Question

What's the next number in the pattern?
0, 0, 0, ?

Answer 1

r the number 000

Answer 2

?

Answer 3

interesting ...

Answer 4

yep

Answer 5

I'll give you a hint, the next number is not 0.

Answer 6

im not expert enough thought :P

Answer 7

\[0, 0, 0, (ﾉ◕ヮ◕)ﾉ*:･ﾟ✧ \]

Answer 8

what r the answer choices

Answer 9

0 man

Answer 10

it could be anything i just need to find anything that doesn't follow ur rule :P

Answer 11

The answer choices are: \[\LARGE answers \in \mathbb{R} \]

Answer 12

this is weird

Answer 13

\[0,0,0,\spadesuit, \heartsuit,\diamondsuit, \clubsuit\] is what i get

Answer 14

I can probably bully my math teacher with this c;

Answer 15

Lol in the mean time, check out my own spin on the dirac delta function/kronecker delta!\[\LARGE \delta(x,y)=0^{|x-y|}\] since \[\LARGE 0^0=1\] and for all other powers\[\LARGE 0^p=0\]
ok, now back to guessing or should I reveal the answer?

Answer 16

Revalation time c:

Answer 17

probably 000111222333?

Answer 18

\[\Huge 119\]


\[\LARGE f(n)=\frac{(n!)!}{n!}-1\]

So n=0, n=1, n=2 all are 0 but n=3 we get 119.

Answer 19

maybe that is the answer

Answer 20

Duh

Answer 21

How come I didn't think of that?!

Answer 22

Maybe if you had given us the equation before though :p

Answer 23

nice :)

Answer 24

micw job

Answer 25

It would be something else for sure had you replied that before ;p

Answer 26

I'll give a medal + fan to someone who can show me a legitimate sequence that has the first 4 or more numbers exactly the same!

lol @ganeshie8 I was hoping that no one would have guessed 119 hahaha, I don't know an alternate one this time around haha but you know how tricky I like to be XD

Answer 27

Here I am supposed to be doing homework and now you're going to make me spend time trying to think of an answer to that question T_T

Answer 28

[n/4] floor function bhahahaha

Answer 29

lets abuse some more familiar functions like logs/exponents/floors/congruences maybe

Answer 30

Wow! thats really a clever trick.. you can generate a sequence with ANY number of same terms in the start xD

Answer 31

0^5, 0^4, 0^3, 0^2, 0^1, 0^0, 0^-1...
Lol, jk

Answer 32

yep i also have this one XD
[n/4] ( 119 ) xD

Answer 33

hello

Answer 34

lol nice XD

Answer 35

just trolling :P i have infinite of ideas xD

Answer 36

try finding this :

?, 0,0,0,0,0,0,...

Answer 37

Maybe I need to put restrictions that after the sequence of 4 repeating terms you have only increasing terms after that and no more repeating terms XD

but very clever, I'll give you a medal and help you get 99 hahahaha @ikram002p that way you can be not just green, but 99 =P

Answer 38

xD so first term is unknown and everything else is 0 or what ?

Answer 39

yes except for first term, all are 0

Answer 40

does this work for f(1) since after that you multiply by 0?\[\LARGE f(a)=\prod_{n=2}^\infty (a-n)\]

Answer 41

[(n)!/(n+1)!] =0 except when n=0 its 1 xD

Answer 42

its 1?

Answer 43

\[\LARGE \frac{(n+1)!}{n!}-n\]

Answer 44

OMG! that PI function works beautifully XD

Answer 45

i had \(f(n) = 0^n\) in mind though

Answer 46

https://oeis.org/A000007

Answer 47

wow !

Answer 48

I solved it thinking you just modified my thing slightly XD

Answer 49

I like that 0^n I have heard people say that 0^0 is undefined so this is very interesting to me thanks for showing me @ganeshie8

Answer 50

ok how about this one :D
11111110000000123456 ? 00000000000000

Answer 51

does this work for generating starting 3 zeroes and strictly increasing afterwards \[\rm \large f(n) = \binom{n}{3}\]

Answer 52

0^0 could be any number u want :3 i saw 3 pages proof its 17 xD

Answer 53

we define binomial coefficient for n<k the usual way :
\[\dfrac{n(n-1)(n-2)\cdots (n-k+1)}{k!}\]

Answer 54

we define 0^0 = 1 in number theory (binomial theorem is important for NT) and when working with sequences and series (think of power series of e^x)

Answer 55

and whenever it makes sense for it to be 1

Answer 56

thats pretty much what i found so far.. but really its a controversial function

Answer 57

Ooooh fancy I guess it could be considered divergent?

Answer 58

yeah its only where u could use it ...

Answer 59

im not able to find the reply in stackexchange but it argues somethign like this :

\[e^x = \sum\limits_{n=0}^\infty \dfrac{x^n}{n!}\]

Answer 60

if you don't define 0^0 = 1, you will need to explicitly define the value of e^0

the same problem will be there for many other series that require you to evaluate 0^0

Answer 61

Right, that's the only argument I know of too for 0^0=1, it is an interesting function to consider since it's like the identity matrix if your have row m and column n then the entries are just\[\LARGE a_{mn}=0^{|m-n|}\]

Answer 62

that looks pretty cool !!! when m=n you get 1, and 0 otherwise