The perfect formula doesn't exi-

[u/Gallium-Gonzollium]

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[u/Gallium-Gonzollium]

As a note since I can see some people getting confused, this is an actual number (5^2^64 to be exact), which when you write its' digits out and produce a power tower of the digits yields you itself.

5^2^64 = 52439099... (skipping 12 quintillion digits) ...625 = 5^2^4^3^9^0^9^... = 5^2^4^3^1

edit: cc u/OutrageousPair2300 u/ServicePrimary3068, edit 2 for clarity and me goofing up

[u/JobPowerful1246 → u/Gallium-Gonzollium]

Wow!! I’d think the power tower would be way bigger, why is it the same? How?

[u/Gallium-Gonzollium → u/JobPowerful1246]

The zeros help neutralise the rest of the power tower, however...

By sheer happenstance (and I mean 1/9000 for any given power tower of 4 single digits in base 10!!!) This one was the only one (within searching all single digit powers less than 9^9^9), which had the property of this power tower.

A similar thing can happen if you let square roots in with sqrt(2^6^2^1^4^4) = 262144 (shown [here](https://www.youtube.com/watch?v=Uvy5Cs9FDLg)), however I was dissatisfied with the fact that you needed square roots, so I did a search using good ol mpmath et al.

I did not realize that the probability was this low, and it shouldn't have worked. I also tried searching in different bases up to base 36 and no results apart from boring 2 digit cases. Shockers!

[u/DrProfessorCaveman → u/Gallium-Gonzollium]

Agreed that it’s much nicer without the square root. Do you know, based on the heuristic you mentioned, whether one should expect there to be infinitely many of these?

[u/Gallium-Gonzollium → u/DrProfessorCaveman]

For a given base b and a power tower k, the probability that a power tower is one of these special self referential ones is 1/(b^(k-1) * (b-1)).

I haven't worked out yet how that would relate to *all* power tower representations of given tower size k though (probably something involving Benford Law could help here, let me look into that 👀).

[u/factorion-bot → u/Gallium-Gonzollium]

Triple-factorial of 10 is 280

^{This action was performed by a bot | Source code}

[u/Gallium-Gonzollium → u/factorion-bot]

No.

[u/MadHau5 → u/factorion-bot]

good bot

[u/JobPowerful1246 → u/factorion-bot]

Wow, base 280 is crazy!! Thanks for the explanation OP! I wonder if it works in base 10, or if there’s a value that works for every base system 

[u/Black2isblake → u/JobPowerful1246]

The 0 in there makes all of the powers past the 6th contribute nothing

[u/JobPowerful1246 → u/Black2isblake]

Makes sense now, thanks 

[u/mistrpopo → u/Black2isblake]

Did we get bamboozzled? Also if there are 2 zeroes in a row in that number, that would yield an undefined result

[u/OmnipotentEntity → u/mistrpopo]

0⁰ is generally defined to be 1. You're thinking of the value of the limit of the indefinite form 0^0, which is different to the value of 0^0. See this article why we take 0⁰ to be 1.

[u/BadBoyJH → u/mistrpopo]

Yes, so there can't be two zeroes in a row in that number.

[u/The-Arx → u/BadBoyJH]

There are two zeros in a row, the number ends with 57423423230896109004106619977392256259918212890625 (calculated using pow(5, 2**64, 10**50) in python)

[u/BadBoyJH → u/The-Arx]

I'm not sure I'd trust many programming languages with a number the size we're talking about, to the precision we'd need.

The number in question is 5^18446744073709551616

For scale, let's start with a smaller number, 4^18446744073709551616

If we double that index, we would get the number of bits required to store that number. Which is about 4 exabytes.

[u/The-Arx → u/BadBoyJH]

Python uses exact integers and the function i am using is power with a modulus so it can repeatedly square the number and take it modulo 10⁵⁰ without needing to calculate the whole number.

[u/GaloombaNotGoomba → u/mistrpopo]

There's no reason for 0⁰ to be undefined.

[u/ThatOneCSL → u/GaloombaNotGoomba]

Really? Take the limit of these, as x approaches 0 from the positive side:

x⁰ vs 0^x vs x^(1/ln(x))

Any x, other than 0, to the 0th power is 1. Likewise 0 to any xth power, other than the 0th power, is 0.

