Primes and Squares

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Simply Beautiful Art

00:54

@WheatWizard :I

Wheat Wizard

hm?

Simply Beautiful Art

I just feel like I shouldn't do problems like that because I'd end up doing the math and then just, you know, doing that.

Wheat Wizard

Its not very hard if that's what you mean. Should only take a minute or two to figure out

Simply Beautiful Art

:P I mean, some of us mathematicians think problems like that are child's play compared to the things we do on a daily basis.

Want a problem I find more interesting?

It's not terribly impossible for anyone decent at their math.

Wheat Wizard

I'm very confused as to what you are saying

Simply Beautiful Art

01:00

Well, for your problem, I immediately know the closed form for that sum, which is given by Faulhaber's formula or similar.

Q: Methods to compute $\sum_{k=1}^nk^p$ without Faulhaber's formula

As far as every question I've seen concerning "what is $\sum_{k=1}^nk^p$" is always answered with "Faulhaber's formula" and that is just about the only answer. In an attempt to make more interesting answers, I ask that this question concern the problem of "Methods to compute $\sum_{k=1}^nk^p$ wi...

For a boatload of other solutions using the power of math.

Wheat Wizard

I think I understand

Ok then

I have a harder challenge

Simply Beautiful Art

The problem I find more interesting, though, relates to my recent challenge, but more math based, if you are interested.

Wheat Wizard

Ah yes I read that one

Simply Beautiful Art

@WheatWizard Sure

@WheatWizard My math problem: Find the closed form for how many steps (one step passes each time you subtract 1) it takes for my sequence to reach zero, starting at n.

Wheat Wizard

Lets consider all functions from Z -> N such that for all n a(n+1) = a(n - a(n)).

Prove that all such functions that are bounded are periodic.

Simply Beautiful Art

01:04

:o okay, that sounds interesting enough.

Wheat Wizard

As does yours :)

Simply Beautiful Art

:-/

Wheat Wizard

@SimplyBeautifulArt Its just 2^n-1 innit?

Simply Beautiful Art

@WheatWizard Pretty much yeah. I got 2^(n+1) - 2.

Wheat Wizard

Oh yeah -2 because we start at 2 not 1

That was a pretty fun one

Simply Beautiful Art

01:15

If there exists a large enough n s.t. 0 ≤ k < n → a(k) < n, then it holds

But I can't figure out how to prove it in general yet without assuming this.

I've another possibly interesting problem related to my sequence if you'd care for it.

Wheat Wizard

I don't think I understand your assumption

@SimplyBeautifulArt sure shoot

Simply Beautiful Art

My assumption

Take for example a(0) = 1, a(1) = 2, and a(2) = 0

Then since for every natural between 0 and 3 (including 0, but not 3) we have a(k) < 3, then a(n) is bounded and periodic.

I haven't the details here, pretty late, going to bed soon

Wheat Wizard

That example doesn't work though

if any term in the sequence is zero they must all be zero

Simply Beautiful Art

maybe. I didn't get into it much yet.

Anyways....

Wheat Wizard

Ok no problem

Simply Beautiful Art

01:20

Consider my sequence, but starting at base 2.

And modified so that each number n must be represented as b^k1 + b^k2 + ... + b + c, 0 < c ≤ b.

When in base b

where b^k1 ≤ n < b^(k1+1)

and b^k2 ≤ n-b^k1 < b^(k2+1)

etc.

(k is a sequence)

For example, 10 in base 3 would be written as
10 = 3^2 + 1

14 in base 2 would be written as
14 = 2^3 + 2^2 + 2

Got it?

My sequence says we start in base 2, then change all 2's to 3's and subtract 1, rewrite, change all 3's to 4's and subtract 1, etc.

But don't change the numbers in the exponents for sanity.

Wheat Wizard

@SimplyBeautifulArt yes

Simply Beautiful Art

My problem to you is this: Prove that, given any natural number, my sequence always goes to zero in finitely many steps.

Wheat Wizard

What do you mean by "change" all 2s to 3s

Simply Beautiful Art

Starting with 14, for example.
base on left, numbers on right.
2: 2^3 + 2^2 + 2
3: 3^3 + 3^2 + 2
4: 4^3 + 4^2 + 1
5: 5^3 + 5^2
...

Wheat Wizard

Ok so you change the bases

Simply Beautiful Art

01:29

Yeah

And if you manage to that, then find an upper and lower bound to the length of my sequence (i.e. how long it takes to hit zero)

Well, I guess the lower bound is trivially less than 2^(n+1) - 2

Wheat Wizard

oh boy

Simply Beautiful Art

Heh

Anyways, good night and good luck

Wheat Wizard

good night

Simply Beautiful Art

Oh yes, and one more thing. You might need the Ackermann function for that upper bound...

Wheat Wizard

9 hours later…

Simply Beautiful Art

10:58

::grins evilishly::

2 hours later…

Nathan Merrill

12:59

The Enormous TREE(3) - Numberphile

►

I'm becoming more convinced that some Numberphile people are active on codegolf.se

given the recent activity of codegolf.meta.stackexchange.com/questions/14057/…

flawr

13:15

Why don't you ask him?=)

Nathan Merrill

sure, let me pull up his email right here...

flawr

nottingham.ac.uk/physics/people/antonio.padilla

Nathan Merrill

huh. Professor of physics

I thought he was in the math department, but not it makes way more sense why he was trying to compare the number to the size of the universe

1 hour later…

El'endia Starman

14:37

@NathanMerrill Still waiting for a video on Tetris in GoL...

Nathan Merrill

15:18

lol

I could totally see that being one on computerphile

3 hours later…

Nathan Merrill

18:19

Joan Murray (skydiver)

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El'endia Starman

Oh, I've heard of this.

3 hours later…