Question

Sum of squares question.

Answer 1

For \(a,b,c \in \mathbb{N}^*\) (a,b,c are positive integers not including 0)

We've got this function here:
\[f(a,b,c) = a^2+b^2+c^2\]

The first and smallest value it can take is \(f(1,1,1)=3\). What's the 14th largest value it can take?

Answer 2

don't you mean 14th smallest value ?

Answer 3

Yeah haha

Answer 4

presumably (a, b, c) is an ordered pair
so (1, 2, 2) is different from (2, 2, 1)

Answer 5

I guess you can do this by just incrementing

Answer 6

f(1,2,2)=f(2,2,1) so it will be indistinguishable

Answer 7

yeah you're just looking for the values it can take

Answer 8

I think there's a way with complex numbers to extract the terms from this generating function of the partition function:

\[\left(\sum_{a=1}^\infty x^{a^2} \right)\left(\sum_{b=1}^\infty x^{b^2} \right)\left(\sum_{c=1}^\infty x^{c^2} \right) =\sum_{a=1}^\infty \sum_{b=1}^\infty \sum_{c=1}^\infty x^{a^2+b^2+c^2} \]
Since \(f=a^2+b^2+c^2\),
\[=\sum_{f=3}^\infty P(f) x^f\]

The 14th nonzero P(f) means f is the answer. :X

Answer 9

This is by no means the answer, but it's as far as I can get :o

Answer 10

n+n*((n-1) choose 2)+ n*(n-1)*(n-2)

Answer 11

:)

Answer 12

i think solving `n+n*((n-1) choose 2)+ n*(n-1)*(n-2) = k`
gives the max(a,b,c) in the kth term

Answer 13

Yea

Answer 14

Oh, how come?

Answer 15

The sum of two squares is 2square

Answer 16

scratch that, doesn' twork...

Answer 17

Why did we have to go through with the partition function etc? Cant we just generate the first 14 values easily? Is it that difficult?

Answer 18

Nah we didn't have to I just thought it might have ended up with a closed form for fun.

Answer 19

couldn't that be any permutation of a,b,c = 1,2,2

Answer 20

ah the 14th

Answer 21

1. 1-1-1
2. 1-1-2^2
3. 1-2^2 -1
4. 2^2 -1-1
5. 2^2 -2^2 -1
6. -------

this is the way till 14th term ?

Answer 22

No, the way to the 14th term is like:

1: (1,1,1)
2: (1,1,2), (1,2,1), (2,1,1)
3: (1,2,2), (2,1,2), (2,2,1)
4: (1,1,3), (1,3,1), (3,1,1)
...

Answer 23

1 1 1

1 1 2

1 2 2

2 2 2

1 2 3

2 2 3

2 3 3

3 3 3

2 3 4

3 3 4

3 4 4

4 4 4

3 4 5

4 4 5


so it's any permutation of a,b,c = 4,4,5 :D

Answer 24

Maybe I've not been clear enough in asking the question, sorry!

What I mean is that I am only looking for increasing values of \(f(a,b,c)\) regardless of the permutations of values of a,b, and c.

@Christos you've skipped the (1,1,3) term and some others too I think. It's sorta hard cause the order gets lost, it's not exactly clear without evaluating what's larger, for instance, \(f(1,1,3)\) or \(f(2,2,2)\).

Answer 25

nah its wrong

Answer 26

It turns out the answer is that the 14th smallest one is when f=27. It's interesting because it's the first value that can be expressed as a sum of 3 squares in more than 1 way,

\[f(3,3,3)=f(1,1,5)\]

Answer 27

from itertools import combinations_with_replacement

a = list(combinations_with_replacement(range(1,5),3))
b = []


for i in range(len(a)):
b.append(a[i][0]**2 + a[i][1]**2 + a[i][2]**2)


g = map(None,a,b)
g = sorted(g, key=itemgetter(1))



answer = g[13]




>>> print answer
((3, 3, 3), 27)

Answer 28

for a^3 b^3 c^3 ---> ((1, 3, 4), 92)

Answer 29

Oooh cool :D You might like this site https://repl.it/ you can use it to send code to people with the program ready to compile.

I did something similar to you to brute force my answer, but I used Java.

Answer 30

((2, 3, 3), 3)
((1, 4, 4), 9)
((3, 3, 3), 27)
((1, 3, 4), 92)
((1, 3, 4), 338)
((1, 3, 4), 1268)
((1, 3, 4), 4826)
((1, 3, 4), 18572)
((1, 3, 4), 72098)
((1, 3, 4), 281828)

Answer 31

oh I see :D:D

Answer 32

https://repl.it/CBoh

Answer 33

first result is buggy

Answer 34

actually first and second

since first ---> 0
second --->1

Answer 35

not sure about the fourth and onwards

Answer 36

my approach is sorta less cool as yours since you're using a map, so I am not able to use it to do as much as you, but here it is, hit run https://repl.it/CBom

Answer 37

i see

Answer 38

hey do you do project euler?

Answer 39

Yea I used to solve problems there, what's your username ?

Answer 40

https://projecteuler.net/progress=Kainui

Friend key
515695_59af051bf075df0d0763d75d3df7cf69

Answer 41

funny thing I remember I solved a lot of those first problems last year , may be in other accounts bah

Answer 42

Yeah I've solve a handful of them but didn't get around to programming the thing, I really just like the math part more haha

Answer 43

xd