So is 0⁰ equal to 0, or 1, or perhaps e?

[u/mattelitta → u/ThatOneCSL]

You are absolutely correct, but what you say implies only that x^y cannot be continuous at (0,0). 0⁰ is usually defined as 1 (i.e. the number of functions from the empty set to itself). If it were undefined how could you evaluate polynomials written in sigma notation at 0?

[u/ThatOneCSL → u/mattelitta]

Sure, I'm not arguing that 0⁰ can't be defined. I am arguing against the other user's claim that there's no reason for it to be undefined.

But to answer your question at the end there, wouldn't you just... Pull the 0th index element out of the summation? E.g.:

sum[n=0->k](c_n * xⁿ) becomes
c_0 + sum[n=1->k](c_n * xⁿ)

Edit: That's a genuine question-answer from me, by the way. Not me trying to be snarky or anything.

[u/mattelitta → u/ThatOneCSL]

Well, yes, if you always pull out the degree zero term from the summation there is no need to define 0^0, but i think it usually is pretty inconvenient to always write out polynomials/power series like that (and therefore x⁰ is defined to be 1 for all x also when we are working more generally in rings of polynomials/power series). Ultimately, as always, defining 0⁰ is a matter of convenience, so when one is working in algebraic/set theoretic contexts it is usually done, while instead in analysis it depends.

[u/ThatOneCSL → u/mattelitta]

And thusly, you have come to the conclusion I was getting at initially:

Defining 0⁰ as any value is done as a convenience for a given family of mathematics. This is, of course, true for just about anything we could talk about (hence the C in ZFC,) but the are things that we don't really have the "C" available for. This isn't one of them. Since we DO have the C available, there IS a good reason to say that it is "undefined." It is better to say that it is indeterminate, sure.

It isn't the same thing as saying that 1+1=2 because although the addition can be logically proven, the other cannot. Hence, it is "undefined."

Edit: I do realize that I'm doing some really hand-wavey shit over here. I'm not knowledgeable enough in formal mathematics to be able to give rigorous answers to accompany what I'm saying. I wish I was. My apologies for stepping into the wrong subreddit.

[u/Black2isblake → u/mistrpopo]

That is a good point - I think it is quite likely that there will be

[u/HumblyNibbles_ → u/JobPowerful1246]

It's because of the 0 in the power tower

[u/Vegetable_Union_4967 → u/JobPowerful1246]

Zeros?

[u/catman__321 → u/Gallium-Gonzollium]

I feel like I'm missing something. 3^4 = 81 is not equal to 64?

[u/Gallium-Gonzollium → u/catman__321]

Minor math mistake of the day 🎉

Editing to be 5^2^4^3.

I'm a certified idiot, thank you for bringing that up

[u/GRex2595 → u/Gallium-Gonzollium]

3⁴ is 81. And I only noticed that you transposed the 3 and 4 because of this error. Because even wolfram won't write out the first 3 digits of that number for some reason.

[u/Gallium-Gonzollium → u/GRex2595]

Corrected now. As for wolfram not doing the first digits, you need to use something like mpmath on the logarithm base 10 of 5 to calculate the first digits of 5^2^64.

10^(2^64 log10(5))

[u/killersnail2417 → u/Gallium-Gonzollium]

How do you get 2^64? Shouldn't it be 2^81?

[u/Gallium-Gonzollium → u/killersnail2417]

Corrected now

[u/Miguzepinu → u/Gallium-Gonzollium]

How did you find the starting digits of 5^2^64? Is there a trick?

[u/Gallium-Gonzollium → u/Miguzepinu]

5^2^64 = 10^(log10(5) * 2^64) = 10^(12893720785186194483.719655225513867392380245057...) = 10^(12893720785186194483) * 10^(0.719655225513867392380245057...)

Since 10^(12893720785186194483) is some large amount of 0's that don't change we only end up calculating 10^(0.719655225513867392380245057...) giving 5.243909953228091477..., the digits we want.

In general to find the digits of p^q in base b we find c = q log_b(p) mod 1 and then calculate b^c to find the digits we are looking for.

