20180227, 13:55  #1 
Feb 2018
5·19 Posts 
A useful template for functions gcd() and totient().
For ANY function f(),
SUM( for i on [1,n] , of f(gcd(i,n) ) ) = SUM( for all dn, of (f(d)*totient(n/d)) ). great! JM M Last fiddled with by CRGreathouse on 20180227 at 14:36 Reason: mcd > gcd 
20180227, 15:02  #2 
"Curtis"
Feb 2005
Riverside, CA
5623_{10} Posts 
If you'd stop calling your trivial observations "useful", you (and the forum) would be much better off.

20180227, 15:03  #3 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
19·23^{2} Posts 
I am nominating this user to have his own subforum.
JM Montolio A,  you are in luck! This forum has the special area for microblogs. Your posts will be perfect there. 
20180227, 15:09  #4 
Aug 2006
1011101100011_{2} Posts 
Neat!

20180227, 15:26  #5 
Feb 2018
5×19 Posts 
JA JA!
But, backing to work, ¿ "the observation" is really trivial ? JM M 
20180227, 16:13  #6 
Dec 2012
The Netherlands
11100001000_{2} Posts 
It follows from proposition 11 and proposition 88/corollary 89 in our Number Theory Discussion group.

20180227, 16:23  #7 
Feb 2018
1011111_{2} Posts 
Oh, sorry.
JM M 
20180228, 11:58  #8 
Feb 2018
5×19 Posts 

proposition 11 and proposition 88/corollary 89
sorry, i think your work only is for f(m)=m. I tested my "template" with many functions f().Look: f(x)  1 1 1*totient(N/d) x mcd(N,i) d*totient(N/d) N/x N/mcd(N,i) d*totient(d) #d(x) #divisores(mcd(N,i)) #divisores(d)*totient(N/d) sigma sigma(mcd(N,i)) sigma(d)*totient(N/d) Nota. #d():#nro. de divisores. sigma():Suma de divisores. Ejemplo. f(x)=x. SUM mcd(N,i). SUM (d*totient(N/d)).  n 1 s 1 t 1 n 2 s 3 t 3 n 3 s 5 t 5 n 4 s 8 t 8 n 5 s 9 t 9 n 6 s 15 t 15 n 7 s 13 t 13 n 8 s 20 t 20 n 9 s 21 t 21 n 10 s 27 t 27 Ejemplo. f(x)=(N/x). SUM (N/mcd(N,i)). SUM(d*totient(d)). n 1 s 1 t 1 n 2 s 3 t 3 n 3 s 7 t 7 n 4 s 11 t 11 n 5 s 21 t 21 n 6 s 21 t 21 n 7 s 43 t 43 n 8 s 43 t 43 n 9 s 61 t 61 n 10 s 63 t 63 Ejemplo. f(x)=#divisores(x). SUM #divisores(mcd(N,i)). SUM (#divisores(d)*totient(N/d)). Sigma(N). n 1 s 1 t 1 S 1 n 2 s 3 t 3 S 3 n 3 s 4 t 4 S 4 n 4 s 7 t 7 S 7 n 5 s 6 t 6 S 6 n 6 s 12 t 12 S 12 n 7 s 8 t 8 S 8 n 8 s 15 t 15 S 15 n 9 s 13 t 13 S 13 n 10 s 18 t 18 S 18 Ejemplo f(x)=sigma(x). SUM sigma(mcd(N,i)). SUM(sigma(d)*totient(N/d)). #divisores(N). n 1 s 1 t 1 1 n 2 s 4 t 4 2 n 3 s 6 t 6 2 n 4 s 12 t 12 3 n 5 s 10 t 10 2 n 6 s 24 t 24 4 n 7 s 14 t 14 2 n 8 s 32 t 32 4 n 9 s 27 t 27 3 n 10 s 40 t 40 4  
20180228, 15:23  #9  
Dec 2012
The Netherlands
11100001000_{2} Posts 
Quote:
All you have to prove is that the number of times that the function is called with a particular parameter is the same on both sides of the equation. The two key facts are: 1. for all positive integers a,b,d,n, we have gcd(a,b)=d if and only if gcd(na,nb)=nd. 2. for any positive integer n, the number of times that gcd(i,n)=1 as i runs from 0 to n1 (or from 1 to n if you prefer it) is given by Euler's totient function \(\phi(n)\). 

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