing a physical manifestation of it
Agreed, Edmund Harriss may be a gifted mathematician but he does not communicate well
Steve Mason how does the complicated shape exactly relate to the sequence?? it's unclear
3blue1brown, care to explore this in the level of detail this deserves?
Well, one cool fact is that the n-bonacci constant tends to 2 as n goes to infinity (which is very easy to see).
6:09 I noticed it has property of high double-number frequency in the Tribonacci constant. I can see 55, 11, 55, 6600, 22, 77 and already in this short part of the sequence.
Numberphile makes me feel way too smart. If only it would translate into math class
It kinda does, you just need to watch a lot of their videos
Anton Ladan Not really, they usually cover more interesting subjects than would be shown in a math class
@@everlast282 yes, but it helped me a lot in math class, more than i thought it would
If this video left you anything but confused, you know either so much that you don't need math class or so little that you didn't even notice how they rushed through half a dozen fields of mathematics without even trying to explain anything.
L
Number of different sized shapes in each shape:
Small: 1 0 0 1 1 2 4 7 ...
Medium: 0 1 0 1 2 3 6 11 ...
Large: 0 0 1 1 2 4 7 13 ...
The Small and Large seem to follow the same sequence (but at a shift), but the Medium is different.
And all of them are a sort of Tribonnaci sequence but with a different start than 1 1 1...
I was expecting the outcome at 3:07. But it still blew my mind
this is the type of the video that you can understand only if you knew all this before. x2 x 1 and x3 x2 x 1 - where they come from for example? yeah, I can google the connection, but without it the video looks like set of random facts and coincidences
i wonder
if you flip a tile on its side (standing up) and you would measure the circumference (1d)
how would it relate to the tiles getting larger
and also
how would the circumference, measured going away and/or perpendicular hold up vesus the the distance in the circumference going towards you
I love the topic of fractals. So this video is very interesting to me.
No joke I was actually just watching James grime on numberphile.
Se se
I really want some of those tiles now!
I find it interesting how most of the things talked about in this video are kind of summations of things I find on other channels, such as 3blue1brown, Vsauce, etc.
The 6 degrees theory idea thing continues to govern my life.
What 6 degrees theory? Could you please elaborate?
Oren Rosenman it postulates that everything is connected in some way through a somewhat significant link. It doesn’t exclusively pertain to only 6 connections. If you look it up on RUclips, you can find many, many videos talking about the subject. It’s quite fascinating.
Let's broaden out this frequently raised question of what permutation of three 0s and/or 1s the tribonacci should start with. Let's consider all eight of them: 000, 001, 010, 011, 100, 101, 110, 111, and see what and how many distinct number sequences using the tribonacci recurrence relation generates. I find there are 4 (or 5 if you count 0000000...) but please check anyone. Fibonacci seems to be a special case in that whatever permutation of two 0s and/or 1s you start with, 00, 01, 10, 11, you still only ever get the one we're all familiar with (except for the 00 case again). What's the formula relating the number of summands (2 for Fib, 3 for Trib, etc) to the number of distinct possible number sequences?
The reason why the Ancient Greeks didn't consider powers higher than 3 was because they considered numbers as ratios compared to a unit length hence why π, τ and φ were defined the way they were. It was also that reason why they avoided sqrt(2) like it was an elephant in the room. Square numbers and cubic numbers were visualised as unit squares arranged to form a bigger square and cubes that assembled a solid cube respectively-higher powers would have involved hypercubes.
What about starting with zeros (and a single one): 0,0,1,1,2,4,7,13,... With Fibonacci it works out to be the same.
With the Tribonacci however it doesn't work out the same. I tried starting with all the 8 possible permutations of zeros and ones in groups of three: 000, 001, 010, 011, 100, 101, 110, 111 and think I got at least 4 distinct sequences (5 if you include 000 going with just 0s forever of course). But please check. If n is the number of 0 and 1 summands (2 for Fib and 3 for Trib) what's the general formula for n that gives the number of distinct sequences?
Fourbonacci: 1, 1, 1, 1, 4, 7, 13, 25, 49, 94, 181, 349, 673, 1297, 2500, 4819, 9289
Ratio: approximately 1.9275
I demand that this gets used as a basis for a board game!
What kind of paper do you use?I need answer please.Thanks
Parcel paper
Thank you
These shapes look like me drawing fantasy maps before realizing I again somehow just drew the same shape again and again in different sizes
Now whenever you guys run out of topics you can just do "The *n*bonacchi" numbers
next coming up - "The quadribonacchi" numbers
ka hei chan called tetranacci numbers
That's on the n-bonacchiphile channel.
Here’s the monobonacci sequence.
1, 1, 1, 1, 1, 1,....
Very fascinating right?
and the nonbonacci?
Justin _
The nonbacci is.... I don’t know what it is. It could be,
0, 0, 0, 0, 0,...
Or
, , , , , , ,...
Both are very interesting.
@@encounteringjack5699 It pretty much has to be 0, 0, 0, 0, ...
You start with n seeds of 1 or some value - so no seeds in this case since n is 0. Then for every term you add the last n terms - so no terms here.
