Random Fibonacci Numbers - Numberphile

[@numberphile]

Extra footage: kzbin.info/www/bejne/fGGmZYhthpqsjrc Fibonacci Numbers in the Mandelbrot Set: kzbin.info/www/bejne/an20p52Jm5uGidk More James Grime videos: bit.ly/grimevideos

[@TheRealGuywithoutaMustache]

I find your videos really fascinating

[@prunabluepepper]

Welcome back Dr. Grime. Hope you're well. Have a great start into the coming week :)

[@aswinkumar6052]

Bro I got an idea if any prime number squared itself and subtract 2 can be a prime. (Prime)^2-2 =may be prime

[@Zoxesyr]

which video did that framed brown paper come from?

[@robertlozyniak3661]

How about a link to the paper?

[@rohansastry]

Every year, I throw a fibonacci party. Each party is as big as the last two combined.

[@geomochi4904]

Rohan Sastry lol

[@miramosa7768]

I do the same, but the first two had zero attendees so I'm not getting anywhere :(

[@randomdude9135]

@@miramosa7768 You take the factorial of them and proceed ;)

[@TuberTugger]

@@miramosa7768 You didn't show up to your own party?

[@chenugent]

XDDD

[@HonkeyKongLive]

I'm a simple man. I see James on Numberphile, I'm in.

[@NightBlado]

Came scrolling here for this

[@galacticbob1]

I see a new Numberphile video and I'm N 😋

[@exoplanets]

Me too

[@wingsofmathematicsbytanush2507]

@@exoplanets me to also.

[@obaroajeke6080]

I signed in to like this comment

[@johnchessant3012]

Fun fact: The author of the paper at 9:20 discovered a proof that there are infinitely many primes using basic topology.

[@noidea2568]

A proof that there are infinitely many primes ysing basic TOPOLOGY? I need to see this!

[@spyfox260]

No Idea Yes, a proof that there are infinitely many primes using basic topology. You are correct

[@johnchessant3012]

No Idea Yes! Look up "Furstenberg's proof" on Wikipedia. You'll need to know the definition of a topology.

[@yusuf-5531]

@@johnchessant3012 is there a proof that there are infinitely many primes using basic topology?

[@jingermcblabbersnitch7162]

Read it. Neat

[@harishchalwadi]

James Prime is back 🙂

[@lindhe]

He's in his prime.

[@exoplanets]

Yay

[@maxonmendel5757]

yeah what had happened with that?

[@DanDart]

Prime is my ship name for Parker + Grime. It's the perfect mathematical ship (of Theseus).

[@Triantalex]

false.

[@silverette666]

i haven't checked this channel in a long while (a few years) and i'm glad to see james is still doing videos with you. he was my favorite part of your channel back when i watched your videos regularly

[@Mu_Lambda_Theta]

"Nobody will make a computer simulation." I'll take that as a challenge. I got some results: I did 45000 random sequences, and ended up with 0,98721748 It seems like 64 bits are not enough for enough terms, and enough precision for calculating averages And I can't be bothered to code more. BUT! While doing that I figured out some cool identities: ( F(2^k) )² + ( F(2^k+1) )² = F(2^(k+1)+1) F(2^k) * (2*F(2^k+1) - F(2^k)) = F(2^(k+1)) Example: (F9)²+(F8)²=F17; F8*(2*F9-F8)=F16 Where Fn is the nth Fibonacci number, of course

[@giladkay3761]

how did you do the program, I was going to start when I realized the second I try to find the value of something divided by 0 the program will crash

[@DehimVerveen]

@@giladkay3761 Why would you divide by zero? You can use the nth root. Also, you can check if the value you're dividing by is zero before doing so.

[@tinydong4586]

Is it the nth term, nnd term or nst term?

[@DehimVerveen]

@MauLob, Would you reckon 128 bits are enough? How many bits do you think are necessary?

[@Mu_Lambda_Theta]

@@DehimVerveen I do not know. With 64 bits you can, worst case, only get to about 90 terms. With 128 terms you could do a little more than double. For the normal fibonacci sequence, 90 terms is enough to hit the golden ratio with double precision. However, I am starting to think that the problem is not integer over/underflow, but floating point precision. And at that point, I have absolutely no clue.

