I don't know how this video only has a thousand views. This is probably one of the best explanations for so many aspects in mathematics in an understandable way for pretty much any audience, which basically no mathematics is like. Not only is this something beautiful, it's also extremely simple to understand even though I've never seen something like this before. Fantastic work!
This episode of ‘Andres tries to play Minecraft’ took an unexpected turn. In all seriousness this video is quite amazing! To start off with, was this all recorded in one take? Either way, I personally find that what makes a good RU-vid video is the passion behind it. And even though I previously had no interest in the mysteries of the Fibonacci sequence, the clear purpose and work put into this video has emanated unto me for an intrigue that’s sustainable enough to look into the subject beyond this video. If that’s not a good video, I’m not sure what is. I hope you continue with these, not only because of our shared early intrigue into the randomness of a single word, but the clear dedication into the video. Great work!
mod 2 gives [0,1,1],0,1,1,... mod 5 gives [0,1,1,2,3,0,3,3,1,4,0,4,4,3,2,0,2,2,4,1],0,1,1,... 3 terms before it repeats mod 2, 20 terms before it repeats mod 5, and since 3 and 20 are coprime, 3×20=60 terms before it repeats mod 10.
Great rule. Another example I can see is mod 7 gives 16, so 3x16=48 terms before it repeats mod 14. An apparent rule I found by myself thus far applies just to integer powers of the modular base. For example mod 2^1 gives 3 terms as you say, 2^2 gives 6, 2^3 gives 12, so increasing by a factor of 2 each time. For mod 3 it increases by a factor of 3 each time, and so on (for prime modular bases anyway). I don't know how to put this algebraically. Does it derive from yours? The Pisano periods for modular bases is given by OEIS A001175. A rule stated there is "Index the Fibonacci numbers so that 3 is the fourth number. If the modulo base is a Fibonacci number (>=3) with an even index, the period is twice the index. If the base is a Fibonacci number (>=5) with an odd index, the period is 4 times the index. - Kerry Mitchell, Dec 11 2005" Another intriguing pattern in that list is that prime modular bases ending in 1, that is 11, 31, 41, 61 each give just 1 fewer terms.
I just found the coprime rule also seems to hold for the tribonacci sequence (where you add not 2 but 3 numbers together). For example if the first 3 digits are 001, then in mod 2 it repeats after 4 terms, and in mod 3 after 13 terms, so in mod 6 after 4x13=52 terms. If the first 3 digits are 010, then in mod 2 it repeats after 2 terms and after 13 in mod 3, so 26 in mod 6. I verified this using "brute force".
i did some code and some searching, these are called Pisano periods (OEIS A001175). for small numbers at least, if x is not divisible by y, A(x*y) is a multiple of the lcm of A(x) and A(y) edit: someone found A001175 before me, i should’ve read
@@coppertones7093 You're the first person I've come across besides myself in RU-vid discussions of Pisano periods who's also mentioned OEIS, even though it's such a great resource. They've also got one on the Tribonacci (where you add 3 numbers in a row, not two), A046738, not to mention other sequences. I hope to find some more interesting patterns of the sort you describe Going back to the Fibonacci, as you probably know, the numbers in OEIS A001175 are only for those sequences beginning 0 1 in any given modular base (MB), because that's how the Fibonacci begins. But in MB 10 for example there are other, entirely separate, sequences that begin with different number pairs, and have different Pisano periods. There are 6 such sequences in all, beginning 00, 01, 02, 05, 13, 26. I constructed a table showing how the digits 0 through 9, down the left hand side, are distributed over these sequences. 00 01 02 05 13 26 0 1 4 4 1 0 0 1 0 8 0 0 2 0 2 0 4 4 0 1 1 3 0 8 0 0 2 0 4 0 4 4 0 1 1 5 0 8 0 2 0 0 6 0 4 4 0 1 1 7 0 8 0 0 2 0 8 0 4 4 0 1 1 9 0 8 0 0 2 0 If you're a coder (which I'm not I'm sorry to say) it would be an interesting project do the same thing for the various sequences of the Trbonacci, MB 10, for which there are 20 separate sequences. The one beginning 001 has a Pisano period of 124, as listed in OEIS. I figured out the Pisano periods for the other 19 myself, but not yet a table like the one above.
This also means the fibonnacci has to repeat in base googol. That means, in base 10, the last 100 digits eventually repeat. That means there exists a fibonacci number such that: Last few digits ...00000000 Next fibonacci's last few digits ...00000001 Next fibonacci's last few digits ...00000001 Next fibonacci's last few digits ...00000002 Next fibonacci's last few digits ...00000003 Next fibonacci's last few digits ...00000005 Next fibonacci's last few digits ...00000008 Next fibonacci's last few digits ...00000013 And so on
wait, why do numbers and their corresponding number on the opposite side of the circle always add to 10? (probably something to do with base 10, but like, what lol?)
