You should be able to solve this

But if you remove a digit from these numbers, you don't get 0

Square numbers whose root isn't divisible by 5. ezpz

>6 is divisible by 5.

Just in case someone doesn't know:

oeis.org/search?q=1,4,9,16,49&language=english&go=Search

Easy. Those are the first roots of the polynomial (x-1)(x-4)(x-9)(x-16)(x-25)(x-36)(x-49)(x-182737781818)

So the missing number is 182737781818

Every even term is 4^n
Every odd term is the square of the difference of the previous 2 terms

Sure you do. We just don't write leading zeroes of whole numbers

this. or absolutely any other number. people who come up with these stupid problems are low iq illiterates who should take up pig farming to have an ever so slim chance of producing value.

Is there a good system of assigning information value to a description? It seems obvious, that "(x-1)(x-4)(x-9)(x-16)(x-25)(x-36)(x-49)(x-C) for x natural number" is a more complex description than "trunc(a2/10)=b2, a,b natural numbers"

nope, not in the general case. we could handwave around Kolmogorov complexity, but it is useless in actual cases. apart from that weak attempt, the question of which solution is simpler usually becomes a combat of argument from ego or even worse, a decision by retarded majority. I don't see a formula containing trunc(.) as simpler than one which only has addition and multiplication, for example.

also, no one said anything about simple. "what is THE next number" was the problem. anyone who comes up with the quintessentially extrovert "it was implied" bullshit should take up, you guessed it, pig farming. and the omission of the condition of simplicity is due to good reason, otherwise the burden of explanation would fall on the author and not on the wannabe solver.

tl;dr these problems are a form of trolling.

So it looks like we need a function
f(n) such that f(n) = 1 for n in {1,2,3,4}
and f(5) = 3

Just realised there's a better solution using similar logic. Next number is 91. Squares whose first digit is square.

>1 is actually 01 (9 actually 09, etc)
>remove last digit
>0

only good answer
keep the facebook bullshit out of Veeky Forums please

91 is wrong. 169 is the thing.

Why is it wrong? I suppose I was sort of silly to call one answer better because both fit the available data, but the other answer did create this argument about 1 = 01

91 is not a square. I suppose you meant 81.

^
Answer

[math]
11^{2}
[/math]

Does Latex not work anymore?

Oh yes I am retarded. Wow. Well then the next number is 100

oh, crap. you're right

Jesus fucking christ why do people unironically reply to these shitty threads thinking they're funny
nega-bump

Yay! I still feel like a brainlet for thinking that 9^2 is 91 though.

Here's your value: 156

> OP is a faggot

xD LOL! Reposting this epic funny to r/Veeky Forums lololol!

...

...

>someone posts a silly but slightly challenging maths question on a maths and science board
>user decides to just do it for fun
>lol what a faggot xD he should go to [place I don't like]
kys yourself

1^2 = 1
2^2 = 4
(1 + 2)^2 = 3^2 = 9
(1 + 3)^2 = 4^2 = 16
(4 + 3)^2 = 7^2 = 49
(7 + 3)^2 = 10^2 = 100

>low iq illiterates who should take up pig farming
One might even call them "piggots."

>given a sequence to explain the rule
>gives a finite set of numbers
truly a brainlet

Got no problem with the solution, but damn my neck hurts now.

>assuming the sequence is infinite when given nothing that implies it is
sub-brainlet behaviour

>91 is a square number

you had to be corrected by someone who thinks 91 is a square number

And you jerk off to pointing that out.

Obviously it is 5 since it is just the terms of the polynomial [math]\frac{1}{120}(-151x^{5}+2385x^{4}-14035x^{3}+38295x^{2}-47374x+21000)[/math].

I like this answer

It is obviously 156.

Your points are uniquely described by the polynomial of degree 4: x^4 - 10*x^3 + 36*x^2 - 50*x + 6755399441056031/281474976710656, so that is the only valid answer.

damn this guy is the only real mathematician in this thread

This guy knows what's up

it should be 121

>1,2,3,4,7,...

This is a fibonacci recursion with starting values of 1 and 2.

The next value should be 11, so 11^2 = 121

The next ones are clearly:
{ x | x is the rest of this sequence. }

Nice try faggot.

81?
......
1*1
2*2
3*3
2*2*2*2
5*5
7*7

3*3*3*3

there is no 5*5

>2 + 3 = 4

nice try, Radiohead

Its a portmanteau

3 number sets:

First set: 1 , 2^2 , 3^2 (Every number increases by one)
Second set: 4^2 , 7^2 , 10^2 (Every number increases by 3)

Third set: 11^2 , 16^2 , 21^2 (Every number increases by 5)

etc...

The answer is obviously 4, proof:

import random
random.seed(236012)
x = [random.randint(1,10)**2 for i in range(6)]
print(x)

> [1, 4, 9, 16, 49, 4]

Using python 3.5.0

1089

OP:

n^2 + (n-1)(n-2)(n-3)(n-4)