[@banjofries]

I was about to say something funny about this but then I tried it, and it's actually impossible for the sequence to have 2 zeroes in a row, because there will always be at least a 1, 2, -1, or -2 somewhere behind or in front of the zero due to zero not being able to change the value of the previous number in the list to zero since the sequence starts with 1. Oh well, I guess there CAN'T be a list that's a few numbers then just infinite zeroes...

[@alvarol.martinez5230]

a slightly higher level explanation of this: the vector (a_n, a_(n+1)) is the product of a 2x2 matrix A = (0, 1; +1 or -1, 1) by the nonzero vector (a_(n-1), a_n). Since A is invertible, the result can never be the zero vector

[@mirabilis]

Unless it's all zeroes.

[@mirabilis]

Easier: Two zeroes in a row would mean the number before those were a zero and the number before that was a zero and so on...

[@Shenron557]

@@alvarol.martinez5230 Wow, beautiful explanation. Thanks!

[@BlazingshadeLetsPlay]

Álvaro L. Martínez as someone who is just learning linear algebra rn my mind is blown. I thought matrix algebra was just a boring strategy to compute stuff but once again math blows my mind

[@albinoagellar268]

James on thumbnail = instant click

[@Reneofficialid]

Me tooooooooo! Hahaha

[@Rickety3263]

Yaaasss😂 Why do I find this so fascinating 😅

[@Martin-qb2mw]

Agreed

[@exoplanets]

Yes

[@Triantalex]

??

[@fmakofmako]

Surprised James didnt say anything about the 1/sqrt(5) in the first few minutes. It would have made his approximation extremely accurate.

[@kaaiplayspiano7200]

(1+sqrt(5))/2*

[@fmakofmako]

@@kaaiplayspiano7200 probably I should have clarified, because I assumed everyone knew this. The formula for large n for fibonacci is phi^n/sqrt(5). If you watched the first few minutes of the video, then there is a moment when he takes Phi to a power and compares it to the actual value and then hand waves the differences although notes the magnitude is the same. That's what would have been a lot closer if he divided by root 5.

[@ashtonhoward5582]

@@fmakofmako it took me a bit to figure out what phi was. Also, for that you need to round to the nearest integer.

[@luisliquete9027]

fmakofmako it works perfectly!! Now, im wondering if the 5 from sqrt(5) may have any relation with the golden ratio that appears in PENTAgons (i dont remember in which exact video i saw this). I might be setting my expectations a bit too high on this one, but who knows??!! 🧐🧐

[@mastod0n1]

@@luisliquete9027 I assume the square root of 5 is coming from the fraction representation of the golden ratio, which (1+sqrt(5))/2

[@rattyoman]

the way he said "do you wanna hear about applications of fibonacci sequences?" at the end was so precious

[@leonhardeuler9839]

I see the word “random” and James, I click instantly.

[@danfg7215]

it had me at Fibonacci

[@tomaszkantoch4426]

Why. James is not just a random guy :D

[@starsian]

I see Euler, I like instantly.

[@leonhardeuler9839]

Jay N Who?

[@leadnitrate2194]

Jay N but not as proficient.

[@ItachiUchiha-ns1il]

I’m curious as to whether or not the growth rate is transcendental or algebraic for the random Fibonacci.

[@Gooberpatrol66]

9:53 I love that they defined an IEEE double float in a mathematical paper

[@otakuribo]

Audrey: 🐶 *casually enters room James, excitedly: *Do you want to hear about applications of the Fibonacci sequence?*

[@exoplanets]

.

[@Triantalex]

??

[@danieloh6782]

Was just watching James Grime! Always look forward to his videos :>

[@Exeedo.]

What I have been thinking while watching this video is: what would happen if the probability between getting a plus or a minus is not 50:50? I think that will be interesting to find a formula/algorithm to find what number the growth rate approach depending on the probability.

[@ragnkja]

The “almost surely” looks a lot like the “almost all” in the video about how almost all numbers contain the digit 3.

[@karapuzo1]

That was some exemplary handwaving, I hope the original paper is more rigorous.

[@calculator7774]

@@karapuzo1 Actually, something happening "almost surely" is a rigorous mathematical phrase. It means the odds of it happening are 100%. Even though it is somewhat counterintuitive, this is not the same as saying that it will always happen, hence the "almost".

[@karapuzo1]

@@calculator7774 Yes, I am aware. This still requires rigorous proof that the probability of the examples where the ratio does not converge is 0.

[@OlliWilkman]

"Almost all" in terms of real numbers means that the set of counterexamples have a measure of zero. Since probability theory is usually formulated in terms of measure theory too, I suspect the analogy is that "almost surely" means that the probability measure of the inverse statement approaches zero?