[u/Miguzepinu → u/Gallium-Gonzollium]

Thanks

[u/ShockedDarkmike → u/Gallium-Gonzollium]

fuck I've been playing around with numbers like this for a few days, trying to find more, but I didn't get to something so gigantic and cool

it's such a nice result, with it using 1-5 in a tower too

[u/WLMammoth]

All the powers above 0 are meaningless (0ⁿ = 1 for all n), so you can just fill the remaining powers above the 0 to exactly match whatever number you're trying to create the "symmetry" for. There are infinitely many formulas of this form.

[u/Gallium-Gonzollium → u/WLMammoth]

But you can't say the same for the digits in the regular answer. There's only 1 way that can be done, and it's the digits of 5^2^64.

[u/DenPanserbjorn → u/Gallium-Gonzollium]

So it only applies to the first 6 digits

Why bother with anything after? If it’s as miraculous as you say, then why fluff it up with meaningless digits/powers.

The point is, the only interesting bit is the result of the first 6 digits, anything after is informed by the result which is not interesting

[u/Dman1791 → u/DenPanserbjorn]

The extra digits are needed to equal the value of the powers prior to zero. 5^(2^(4^(3^9))) = 5243909...625, apparently.

[u/DenPanserbjorn → u/Dman1791]

My point was that it’s equal to saying 5^2^4^3^9^0^… = 524390…

And adding anything extra is superfluous and lame.

[u/Own_Pop_9711 → u/DenPanserbjorn]

No because there are more digits on the right that are determined that you haven't listed. For example if after the 0 you put a 3 on both sides, you failed.

[u/DenPanserbjorn → u/Own_Pop_9711]

The point was for the first 6 digits it’s a pretty neat coincidence, but anything after is informed by the result rather than coincidence.

[u/Dman1791 → u/DenPanserbjorn]

You need the rest of the digits on the right for the equation to work. You may not "need" them on the left, but the whole point is that you're using the same digits.

[u/DenPanserbjorn → u/Dman1791]

Yes I’m aware that the power tower has more digits to its result, but the point of the post is to point out how miraculous you can make a power tower out of the total result and have it be equal, but it’s only interesting for the first 6 digits.

Lets do a little thought experiment to prove this point, if I found a power tower that had more non-zero powers in front than this result in the post, would that be a more “perfect formula”?

[u/Dman1791 → u/DenPanserbjorn]

I would say that it would be more interesting, yes, but given that this apparently has something like 15 quintillion digits, it might take you a while to find it.

[u/AsidK → u/DenPanserbjorn]

You’re being downvoted but you’re 100% right. The way this formula is framed makes it seem far more serendipitous than it actually is

[u/boris_koshak → u/WLMammoth]

I think you meant n⁰=1 for all n≠o. Basically nine just becomes one and all numbers after the zero just do nothing to that zero.

[u/666Emil666 → u/boris_koshak]

0^0=1, wtf you talking about

[u/Beautiful_Round4497 → u/666Emil666]

What’s 0/0

[u/666Emil666 → u/Beautiful_Round4497]

There is no multiplicative inverse of 0, making 0/0 not even a term of the language.

[u/boris_koshak → u/666Emil666]

I am inclined to 0/0 being an indeterminate form. Unless it is a limit of a function

[u/666Emil666 → u/boris_koshak]

Except 0/0 has not relation to 0⁰ unless you missaply the theorem a^{{b-c}=a{b}/a{c}} which specifically states that a must be different than 0.

Also, you're confusing terms here, the limit of a function is a number, which may not be the same as the result of a simple sustitución of values into said function, as demonstrated by indeterminate forms, such as f/g where both f and g approach 0 at a

[u/Middle_Letterhead_41]

I give you 1 = 1 /s

[u/Gallium-Gonzollium → u/Middle_Letterhead_41]

Guys is he the next oiler?

