3
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This one seems counter intuitive to me but I am not seeing a mistake in my reasoning.

Please let me know if you find one.

Let:

  • $x > 0$ be an integer
  • $\mu(x)$ be the möbius function
  • $x\#$ be the primorial for $x$
  • $r(m,d)$ be the remainder when $d$ divides $m$.

Does this follow:

$1 + \sum\limits_{i|x\#}\dfrac{r(x,i)}{i}\mu(i) = \prod\limits_{p\text{ prime, } p \le x}\left(\dfrac{p-1}{p}\right)x$

Here's my thinking:

(1) $\sum\limits_{i|x\#}\left\lfloor\dfrac{x}{i}\right\rfloor\mu(i) = 1$

This follows from the inclusion-exclusion principle and logic found in step 1 here

(2) $1 + \sum\limits_{i|x\#}\dfrac{r(x,i)}{i}\mu(i) = \sum\limits_{i|x\#}\left(\dfrac{x}{i}\right)\mu(i)$

$1 + \sum\limits_{i|x\#}\dfrac{r(x,i)}{i}\mu(i) = \sum\limits_{i|x\#}\left(\left\lfloor\dfrac{x}{i}\right\rfloor+\dfrac{r(x,i)}{i}\right)\mu(i)$

$= \sum\limits_{i|x\#}\left(\dfrac{x}{i}-\dfrac{r(x,i)}{i} +\dfrac{r(x,i)}{i}\right)\mu(i)$

$=\sum\limits_{i|x\#}\left(\dfrac{x}{i}\right)\mu(i)$

(3) $\sum\limits_{i|x\#}\left(\dfrac{x}{i}\right)\mu(i) = \prod\limits_{p\text{ prime, } p \le x}\left(\dfrac{p-1}{p}\right)x$

$\sum\limits_{i|x\#}\left(\dfrac{x}{i}\right)\mu(i) =\prod\limits_{p\text{ prime, } p \le x}\left(1 - \dfrac{1}{p}\right)x$

$= \prod\limits_{p\text{ prime, } p \le x}\left(\dfrac{p}{p} - \dfrac{1}{p}\right)x = \prod\limits_{p\text{ prime, } p \le x}\left(\dfrac{p-1}{p}\right)x$

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  • $\begingroup$ Use the math softwares. p = primes(10); L = length(p); D = []; for m=0:2^L-1, d=; for l=1:L, b = mod(floor(m/2^(l-1)),2); d = d* p(l)^b; end; D = [D,d]; end; D N = 10; mu = [1,zeros(1,N-1)]; for n = 1:N, mu(n+n:n:end) = mu(n+n:n:end) - mu(n); end; mu octave-online.net $\endgroup$ – reuns Apr 23 at 8:08
  • $\begingroup$ @reuns Thanks for your code. I tried running it on octave-online.net and it says that there are syntax errors. I think that , "d=;" is incorrect. I will spend time this evening learning the octave-online.net syntax. When I get past the syntax error, I will post here so that others can see the result. $\endgroup$ – Larry Freeman Apr 23 at 20:00
  • $\begingroup$ @LarryFreeman It all looks correct to me. These are relatively simple calculations based on equations already known to number theorists, and some of which you've also asked about yourself here in other questions. Also, have you tried manually to calculate the values of both sides for a few small values? I've tried it myself for $x$ from $1$ to $4$ to confirm it works for at least those values. Note that I've found for myself that checking these sorts of equations for several small values sometimes helps me to better understand what's going on & why they work. $\endgroup$ – John Omielan Apr 24 at 5:14
  • $\begingroup$ Thanks, John. I wrote a simple java application to sanity check values. I am planning to work with Reuns script as I have time. $\endgroup$ – Larry Freeman Apr 24 at 5:22

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