[@dlevi67]

@@OlliWilkman If the cases in which it does not converge to Viswanath's constant are those and only those with patterns, those can be equated to rational numbers in binary, which have measure 0 in the set of the reals ("all" random binary digit numbers between 0 and 1)

[@bhooshanpandit1344]

Yes!!! Please *MORE JAMES GRIME* & *MATT PARKER* & *TONY PADILLA* !!!

[@gabor6259]

*& Hannah Fry & Tadashi Tokieda & Cliff Stoll & Holly Krieger & Simon Pampena & Zvezdalina Stankova & Ben Sparks & ...*

[@exoplanets]

Yes!

[@shugaroony]

This vid is just like classic Numberphile. James Grime, brown paper, and no fancy graphics with silly sounds. Thumbs up. :)

[@Justuy]

And I am wondering why he isn't calling it 'RANDOMACCI NUMBERS'!

[@johnchessant3012]

Fun fact: The nth Fibonacci number is given exactly by rounding (phi)^n / sqrt(5).

[@shambosaha9727]

Ah, yes. Of course you can exactly figure out what the answer almost is.

[@andywright8803]

@@shambosaha9727 no, you can figure out what it is EXACTLY. It's (phi)^n / sqrt(5) rounded to nearest whole number

[@shambosaha9727]

@@andywright8803 But you're still rounding

[@CauchyIntegralFormula]

Yeah, but (phi)^n/sqrt(5) isn't exactly F_n. The closest integer to (phi)^n/sqrt(5) is exactly F_n

[@shambosaha9727]

@@CauchyIntegralFormula That's what I said.

[@flikkie72]

To get a (random) sequence of this in excel: Put 1s in cells A1 and A2 and put "=A1+A2*(-1)^round(rand(),0)" in A3 and pull it down.

[@alainrogez8485]

0:08 "We have to recap the Fibonacci sequence first". Didn't Numberphile do hundred of videos about this sequence, did it?

[@ragnkja]

Alain Rogez Just in case this is someone’s first Numberphile video about Fibonacci numbers.

[@Qbe_Root]

Each Numberphile video about the Fibonacci sequence is equal to the sum of the previous two

[@yayaaabunni]

I swear if JG was my math professor, I would be front row, every clad and turn in every assignment.

[@Tehom1]

For a moment I thought there was some likelihood of getting a tail of all 0s. Because James only needed to hit 0 once more, and after that he's always adding or subtracting two 0s and getting 0. But on reflection, that's impossible. Once you have a nonzero X followed by a 0, the next one is either X or -X and never 0.

[@shadowhejhog]

oh that’s fascinating! i thought it would end up as 0,0,0,... too.

[@thej3799]

You're moving into the ratio of what defines a dimension

[@TXKurt]

To those writing programs: Look at the expression at time 6:45. |R_n|^(1/n). I believe this is what you want to average. By averaging this over many random series, already with n=40 you should get around 1.125. Going up to n=1000, the result is starting to look familiar: 1.13174.. (400000 series averaged).

[@PC_Simo]

2:16 Of course, it has to be a rough estimate; since, for any finite n (i.e. any n that you can actually pick), F(n+1)/F(n) ≠ φ. That’s, because φ = (√5 + 1)/2 is an irrational number, and F(n+1)/F(n), by definition, is rational.

[@senororlando2]

Damn Dr. Jimmy hasn’t aged a day in 10 years

[@martinbergman7693]

Are those framed pictures in the background (which has been standing on the floor for many years, it seems) ever going to be hung on a wall? That's what I want to know.

[@Danicker]

I love how "almost surely" is a techincal mathematical expression. Great vid as usual!

[@sheerrmaan]

I love his passion and his way of communicate things

[@123amsterdan456]

Love the Grahams number paper on the wall :)

[@someoneunknown6553]

I've been binge watching the James Grime playlist

[@AaronRotenberg]

7:45 is a little misleading. "Almost surely" means the probability is _exactly_ 100%, not "almost" 100%. The catch is that, for infinite sequences of events, 100% probability ≠ guaranteed.

[@ragnkja]

Aaron Rotenberg This is explained very well in “3 is everywhere”, one of the first videos on this channel.

[@flexico64]

I was thinking of the explanation involving a dart board that explained some things are "possible but have 0% probability." *brain starts smoking*

[@delores1656]

I could listen to James explain anything for hours.

