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#1 |
Feb 2018
5·19 Posts |
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For ANY function f(),
SUM( for i on [1,n] , of f(gcd(i,n) ) ) = SUM( for all d|n, of (f(d)*totient(n/d)) ). great! JM M Last fiddled with by CRGreathouse on 2018-02-27 at 14:36 Reason: mcd -> gcd |
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#2 |
"Curtis"
Feb 2005
Riverside, CA
562310 Posts |
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If you'd stop calling your trivial observations "useful", you (and the forum) would be much better off.
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#3 |
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
19·232 Posts |
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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. |
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#4 |
Aug 2006
10111011000112 Posts |
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Neat!
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#5 |
Feb 2018
5×19 Posts |
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JA JA!
But, backing to work, ¿ "the observation" is really trivial ? JM M |
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#6 |
Dec 2012
The Netherlands
111000010002 Posts |
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It follows from proposition 11 and proposition 88/corollary 89 in our Number Theory Discussion group.
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#7 |
Feb 2018
10111112 Posts |
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Oh, sorry.
JM M |
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#8 |
Feb 2018
5×19 Posts |
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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 -------------------------- |
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#9 | |
Dec 2012
The Netherlands
111000010002 Posts |
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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 n-1 (or from 1 to n if you prefer it) is given by Euler's totient function \(\phi(n)\). |
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