@@encounteringjack5699 The recurrence relation for the tribonacci is A(n) = A(n-1) + A(n-2) + A(n-3); for the fibonacci it's A(n) = A(n-1) + A(n-2); for the monobonacci it's A(n) = A(n-1); which is yes, a sequence of repeated 1s or whatever number you start with, including 0. So respectively 3 terms, 2 terms, and 1 term on the right. As for the nonbonacci it has to be A(n) = ; which has 0 terms on the right, a blank. So your sequence of commas with nothing between them, not even 0, is just about correct.
Why are the first three tiles in the sequence which represent the numbers 1, 1, and 1, all different sizes?
First time I've ever seen two parentheses used back to back to write an "x".
Some armchair speculation: Tribonacci doesn't occur in nature because the simplest form of division is into halves, not thirds. For example mitosis produces two identical cells, not three.
Look up the snub cube, a shape whose surface area and volume use the tribonacci constant in their formulae
Extremely simplistic view
I like it.
the universe takes the path of least resistence
@@EyeOnTheTV Let's say you're right and mitosis produces two cells. How can that result in a Fibonacci reproductive rate? 1,2,4,8, ... isn't a Fibonacci sequence. But imagine 1 cell dividing into 2, but only one of those daughter cells subsequently divides into 2, while the other remains quiescent so now we have 3. Then in the next generation the previously quiescent cell divides into 2, while of the 2 daughter cells only 1 divides into 2 while the other remains quiescent, then we have 5. Keep going like that and a Fibonacci develops. The path of least resistance here looks like a compromise between not dividing at all (or remaining quiescent as I put it) and submitting to the need to reproduce. Does such a pattern exist in nature?
I'm not sure how to word this, but here we go. I wonder what the ratio between n-nacci ratios approaches. I wish I had the math ability to work that out. I could call it the Seiden Number.
Rick Seiden Figure it out and write yourself a check for $1M.
0:35 Than noise is so satisfying
flurp
Sweet! Subscribing to podcast!
This is awesome! 🤯
By the way the sequence given in the video thumbnail (1, 1, 2, 4, 7, 13, …) is not the tribonacci sequence
It is. As with the Generalized Fibonacci Numbers, It uses the same general algorithm with different initial values - 0,1,1
I thought I already loved math. How is it that I'm loving math more and more.
He says at 1:43 that the next term is the sum of all the previous terms, 1+x+x^2 = x^3. But that's not how summing works, this would just be 1+x+x^2. Can anybody explain what he's on about with this x notation?
Fun fact, if you start a fibonacci sequence with ANY two numbers, they will always normalize to the golden ratio.
Yeah that fun fact is called math
@@sardonyx001 A lot of fun facts are called math. Math has a lot of stuff. Most of which is fun and factual.
What's the difference between
1 1 1 3 5 9 17...
and
0 0 1 1 2 4 7 13...
?
If you do the same thing (i.e. "seed" the fibonacci sequence with 0 1...) you end up with the same series, but obviously there's two different "tribonacci" sequences here. Maybe interestingly, starting 0[0] 0[1] 1[2] or 0[0] 1[1] 1[2] where [n] is the nth term in the series doesn't affect null initialed seeds of the 3bonacci sequence...
What if you did Fibonacci, except you left a gap between the numbers you add together?
(1)1(1)
Then add the numbers in parentheses
1(1)1(2)
11123
11123469...
Would that have any interesting properties?
Apparently the ratio is the root of x^3-x^2-1=0. Weird.
It's known as Narayana's Cows (OEIS A000930), an allusion to Fibonacci's rabbits no doubt. I've been studying it on and off for a few years. One interesting property is -((Cn)^2) + (Cn+1)^2 + (Cn+3)^2 = C(2(n+2)). I'm not sure if I've expressed that generalization correctly, but examples are -(9^2) + 13^2 + 28^2 = 872 and -(13)^2 + 19^2 + 41^2 =1873. It's also a kind of Tribonacci, except when you add three consecutive terms the sum is the next but one along, for example 2+3+4=9. I call the ratio the Bovine ratio symbolized by the Geek letter M (pronounced Moo) and is the root of x + 1/x = x^2, that is Moo + Baby Moo = Moo squared. I wonder if Narayana was thinking of that.
The thumbnail of this video reminded me instantly of the three kingdoms.
Okay. Where can I buy these Rauzy fractal puzzle pieces?
I want them as floor tiles.
Wait, where did those polynomials come from?
For Fibonacci: 1 is the first term, x is the second term and x^2 is the third term. Thus, 1 + x = x^2. Solve for x and you get (1 + sqrt(5))/2 which is the golden ratio.
@@HighTech636 The thing that kinda confuses me about this explanation is it's not really clear to me why we're allowed to assume the three terms are a geometric progression. In fact, even just looking at the first three terms shows that it isn't. I get that the idea is consecutive terms in the sequence will approach a geometric progression the further you go down but no one ever explains how we can be certain that will happen :/
@@snillie x is the golden ratio. The golden ratio squared is 1 + the golden ratio. The golden ratio cubed is the golden ratio + the golden ratio squared. The golden ratio to the 4th power is the golden ratio squared plus the golden ratio cubed etc. That can be represented as 1, x, x^2, x^3, ...., x^(n-1).