[@randomdude9135]

Divakar Viswanath It's rare to see a Mathematician with an Indian name on Numberphile. And I'm happy to know about his finding 😊

[@erickcapitanio1957]

and an awesome sounding name on top of that

[@harikishanrakhade6108]

And what about Kaprekar who gave the Kaprekar constant?

[@DeeEm2K]

@@harikishanrakhade6108 alright calm down this isn't a contest

[@davidgillies620]

Off the top of my head I can think of Bose (Bose-Einstein distribution), Chandrashekar (white dwarfs), Varadarajan (supersymmetry) and Agrawal (AKS primality test). India is rightly celebrated for the calibre of mathematicians it turns out.

[@btf_flotsam478]

@@davidgillies620 I could think of Rao off the top of mine; I think it says something about my time studying statistics that I didn't think of the others, despite having heard of the first two (I also think you not naming him says something about statistics in mathematics, but that's a different kettle of fish).

[@romekhanna]

He's back... Love his passion

[@danuttall]

9:30 "Wouldn't the pluses and minuses just cancel out over the long run? No they don't." Cosmologists have been asking that question too, when it comes to matter and antimatter. Because they don't quite cancel out, we have lots more matter in the universe than antimatter.

[@UnkleRiceYo]

1:14 why have you included almost sure convergence? Surely that should just be normal convergence for sequences?

[@TheKemalozgur]

2:20 Actually you can increase the accuracy dramatically by not using f(1)*g^1000000. Because obviously in the first ones golden ratio is nearly meaningless. Lets say, starting from fifth fibo; f(5)*g^999995. Result becomes 1.9691 x 10^208987, which is very close to actual. Or; f(10)*g^999990 ~= 1.9531 x 10^208987.

[@scottanderson8167]

Yay!! A Grimy video!! Love James Grimes!

[@ayaipeeoiiu8151]

Is there a curve to know what’s the ratio for any randomnacci sequence (i mean for the regular fibonacci it’s (0;1)->1,6180339887... and for (0,5;0,5)->Viswanathan’s constant) and we can try to understand the pattern and maybe find a formula for these constants

[@LelouchLothric]

James is so amazing!

[@nO_d3N1AL]

It's amazing how quickly Fibonacci tends towards the Golden Ratio: 1, 2, 1.5, 1.667, 1.6, 1.625, 1.615...

[@KevsCoolProductions]

google binet's formula if you wanna see why

[@starsian]

yes, because the growth is exponential

[@aldobernaltvbernal8745]

pick 2 random numbers and use them as your starting point, then do the same thing as you would do in the fibonacci sequence, it will still converge to the golden ratio.

[@zanedobler]

It actually doesn't approximate it all that quickly. The golden ratio has the simplest continued fraction, making it the most irrational number.

[@arcanely]

What if the chance of a + was 2/3 and - was 1/3? What would the new growth rate be? Is there a way to generalize the growth rate for different probabilities?

[@GoingsOn]

This Viswanath’s constant kind of reminds me of Mills’ constant θ, being hard to calculate and also being based on integer sequences.

[@Alytreta]

469 Tibia language (or Bonelords language) Please we need your help to solve it

[@borisvaleev4268]

Question. If there is no difference between 1st 3:30 and 4th 3:57 iteration, how you can predict 1000000th?

[@LaytonBehelit]

He means there is a trend (an expected value) which you can predict, even if in the short term this trend isn't necessarily visible because of randomness. So predicting the e.g. 15th term using this method is much less useful/accurate than predicting the millionth term.

[@mementomori7160]

I have a question related to this and another one of your videos In one video you showed up that no matter with which numbers you start if you stick to the fibonacci sequence's rule, you will still get the golden ratio. My question is: does it work with random fibbonacci too? Do you get the same ratio?

[@seracol1888]

I ran this in a script, and picked the numbers 54 , 78 as starting points and the ratio still seemed to converge towards 1.131

[@ianflanagan209]

1, i, 1+i, 1+2i, 2+3i, 3+5i, 5+8i....creates 2 fib sequences simultaneously as the real and imaginary parts follow the fib sequence. this could be seen as the sum of 2 separate fib sequences, fib real+fib imaginary which is (f_n+1)+(f_n+1)i=((f_n-1)+(f_n)+(i*f_n+1)+(i*f_n)). The cool thing about this is we get a+bi format which can be represented as re^i*t and the limit of the respective sequences is phi and i*phi so we have a system that combine 3 of the most important constants in math phi, e and i.

