Caboose Numbers - Numberphile

NUMBERPHILE

Videos by Brady Haran

Thanks to fact checkers from The Numberphile Society, including Debbie Chakour and CJ Smith.

[descuddlebat]

"Honorary, near caboose" - definitely what people will call a near miss found by Matt

[Neefew]

5:14 Matt can join a long line of mathematicians who when studying an interesting piece of maths discover that the work has already been done extensively by Euler

[AndrewVanner]

101 can be the first and only Parker Caboose Number.

[physics]

10^n + 37 is prime for an inordinate number of integers n. My favorite prime is 10^39+37 which is a one followed by 37 zeros and then the number 37. If found this incidentally in the context of some research I was doing.

[DirkThys]

The dog in the background is thinking: "this Python code is gonna take ages - time for a nap"

[shambobasu]

Using red pen to mark correct while the green pen is right next to Such a Parker move 👏.

[JohnJohn]

Hi. Train enthusiast here. 2 issues with the animation:
1. Your boxcars shouldn't have 8 axles. Probably 4 axles is plenty for the sort of train you're drawing.
2. Wheels on a steam locomotive have rods connecting them to the pistons, but wheels on the cars don't.

Okay, I'll sit down now.

[Tanmark]

This would be a good moment for a sequel to "people online made my code 40 million % more efficient".

[mstmar]

any caboose number has to be prime. if it's a composite number, say a*b, then a^2-a+ab = a * ( a - 1 + b ) which is 2 factors greater than 1 (since a and b are greater than 1).

[TheRubySpider]

Two small observations that Matt didn’t outright say.
A caboose number has to itself be prime, because the caboose outputs itself for n=0 and 1.
And to generalize the logic about n=42, any n greater than the caboose by a square number would create a difference of squares, and therefore output a non-prime.

[richardfarrer]

One way to speed up the program. n^2 - n + c is not prime for (nearly all) number which are not coprime with c. So check c for primality first and don't try to calculate the rest if c is not prime. in addition, (n+1)^2 - (n + 1) + c - (n^2 - n + c) = 2n. So, rather than recalculating the whole formula, just add 2n on to the nth result to get the (n+1)th.

Also, when asking, "when does this fail?" for the original formula, my immediate thought was that the answer was 42, of course.

[darkpulcinella]

there is a "Old Numberphile videos" vibes in this new one, i love it

[DjVortex-w]

My brain hurts trying to unravel
"for i in [i for i in range(3, n)]"

[proxyprox]

I love how Matt explains things in the most roundabout way possible.

For example, at 1:49 he could have just cancelled -41 and +41.

[japanada]

It's been proven that 41 is the last caboose number. Rabinowitz showed that c is a caboose number if and only if 4c-1 is a Heegner number, and the Stark-Heegner theorem proves that the largest Heegner number is 163. See the Wikipedia page for "Heegner number" for more information about all these points.

[blaketheory]

Checking if a Boolean is "== True" is certainly a Parker way of programming.

[davidappelgate]

Where I live in Northwest Oregon, our two area codes are 503 and 971, both of which I already knew were primes, but I didn't know they were both primes in 41's sequence! Even prouder :)

[GeHeum]

I feel like there should definitely be an "easy" upper bound on this. If a number C is Caboose that means that there are C primes between C and C^2 - 2C (the values of n=1 and n=c-1). Now use any bound you like on the amount of prime numbers within a region, and you have your easy bound on the largest possible C. Then use a computer to hopefully check the remaining small C.

[alanturing]

Nice Video.
I ran some code for my self that confirmed that there are no other Caboose Numbers up to 100 million.

[sbares]

What c=3, 5, 11, 17, 41 (and also 1 and 2 and no other numbers with 4c-1 square-free) have in common is that extending Q by a root of the polynomial x^2 - x + c gives a quadratic number field of class number 1. The near-examples at 7:56 give number fields of class number 2, except for x^2 - x + 7 whose discriminant -27 is not square-free (in this case Q(sqrt(-27)) = Q(sqrt(-3)) has class number 1).