Largest Number Game - Start at 1!

[AndreasC]

Summary:: This is a thread about notating a larger number than the previous poster. That's it.

Well, there's a few more rules.
1) You don't have to actually write it down. You just have to appropriately describe it.
2) Your description must correspond to one and only one natural number.
3) You can only use established mathematical notation to describe the number and notation ONLY, no natural language (except for explaining what your notation means) and definitely no silly stuff like "the largest number anyone comes up with here +1".
4) You should prove that your number exists, is a natural number, and is larger than the previous one. For some really large numbers, it may be very difficult to figure out which is larger. You can just nominate your number and the rest of us will try to figure out if it really is bigger than the previous one.

If this thread ends up being long (and I hope it does) we could declare winners of the week, or winners of the month, or winners of the first x number of posts or something like that. A winner should be someone who provided a larger number that was larger IN AN INTERESTING WAY. Now that is pretty subjective, but some examples of numbers that weren't larger in an interesting way would be (previous number)+1 or (previous number)*10. Such entries are allowed but are better avoided. Another number which wouldn't be larger in a very interesting way (but would still be allowed) would be doing exactly what the precious poster did, but once more. For instance, if I post the number 10^10 and you post the number 10^10^10, that's not particularly interesting because it's just what I did, again. Nothing wrong with that, it's just not going to win a prize.

One last thing: If you already know a number that would vanquish most competition, maybe wait a bit before posting it so that people have the chance to think a little bit and find interesting new big numbers, and you could try to find new interesting ways to write down big numbers even if they are smaller than the one you are already aware of.

I'll start: 1

 

[PeroK]

I'll go for $2$.

 

[Ibix]

$(1+i)^8$

Sixteen, in plainer notation.

 

[gmax137]

$(1+i)^8$

Sixteen, in plainer notation.

What is "i" here? apparently $(1+i) = \sqrt2$

 

[Ibix]

What is "i" here? apparently $(1+i) = \sqrt2$

$i=\sqrt{-1}$ - I assumed that was standard notation. Hence $(1+i)^8=(\sqrt 2e^{i\pi/4})^8=(\sqrt 2)^8e^ {2\pi i}=16$.

 

[PeroK]

What is "i" here? apparently $(1+i) = \sqrt2$

It's a complex number.
There are eight complex solutions to $z^8 = 16$. Two of those are $\sqrt 2$ and $1 + i$ where $i = \sqrt {-1}$.

 

[etotheipi]

$F_2 = 2^{2^2}+1$

 

[Vanadium 50]

The largest number the forum let's me write is 18.

 

[hutchphd]

I fold. The pot is too rich for my blood.

 

[AndreasC]

Speeding things up a bit, we got a long way to infinity:
18^2

 

[PeroK]

Speeding things up a bit, we got a long way to infinity:
18^2

In 1930 at Sabina Park, Jamaica, Andrew Sandham scored the first triple century in test cricket. Playing for England against the West Indies he scored 325 runs, which is one more than $18^2$.

 

[AndreasC]

In 1930 at Sabina Park, Jamaica, Andrew Sandham scored the first triple century in test cricket. Playing for England against the West Indies he scored 325 runs, which is one more than $18^2$.

1930 you say? Did he do anything interesting in 1931?

 

[PeroK]

1930 you say? Did he do anything interesting in 1931?

That was his one claim to fame. He never played for England again after that match!

 

[Ibix]

Did he do anything interesting in 1931?

1931 was 5692 in the Hebrew calendar (well, technically it was 5691/2, but let's use the larger number).

 

[AndreasC]

1931 was 5692 in the Hebrew calendar (well, technically it was 5691/2, but let's use the larger number).

5691/2? Why not 5691^2 instead?

 

[Grinkle]

PSMB4 proteasome 20S subunit beta 4

is gene id 5692 out of a list estimated to be approximately 30000 genes long.

edit: I am too late - entry retracted.

 

[AndreasC]

PSMB4 proteasome 20S subunit beta 4

is gene id 5692 out of a list estimated to be approximately 30000 genes long.

So your number is 30000? I went past that with my last post.

 

[mathman]

$i^i$ for large enough $n$.

 

[AndreasC]

$i^i$ for large enough $n$.

What?

 

[mathman]

What?

$i^i=e^{iln(i)}=e^{2n\pi-\frac{\pi}{2}}$ for all integer $n$.

 

[AndreasC]

$i^i=e^{iln(i)}=e^{2n\pi-\frac{\pi}{2}}$ for all integer $n$.

