Featuring Neil Sloane from the OEIS. Full “Amazing Graphs Trilogy” and extras at:

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Neil Sloane is founder of the OEIS:

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Amazing Graphs

Part 1:

Part 2:

Part 3:

Extra Bit:

Sequences featured in this video include…

Stern’s:

Hofstadter’s:

And Remy’s:

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Watch the full Amazing Graphs Trilogy (plus an extra bit): https://www.youtube.com/playlist?list=PLt5AfwLFPxWLkoPqhxvuA8183hh1rBnG

what are the odds that a given sequence will be satisfying?

That last sequence graph looks a bit like the first ionisation energy for the elements in sequence. They increase exponentially from Hydrogen, helium, lithium gets a slight drop as you have gone from 1s to 2s. Beryllium, boron gets a drop as you go to 2p, carbon, nitrogen, oxygen gets a drop as 2px now has 2 electrons, fluorine, neon, sodium gets a slight drop as you move to 3s, calcium and on it goes. The slight drops get longer in interval.

What creates the repeating structure of the Q-sequence?

1:47 Sagrada Familia, anyone?

8:15 It's probably not the same one, but I wouldn't be entirely surprised if that was "the" Adam Savage, of Mythbusters, as a Patreon supporter.

Such magic – thank you for a great introduction. Off to play with overlapping binary place holders 🙂 brilliant stuff

Please tell me Neil did the artwork for this series. Those stylised portraits constitute some amazing graph-ics.

"ill put down a 0, no-ones is going to object to that"

The subtle sarcasm is strong with this one.

MORE MORE

Love this little series

That laptop balanced on top of a pile of books is so upsetting.

This man is now 80 years old. Incredible.

MORE AMAZING GRAPHS PLEASE

"You could die at any moment."

Momento Mori

7:45 this one has like a shadow offset of the Sierpinski triangle but infinitely repeated…

More Neil, please. He’s always interesting and he’s a very nice person!

adam savage is one of your patreon what the heck

At 1:52 looks exactly like Sagrada Familia, take a look!

What program do they use to make these graphs?

brilliant, thank you!

This might be my favorite playlist on YouTube

4:50 fibonachi exists in negative values, and so we wouldn't have any problems if we needed the -3rd number of the fibonnachi sequence, for example

"it could die at any moment… beautiful"

haha

Who else is watching at 3 am?

Neil sloane is my new favourite numberphile appearance 🙂 I can listen to him for hours! Not that I understand 90% of what he says, but I still like to listen 🙂

The graph for the Rémy Sigrist function appears to have curvy Sierpenski-looking objects…

Wow it sounds very ASMR

He should do a math ASMR channel

it s like pascals triangle… no no it s actually like fareys series 😀

But have you checked past 10^(10^40) where the Mertens Conjecture breaks down?

*****as it can be seen, basically all the graphs have fractal character – did someone try to calculate the ratio between the size (or beginning of the location) of the repeated parts of the graphs? is it something like feigenbaum constant ??? does the ratio converges in a such graph? this wants the sequences to be calculated very far away (some to milions, some to bilions terms), but the ratio in the fractal graphs could be interesting, is it universal? is it special for every sequence with this kind of graph with fractal character?

*****You can see versions of Pascal’s Triangle in the Alps graph.

Did I see a serpinski triangle in the 'snow' of the remi segrist graph?

Damn. that last one be looking like a Bob Ross painting

The alps are my favourite graph of all the videos, very cool (no pun intended)

@ 8:36 “plots…in yo asszzzz”

this is my favourite channel to watch at 3am

0:12 No U

My life is complete 7:40

Neil Sloane is such an inspiration.

Hofstadter's Q Sequence bothers me. I've programmed and found that when it "rescatters" on the graph, there is a high point and a low point, before the graph comes back together again along a slope of approximately y=2x. That is, for every n'th time the graphical represenation of the data rescatters, the highest point of that scatter will predictably be at (2^n + 2^(n-1), 2^n). The lowest point of the n'th scatter will also be found at (2^n + 2^(n-1) + 1, f(n) < 2^n)… However I can't find that function for that low point.

7:54

That's in my nightmares now.

His wallpaper reminds me of whataburger…

Its 3 am and I have surgery lecture at 7 am BUT I CAN'T STOP WATCHING I DON'T EVEN TAKE MATHS

Wouldn't it be a nice empirical thing to plot sequences by a computer (in more dimensions) until we find the Riemann Zeta function?

Next 4th of July I'm coming here instead of watching fireworks.

WHY DON'T THEY SHOW THE RELEASE DATE OF THE VIDEOS?