That first 60 Ring showed even numbers (0, 2, 4, 6, and 8) each 4 times, and odd numbers (1, 3, 5, 7, and 9) 8 times each. Not sure what to make of that if anything.
Here's a table I constructed showing the distribution of base ten digits 0 - 9 (down the left hand side) over the 6 separate sequences, identified along the top by their starting pairs of numbers. 00 01 02 05 13 26 0 1 4 4 1 0 0 1 0 8 0 0 2 0 2 0 4 4 0 1 1 3 0 8 0 0 2 0 4 0 4 4 0 1 1 5 0 8 0 2 0 0 6 0 4 4 0 1 1 7 0 8 0 0 2 0 8 0 4 4 0 1 1 9 0 8 0 0 2 0 I think the numbers you're talking about are listed under 01, which is how the standard Fibonacci sequence starts. But there are other possible separate sequences in base 10, such as the one starting 13 which has even numbers (2,4,6,8) each 1 time, and odd numbers (1,3,7,9) 2 times each. Note that each row adds up to 10. Hope that puts your question in context.
What about other sequences, like the Tribonacci for example where you add up three numbers in a row? In base ten starting 0 0 1 you continue 1 2 4 7 and so on, and get a cycle or repeating pattern of 124 final digits. In base nineteen you get a cycle of 360 final digits, which is surely just as interesting as the cycle of 60 in base ten Fibonacci.
I think the more agreed-upon notation for integers mod n is Z_n (where _ represents a subscript) and the elements are written [1], [2], up to [n-1]. Great video nonetheless, you deserve way more views
That's a good point! We stuck with this notation (which is at least as prevalent as ℤ_n) because we wanted to emphasize that, for example, the [2] in ℤ/3ℤ is different from the [2] in ℤ/6ℤ. In retrospect there were probably easier ways to do that... maybe [2]_3 and [2]_6?
At 35:25, isn't it enough that one of the elements on the RHS be (0+NZ)? Since multiplication works within like terms, it seems like (0+2Z)*(a+3Z) should be (0+6Z) for any a. What have I missed?
I watched literally two minutes more of the video and answered my own question. We're just using a different rulebook where what matters is that the isomorphism holds. Otherwise (0+2Z)*(2+3Z) would also be (0+6Z) when we need it to be (2+6Z), and the entire thing just wouldn't work.
Every 60 Fibonacci numbers a pattern of final numbers will repeat. Every 12 Lucas numbers too. And every 20 numbers in the sequence that includes the pair (2,2). Every 4 numbers in the sequence that includes (2,6). Every 3 numbers in the sequence that includes (5,0). Every 1 number in that which includes (0,0). 60 + 12 + 20 + 4 + 3 + 1 = 100, which is the number of possible pairs to be made from the digits 0 through 9 in the decimal system.
@@jocabulous OK I'll try and spell it out. If you begin your sequence with 0, 1, and go on 1, 2, 3, 5, etc then at the 61st and 62nd numbers you'll find that the final digits of the numbers you get start being 0, 1, 1, 2, 3, 5 etc again. The same will happen with the 121st and 122nd numbers. The final digits will begin to repeat the same pattern every 60 numbers in the Fibonacci sequence. That's what the presenter points out at the beginning of the vid. However 0,1 aren't the only two numbers you can start with. There's also a sequence, known as the Lucas sequence, which begins 2,1 and continues 3, 4, 7, 11, 18, 29, 47, 76, 123, 199, 322, 521, etc. As you can see the final digit sequence starts repeating much earlier, every 12 numbers in fact. What about other possible starting pairs of numbers? Well, if we try 3, 1 and start adding to get 4, 5, 9, etc we'll eventually get to numbers ending 0, 1, 1, 2, 3, 5 etc. This means that we've really just been doing the Fibonacci sequence again, not a separate sequence like the Lucas. The same applies to 4, 1 and 5, 1 and indeed to all sequences containing possible pairs of digits in the decimal system of counting, which amount to 100, except for those containing 2,1 (which we've already talked about); 2, 2; 2, 6; 5,0; 0,0. which have cycles of 12, 20, 4, 3, and 1 respectively. (If you start 0, 0 then the final digit starts repeating immediately which means the cycle is just 1.) If you add all these together along with the Fibonacci cycle of 60, you get 100. This means we've accounted for all the possible digit pairs in the decimal base, 00, 01, ....99. I hope that's a bit clearer, and of course please check. This is something I figured out myself, and I may have made a blooper.
As someone who has dedicated dozends of hours to fuigering out the secrets of the pisano periode, fibonacci prims, and more: this is a good video to get into the toppic. I would just recomend unerstanding the formal ways of wrighting Honnycomb and thins like ggt(). Otherwise A verry good video, eventho I dind't learn anything new to me
