How to calculate the GCD?

Question

How to evaluate the following with the help of Mobius function ?

$$\displaystyle\sum_{i=1}^n \sum_{j=i+1}^n \sum_{k=j+1}^n \sum_{l=k+1}^n {gcd(i,j,k,l)^4} .$$

In other words, we have to select all possible quadruplets from (1 ton n), and then sum up their value with power 4.

Example:-

N=4 :

(1,2,3,4) : gcd(1,2,3,4)^4 = 1

Total sum = 1

For Second Case, let N=5 :

(1,2,3,4) : gcd(1,2,3,4)^4 = 1

(1,2,3,5) : gcd(1,2,3,5)^4 = 1

(1,2,4,5) : gcd(1,2,4,5)^4 = 1

(1,3,4,5) : gcd(1,3,4,5)^4 = 1

(2,3,4,5) : gcd(2,3,4,5)^4 = 1

Total sum = 1+1+1+1+1 = 5.

My approach is: To calculate the number of quadruplets with gcd=2,3,4,...till n.

Say, the number of quadruplets with gcd=2 are x1, for gcd=3, its x2...and so on.

Now, l = $$\binom{n}{4}$$-(x1+x2+.......xn-1) = number of quadruplets with gcd=1.

Now, the final answer = l^4+x1^4+x2^4+.....

The only problem I have is how to calculate the number of quadruplets with gcd=x with the help of mobius function ?

metamorphy · Accepted Answer · 2019-06-24T15:58:51.890

3

Consider the sets of quadruples of integers $$Q(n)=\{(i,j,k,l) : 1\leqslant i<j<k<l\leqslant n\},\\ Q(n,d)=\{(i,j,k,l)\in Q(n) : \gcd(i,j,k,l)=d\}$$ for $1\leqslant d\leqslant n$. Clearly we have $$|Q(n)|=\binom{n}{4},\quad Q(n)=\bigcup_{d=1}^{n}Q(n,d),\quad Q(n,d)=dQ(\lfloor n/d\rfloor,1)$$ (where, for the last one, $d(i,j,k,l):=(di,dj,dk,dl)$ and $dS:=\{ds : s\in S\}$ are assumed).

This means that if $F(n)=|Q(n,1)|$, then $|Q(n,d)|=F(\lfloor n/d\rfloor)$ and $$\color{blue}{\sum_{d=1}^{n}F(\lfloor n/d\rfloor)=\binom{n}{4}}\quad\implies\quad F(n)=\sum_{d=1}^{n}\mu(d)\binom{\lfloor n/d\rfloor}{4}$$ by Möbius inversion. The sum we need is $S(n)=\sum\limits_{d=1}^{n}d^4 F(\lfloor n/d\rfloor)$. To compute it, one may avoid the computation of $\mu$, using the "blue" equation above as a recurrence for $F(n)$, and grouping terms with equal $\lfloor n/d\rfloor$ (here are my answers suggesting this too: 1, 2, 3).

edited Jun 24 '19 at 15:58

answered Jun 24 '19 at 08:25

metamorphy

43,591

I am very new to the concept of mobius inversion. Can you explain the part on how to calculate the last summation. And what is the meaning of [n/d]..does it mean the greatest integer function? – Firex Firexo Jun 24 '19 at 11:51
The correct answer for n=8 is 85 but your formula gives 69 :( – Firex Firexo Jun 24 '19 at 15:14
What do you do [n/d] becomes less than 4 ? – Firex Firexo Jun 24 '19 at 15:17
I am waiting @metamorphy. – Firex Firexo Jun 24 '19 at 16:01
F(8)= (8 C 4)1 + (-1)(4 C 4) + 0+0....= 69. Where did I go wrong? – Firex Firexo Jun 24 '19 at 16:03
$F(8)=69$ but the sum you need is $S(8)=F(8)+16F(4)$ (+zeros). – metamorphy Jun 24 '19 at 16:03
Ooh..I got it . Tysm. – Firex Firexo Jun 24 '19 at 16:07
https://math.stackexchange.com/questions/3273070/how-to-calculate-the-value-of-this-function @metamorphy Can you solve this recurrence relation ? – Firex Firexo Jun 24 '19 at 20:05
@metamorphy what if we have to answer the sum for all n from 1 to something like maybe, 1 to 10^3 ? Calculating for every n is gonna take O(n^2). Something better? – resound Sep 07 '19 at 23:57
@resound I would try switching the order of summation, starting with either $\sum\limits_{n=1}^{m}\sum\limits_{d=1}^{n}d^4F(\lfloor n/d\rfloor)=\sum\limits_{k=1}^{n}[\ldots]F(k)$ (with a closed form of $[\ldots]$, perhaps complicated but manageable), or directly with the defining sum at the top. Details are worth a dedicated question I think. – metamorphy Sep 08 '19 at 05:41
1

@metamorphy Have asked the same question as was very curious. Please check it out. – resound Sep 08 '19 at 05:44

score 0 · Answer 2 · answered Jun 24 '19 at 05:20

0

A quadruplet with gcd a multiple of $d$ occurs iff all $i,j,k,l$ are multiples of $d$, for which there are ${\lfloor n/d\rfloor\choose d}$ possibilities. So you'll arrive at $${n\choose 4}\cdot 1^4+{\lfloor n/2\rfloor\choose 4}\cdot (2^4-1^4) +{\lfloor n/3\rfloor\choose 4}\cdot (3^4-1^4)+{\lfloor n/4\rfloor\choose 4}\cdot (4^4-2^4)+\cdots$$

answered Jun 24 '19 at 05:20

Hagen von Eitzen

382,479

3

You should give a formula for the general term, since it's not clear to someone who doesn't already know the answer. – Greg Martin Jun 24 '19 at 06:10
@GregMartin Can you explain the answer in a bit more detail and give me the general formula ? – Firex Firexo Jun 24 '19 at 06:16
@Hagen von Eitzen For n=6, answer is 15, and your formula gives answer as '5'. It is incorrect. – Firex Firexo Jun 24 '19 at 06:24
@Guhfggehlthon Look carefully, the answer does come 15 for $n=6$ – resound Jun 29 '19 at 20:05

How to calculate the GCD?

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