[@MrPictor]

6:55 What's the back story behind this pigeon?

[@SimonTiger]

No. phi^1000000 gives the 1000000th _lucas number,_ not fibonacci number. In order to get the fibonacci number, you have to divide by root 5.

[@antoniodagostino5891]

Your comment isn't completely correct phi^1000000 isn't equal to the 1000000th Lucas number, it gives the 1000001th Lucas number instead. 1.618 x 1.618 x 1.618 x 1.618 for example, doesn't give as result the fourth Lucas number, but the fifth. So in order to get the 1000000th Fibonacci number, you need to divide the 1000001st Lucas number by √5. Your asnwer is true only if we assure that 0 is included in Fibonacci sequence (but the video doesn't include it).

[@diegorattaggi2095]

This is not correct. phi^1000000 is L_1000000 - phi^(-1000000) which is approximately L_1000000.

[@GeorgeVajagich]

Do we know if the number is algebraic or transcendental?

[@nin10dorox]

2:23 The approximation with F can be a lot more accurate. Phi^x becomes proportional to the golden ratio, but doesnt approach it. If you multiply by (phi + 2) / 5, the approximation will be incredibly close.

[@torreywhiting5402]

2:00 to put that into perspective (kinda), the largest number Java bothers going to is 10^1024 (anything more than that and it shows the word Infinity) 10^1024 isn't even *close* to the number he showed

[@jillkitten5388]

Wish they would have more formally mentioned: Fib(n) = Round(Phi ^ n / Sqr(5)) which gives the nth Fibonacci number.

[@bobbycraig2583]

round() is a python function.

[@jillkitten5388]

@@bobbycraig2583 It is not just a python function, it is a function in almost all programming languages [in one form or another], the point is that it is the common round function/procedure which in text is hard to represent the mathematical symbolism in an unambiguous way, so by representing it as a function called "Round()" is the simplest most unambiguous way to represent it.

[@bobbycraig2583]

@@jillkitten5388 i know but i only know python. i use [ ] to show rounding

[@aaaichunder]

Apparently i am the thirteenth viewer.... I wonder where the 1st, 1st, 2nd, 3rd, 5th and 8th is?

[@randomdude9135]

7290 here

[@Tehom1]

Because having two people who are "first!" is just normal.

[@leadnitrate2194]

Tehom it really is, though with the number of people exclaiming “First” in comment sections these days.

[@exoplanets]

:O

[@VidNudistKid]

Your comment had 89 likes before I got to it

[@johannesh7610]

The fibonacci number Fn = (φ^n + φ'^n) /sqrt(5), where φ/φ' = (1 ± sqrt(5))/2. This is the exact formula

[@triskel20]

"And thats as far as we got" - - Amazing ending, for a moment I thought he was going to say they discovered hundreds more digits!

[@orange-micro-fiber9740]

Feels similar to a 1D random walk, but random walks are usually just 1 unit.

[@flytoheights1]

Love your videos as always!

[@kevina5337]

Good to see Dr. Grime is still alive... haven't seen him in a while lol

[@sam111880]

You can also apply this random fib sequences to one dimensional random walks of particular step patterns rules. Pretty cool stuff 👍

[@kajdronm.8887]

Has this something to do with kolmogorov's zero-one law? Is the radius converging to a given value a tail event?

[@btf_flotsam478]

I wouldn't be surprised. I would expect an argument to go something along the lines of how a Fibonacci series with any starting pair of whole numbers would tend towards the golden ratio, and then use that to justify how the end ratio would be independent of the initial terms.

[@omidmirzaeeyazdi6236]

8:40 0-1=-1 and it's not gonna go beck to the first number !

[@joshualowe6950]

2:28 that’s the 1000001th number though. To find the 1st term phi^0 as you take 1 and do not multiple by phi to get it. To get the 2nd term, 1xphi^1 (approx obviously), 3rd term 1xphi^2 so 1000000th term is 1xphi^999999. Fn ~= phi^(n-1)

[@yogipro183]

Fibonacci series is also known as Hemachandra series in India since Hemachandra proposed this series very much earlier with fantastic application in music and architecture of statues.