@@snillie I think the best way to frame this is in terms of linear algebra. The Fibonacci recurrence f(n+2) = f(n+1) + f(n) is equivalent to the matrix equation f(n+1) = [[1 1] [1 0]] * f(n), where f is vector-valued, implying that f(n) = [[1 1] [1 0]]^n * f(0). For the Tribonacci numbers, the recurrence is f(n+3) = f(n+2) + f(n+1) + f(n), equivalent to f(n+1) = [[1 1 1] [1 0 0] [0 1 0]] * f(n).
These look like nations, applications for regional tiling in strategy games perhaps?
I like this stuff👍
superb - thanks
Micheal Stephens talked about snub shapes on D!ng a while back, could you do a video on snubbing polyhedra in the near future?
Fibonacci is a name, which was derived from Figlio Bonaccio and means son of Bonaccio.
"Tribonacci" would be the grandson of Bonaccio. In Italian this is Nipote Bonaccio or short Nibonacci :-).
Never knew Tyler from DP did math... 😂
Will the tiles things work with any bonacci numbers? Or this only works with Tribonacci only?
For 4-bonacci the tiles are 3d, and for n-bonacci there are (n-1) dimensional tiles.
This got me the chills ! Huge thumbs up
I would have started the tribonacci series with 0 0 1. Therefore:
0 0 1 1 2 4 7 13 ...
That is what wikipedia lists it as.
Shouldn't series start with 0? Fibonacci is fine:
0 1 1 2 3 5 8 13 ... (adding 2 numbers, so start with 0 & 1)
but "Tribonacci"?
0 1 2 3 6 11 20 37 ... (adding 3 numbers, so start with 0, 1 & 2)
Was just thinking the same thing...
Fibonacci starts with 1 and 1. The 0 is useless, we usually drop it. It changes nothing to the ratio or the numbers...
I mean in France we do that! Maybe different in your country?
@@elfinan Vive la France!
@@elfinan Fibonacci starts with 0.
When asked to generate the series in algorithms.. a=0, b=1, c= a+b, and it continues.
Shouldn't it start 1, 1, 2? Or 1, 0, 1?
The two 1s at the start of the regular Fibonacci are not arbitrary. You start with 1 and add the number before it. As there isn't one it's just 1.
For tribonacci, ignoring negatives it's 0 + 0 + 1 = 1, then 0 +1 + 1 = 2, then 1 + 1 + 2 = 4
1,1,2,4,7,12,23 etc.
Including negatives it's different still: -1 +0 +1 = 0 (1,0) 0 + 1 + 0 = 1 (1,0,1) ...= 2 (1,0,1,2) etc. and the sequence becomes 1,0,1,2,3,6,9,18
Starting with three 1s just seems plucked out of thin air, or am I missing something?
Legend has it that Abruzzi is still looking for Fibonacci
Please do a series on fractals.
Everything after 5:15 is impossible to follow. Just a jumble of terms.
"Matrices", "Certain properties", "expansion and contraction", "eigenvectors".
Did you change paper companies for your brown paper? The texture looks different.
Mmmm-kay. I'm familiar with the derivation of x^2+x+1=0 for the Fibonacci series, but nothing about that makes it obvious to me that your Tribonacci polynomial should work. I expect you are right, but I'd like to know why. I'd also love to know where the squinkly shape comes from, because surely there's some treasure buried there.
Absolutely mathematically amazing =D
Saw the thumbnail and thought it was a new take on Heesch tiling.
Rod Spade and I would like to know where the x^2 and x^3 came from. It's not explained. It seems contradictory since each number is the sum of previous numbers in the series. In the Fn series, he writes x^2 but that corresponds to '2' which is NOT a square. What's going on here?
RUclips is listening to what I say… I had this idea earlier, but I started with 0,1,1,2,4,7,13,…
Well, you were not wrong. It's just that they arbitrarily started with 1,1,1.
Fibonacci can be started with 0,1
0,1,1,2,3,5,etc
Tribonacci started with 0,1 looks like this:
0,1,1,2,4,7,13,24
It seems there's some disagreement as to what is the real 'tribonacci' sequence.
The constant stays the same however, just like in the fibonacci series
Should start tribonacci with three numbers
Benjamin Wang is right. If you just start with 0,1 then the Tribonacci recurrence doesn't tell you what the next number should be. My preference is to start with 0,0,1
i like it... one question ... how is this gonna help in our everyday life ?
By getting smarter! :-p
I wanted to know if the Tribonacci constant also has a numeric / algebraic expression, as phi = (1 + sqrt(5))/2
Wonderfull point of view of fibonacci
Like 6.8K!!
This is an interesting topic, but without reading the comments I wouldn't have had a clue about why you were raising things to powers, the significance of the shapes' jagged edges, or how the two related. I can normally follow your videos well enough but this one seemed very underexplained.
Muito foda! Parabéns pelo conteúdo do canal. :D