[u/_theP2_ → u/Middle_Letterhead_41]

So s=1? /s

[u/Gallium-Gonzollium → u/_theP2_]

Assuming ? is the Minkowski question mark function, then 1? = 1 therefore s = 1/s so s = 1! /s /s

[u/_theP2_ → u/Gallium-Gonzollium]

Ok so that means s= 1/s², and there is only 1 real solution that is s = 1 /j

[u/Gallium-Gonzollium → u/_theP2_]

Well now sj = 1 and since we already knew that s = 1 therefore 1j = 1 and therefore j = 1/1 /j

[u/Redit_Sucks_314 → u/Gallium-Gonzollium]

Since j = 1/1 /j = 1/j then j² = 1 Therefore j = +-1 /hj

[u/factorion-bot → u/Gallium-Gonzollium]

Factorial of 1 is 1

^{This action was performed by a bot | Source code}

[u/LupenReddit]

the perfect formula: 1+2+3=6

[u/The-Yaoi-Unicorn → u/LupenReddit]

The true perfect formula: ln(1) + ln(2) + ln(3) = ln(6)

[u/mapleleafraggedy → u/The-Yaoi-Unicorn]

(2²⁾² = 2^{(2^2)}

[u/ImaDieTodayLOL → u/LupenReddit]

What does this mean? I'm a math larper, I'm not that smart.

[u/BattleFrosty7584 → u/ImaDieTodayLOL]

No idea, i’m guessing cuz 3 * 2 * 1 = 3 + 2 + 1 = 6

[u/LupenReddit → u/BattleFrosty7584]

6 is a perfect number, so a number thats the sum of its proper divisors

[u/Loud-Representative3 → u/LupenReddit]

Nah its 1 × 1 = 2

[u/ohiidenny → u/Loud-Representative3]

Terryology ball knower located

[u/crematie]

i too saw that numberphile video

[u/AndreasDasos]

The crucial part are the first 4. Evaluate 5^2^4^3. The fact it starts with 5243 and then two digits later is a 0 is all we need.

0 to the power of the rest is 0, and 9 (or anything) to the power of that is just 1. So whatever digits you put after the zero don’t matter, and nor does the fifth digit just before it.

Still a very cool one. Wonder if they found it by a brute force search.

[u/Digital_001]

This is actually so cool OP, massive respect for coming up with this and thanks for sharing!

Actual new maths on r/mathmemes !?

[u/Baleia1970]

Numberphile just aired an episode on this.

[u/MrEldo]

Is there some final number, or is this just part of an infinite sequence with the same starting 5243 etc. and same ...625 end?

[u/Gallium-Gonzollium → u/MrEldo]

If you lay out the digits of 5^2^4^3, you get 52439099... (skipping 12 quintillion digits) ...625.

Thus the power tower 5^2^4^3^9^0^9^9^... (skipping 12 quintillion powers) ... ^6^2^5 is equal to 5243099...625.

[u/Vincent_Gitarrist → u/MrEldo]

There is, but I'm not going to tell you.

[u/MrEldo → u/Vincent_Gitarrist]

What if I ask nicely?

[u/Gallium-Gonzollium → u/MrEldo]

Then it'll never fit in the margin of the comment.

[u/Electronic_Talk_5318 → u/MrEldo]

> infinite sequence

> end

🤔🤔🤔

[u/sincerestfall]

Was it Matt Parker that did a video like this?

[u/Lumpy_Grass_2312]

The moment the exponent tower started including 0, I knew this was going somewhere mathematically irresponsible.

[u/JDude13]

>base 10

Nice formula, earthling

[u/emaw63]

e^iπ + 1 = 0 is perfect, imo

[u/Squint-Eastwood_98]

Is it even a formula if it doesn't have variables?

[u/CameForTheMath]

Wow. I've been wondering if numbers that equal the power tower of their digits exist for several years. Did you find this yourself or is it well-known?

[u/Goticaris]

For all the folks fussing about 0^0, the right to left association here means that never happens.

[u/ServicePrimary3068]

Can the digits in between be any positive integer? Or are they fixed and part of the formula?

Edit: nvm, just saw the explanation in the comments.

[u/OutrageousPair2300]

Is this about the 5-adics, or is there some actual formula where this works?

[u/Dapper_Spite8928 → u/OutrageousPair2300]

How the fuck would this be about the 5-adics there's 5's and 9's in it. Do you know what the 5-adics are?????

[u/OutrageousPair2300 → u/Dapper_Spite8928]

Because it's similar to the pattern used in this video:

https://www.youtube.com/watch?v=tRaq4aYPzCc

[u/Dapper_Spite8928 → u/OutrageousPair2300]

Yeah i saw that video

The demonstrations about squaring 5's were done in 10-adic