[@dskinner6263]

I'm having trouble understanding - at 2:25 he's saying the approximation and the actual number are very close. But isn't one literally more than twice as large as the other? That's like me saying that pi and tau are approximately the same. ;)

[@yvesdelombaerde5909]

Even if you start de Fib. sequence with random numbers you get the Fn/Fn-1 -> phi 1.618… . So phi is the consequence of the cumulative summing process and not really related to the numbers in itself (0,1,1,2,3,5,8,…).

[@klaasbil8459]

One could improve on the formula given at 1:58, by skipping the first part of the Fibonacci series where the ratios betweeen successive numbers are quite far from the Golden Ratio. As an example, the 20th Fibonacci number is 6764. So phi^1000000 could be approximated by 6765 * phi^999980. Skipping the first 1000 is most likely even better. Keep in mind that you need a lot of decimals in your phi value to get close to the correct answer!

[@vsm1456]

What if you have an unfair coin? Can you put probabilities in the equation? At 100% "+", 0% "-" we get 1.618... growth ratio. At 50%/50% we get ~1.132 ratio. At 33,3%/66,7% do we get ~1 ratio which means it barely grows? Or does it only happen in plus-minus-munis pattern and every randomisation makes it grow, even if only slightly? It looks like the ratio starts to grow again if we go further: At 0%/100% (i.e. always minus) we get 1,618... again, but with 3 positions delay (1,1,0 then -1,-1,-2,-3,-5, etc.)

[@fulltimeslackerii8229]

Since the Fibonacci sequence starts with 1,1,2..... wouldnt it be more accurate to do 2 x Golden)^n since 2 is kind of a better starting point? That would have been a much closer approximation for the example

[@androido7487]

I was just wondering, what would happen if we'd gradually modify the probability of the occurrence of the operators (eg.0.6, 0.7,...) . Would we get some other constants? And if so, is there any relationship between these constants?

[@samiramin5895]

So the magnitude of Tn grows exponentially, but what does the probability distribution for a large n look like? Is it bimodal with peaks at +/- 1.132^n ? Second, do the two starting digits make a difference?

[@elliwesishawkins4799]

If you were to solve backwards for Fibonacci sequence, where as knowing that fn=f(n-1)+f(n-2) could be arranged so that determining, say, the numbers before the sequence officially starts and going backward would be finding f(n-2) as the number solved is behind the sequence. So the starting numbers of value 1, the 1 again we know the number before then would be 0, (so that 0+1=1) then before that would be 1, again before would be -1, then backwards to 2, backwards to -3, ect where it alternates between positive and negative numbers. Following this patterns backwards creates a mirror of the Fibonacci sequence where every other number is negative. And by “mirror” I mean reversely ordered from 0 as it comes from, presumably, a negative infinity to add its way down to 0, then back up to positive infinity in the recognized Fibonacci sequence.

[@martinepstein9826]

When you go backwards you get the Fibonacci sequence again but with alternating sign. So the ratio F(n+1)/F(n), where n is large and negative, approaches -0.618... i.e. the negative reciprocal of the golden ratio. This makes sense because 1.618... and -0.618... are the two eigenvalues of the matrix [1 1; 1 0] which is the matrix that generates the Fibonacci sequence.

[@elliwesishawkins4799]

Martin Epstein thank you, I was doing it myself and edited mine only to see yours and so mine now reflects yours. I appreciate you responding lol, I was about to do some crazy math myself.

[@elliwesishawkins4799]

Martin Epstein also my recognition that it was backwards and every other negative wouldn’t have given me the actual ratio, thank you very much

[@DStecks]

It makes intuitive sense that the sequence will (almost) always grow because only very specific patterns will keep the values low, and if the values ever get larger, they compound. So only a tiny sliver of the possible results don't result in growth.

[@vornamenachname5267]

Does the sequence hower around ± constant^n or is that just the magnitude of the peaks? I would imagine that it howers around 0 the entire time.

[@Dalroc]

8:53 There's your answer

[@jeremydavis3631]

I wasn't sure at first, but you're right. The sign is expected to change frequently, regardless of the magnitude. If the previous two terms are a and b and two consecutive minuses appear, the next two become (b-a) and (-a). That has a 1-in-4 chance of occurring for any given pair of consecutive terms, so it won't ever settle on either sign (almost certainly).

[@lee45283]

Hey, I really want that portrait of the pigeon in the background, what’s it called and who’s it by?

[@waynesalvador9925]

So you should be able to discern whether a growth pattern is natural/random or artificial/manufactured by comparing which growth curve is actually followed?