Alright I guess but that is neither a natural number, or any specific number, so as per the rules it doesn't count, so we're still at 5691^2.

 

[Ibix]

we're still at 5691^2.

Which is almost as big as the $2^{13}\mathrm{th}$ triangle number, which is $2^{25}+2^{12}$.

 

[mathman]

Alright I guess but that is neither a natural number, or any specific number, so as per the rules it doesn't count, so we're still at 5691^2.

Choose specific $n$ large enough.

 

[jbriggs444]

Choose specific $n$ large enough.

That's still not a definite n unless one:
1. Nails down what is meant by "large enough" and
2. States how a specific n is to be chosen from the set that qualifies.
Note that "large enough" needs to avoid referencing previously proposed results. Otherwise the "+1" rule comes into play.

 

[mathman]

That's still not a definite n unless one:
1. Nails down what is meant by "large enough" and
2. States how a specific n is to be chosen from the set that qualifies.
Note that "large enough" needs to avoid referencing previously proposed results. Otherwise the "+1" rule comes into play.

What n is necessary can be calculated, but I feel it is a pointless exercise.

 

[jbriggs444]

What n is necessary can be calculated, but I feel it is a pointless exercise.

From what input?

 

[Frigus]

I know it is smaller than previous number but still I will choose the classic 1729.

 

[AndreasC]

Which is almost as big as the 213th triangle number, which is 225+212.

Triangle numbers make me think of Pascal's triangle, in particular a Pascal's triangle with $2^{13}$ rows.
That's an even number of rows, so the two middle elements of the last row will be the largest elements in the triangle, so counting from 0, these correspond to the $2^{13}/2-1=2^{12}-1$th and $2^{12}$th elements of the final row. Let's keep the second one for convenience.
Of course we all know these are given by the binomial coefficients for the $2^{13}-1$th row (again counting from 0).

But you know what, scratch that, that -1 looks icky so I'm just going to go with 2^13.
So my number is:
$$\binom{2^{13}}{2^{12}}$$
Which is the largest element in a Pascal's triangle with 2^13+1 rows. Damn zero counting messing up perfectly nice expressions!

By the way, I used an online calculator and it turns out this number is 9615155570365254047227283265021467711025991035340366300886149550135565691057342466944603929439490383004169534206477572446491582535095837998914705593300555204352193678851740825923377266620092984308755649698063590523969392585613621653256885995062257245551416099846679689038343010920712058163968782232017067433340351596307366535225621267067230480999835610166354269814566942733865950136905311018353023308970209769445198232619684558275284275174618593003596288773757787467658855559895233002210026842266032772261202786109019446662426866029795063909388250125973439328264718636976930181498507864741491000997586646158361961098798977549146197844156540317409826695247163558411170902652056748563987883295363608297575794269305470895659742910448553512746676200342264037267129869589279787693014150377572761383944846547903735233619524050464444640050226614504950200139145485687318150358264371355503932037578413296818920605527048121990484055783923596713949120648671932195299617208909489017018933187731986282053887284398050526475460991382066220665701597070401877627019703437457567133546979044339622463179066504629377354012889154349952629385464825339262612946972650228997680756393014282218764909382470142575063900872677832134194990792190871089999293965694818678166163518250989227873953011556650833554169247455051301576568543953154470139746280835889786447011344440460824524401612099016234703656550455374160714376570323689569161707534178053221275681773123105515245363672218069765122663106399134759557272107059817552312206466970793525321693544528647764948764300998992847316511777509137834954259355968994821328054414320606150000860487219401899330126980880462015376239178865830102283591071402075610445010005425667379821106826725550029532844424492079810766473223845303781350659464050659720514455344801311567922440177925116714197147868668827943105578112696700265755387379116445599917662989386290503464639951174157261969374065849897147924609942688516501465102798787017833189346243891055921084716433338176140840404356286933391889157369609181231091473552803364573940086587860706383887433356093611183930698966662447008092814154083088817053025423779396495364048665487012244059071607002374287432572554452785826794762146016037507745816047249919877154769020846547780291969198898600420010614057730299123690072863658771363362640250416280201266872610282805384485362133933322921562710923524707235492827869464736848935768129206141087018025317154087975230557714962518047532497496473278279861870106525319750
...so I think from now on it's going to get a bit tough!

 

[cbarker1]

(2^25+2^12)^2400^2

 

[Baluncore]

Graham's number