[@lechutl]

1:57 shouldn't it be phi^999999 as it is roughly the 1000000th term of a geometric sequence with first term 1 and common ratio phi?

[@booli8542]

Not if the first term is F0.

[@robin888official]

You have to divide phi^1000000 by sqrt(5) to get the 1000000th fibonacci number. (By an error of only 0.618^1000000, so the formula gets extremely accurate!)

[@mdnpascual]

Can you use this as a measure to quantify how "random" a random generator can be?

[@btf_flotsam478]

In theory, yes. In practice... there are tons of ways to do that, and a lot of them can pick up a lot finer patterns.

[@Veptis]

So any binary number can be made into one of these sequences. And each number has a growth ratio (if you take the rear most element or the moving average). That means you can map integers to real numbers? But you can't as every integer has a limited number of binary digit so it's not an infinite sequence. But if you inverse numbers, you can map reals to reals using this and find some interesting bits of bijictives.

[@kinyutaka]

One thing I see when I make a random sequence is a lot of resets, which is obviously what keeps the sequence within a general boundary that's lower than Fibonacci. But going "only" to R1000, I'm seeing a lot of variance, approximately 5e43 to 5e64 (with exactly half of the results of 100 tests over the expected figure of 5.67e53) Perhaps there is a second, shrinking value to add or multiply to the 1.1319^n equation to better get the "true" upper boundary with better accuracy? Not sure what that would be yet.

[@ghgh6612345]

If exactly half the results are over the expected average, surely it's working perfectly then?

[@kinyutaka]

@@ghgh6612345 Definitely giving an appropriate trend.

[@DiamondzFinder_]

If you had a weighted coin, could you set the likelyhood of heads or tails such that you'd get different growth ratios? And if so, which growth ratios could you get?

[@j.vonhogen9650]

So, this random Fibonacci growth ratio, will it give you a minimum Fibonacci number, or an average Fibonacci number? I'm afraid I don't quite understand the meaning of this ratio in terms of practical use.

[@j.vonhogen9650]

@King Pistachion - Not sure if you were trying to answer my question, or if you wanted to tell me something else.

[@laojackos]

James Grime yessssss

[@kyraaa__]

I’m going to code this :D

[@numberphile]

Give us some examples of what you get for n=1999,999 and 200,000 and how close it is to the constant!?

[@bananoramatfw]

@@gonzalogarcia6517 bro what are you on about

[@Maniclout]

@@gonzalogarcia6517 * visible confusion *

[@bananoramatfw]

@@gonzalogarcia6517 this comment is pristine

[@bhooshanpandit1344]

**shook**

[@beliasphyre3497]

How about a histogram depicting the possibilities at the nth iteration?

[@krishnachattopadhyay3251]

There's no need of doing a toss there(3:38). The next term will be 1 in both the cases

[@ragnkja]

Abhirup Chattopadhyay No, it can be -1.

[@randomxnp]

So what proportion of sequences are going to diverge to positive numbers and what proportion diverge to negative numbers? Given you start with positive numbers you're more likely to end up positive.

[@vadimmy]

Починали завжди з 1? Чи буде ріст якщо починати з -1?

[@DanDart]

Did you forget to divide by root5 there at the beginning?

[@Frownlandia]

I wonder what happens if you reverse the order, where it's R(n+1)=R(n-1)±R(n)? My instinct is that the constant would be smaller, but would it be less than 1?

[@Maharani1991]

Easily the most mindblowing fact in this video to me is that 1999 was twenty years ago...

[@Johan323232]

A fun puzzle with Fibonacci numbers. If you take the Fibonacci recurrence and start with two positive numbers, it goes to infinity. If you start it with two negative numbers it goes to zero. However there are numbers you can start the sequence with and have it go to zero. Finding them is a fun way to practice computing recursion limits.

[@byteridr]

What is if you do that with complex numbers. At first you take the length of one with a random angle and take this as your first complex number. Then you take the length of one with another random angle as your second complex number and add them to get the absolute value of your third complex number. You then cohoose another random angle to have your third complex number complete. Then add second and third and so on...

[@mitchellboyce9853]

Does James have the brown paper from Ron Graham explaining Graham's Number framed on his wall? Or is this not filmed at James's place?

[@AndogaSpock]

Hello Numberphile team, could you make a video about "minimal covering of pairs by triplets" may be work your way up to "minimal covering of pairs by octets".