Calculating $sum_{substack{k|r \ k leq n}} mu left({ {k}}right)$?












3














Background & Question



I recently thought of a combinatoric method to get an interesting result:



$$ sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! O(frac{1}{sqrt n}) -1 $$



I was wondering how other methods compared to mine in calculating the quantity:



$$sum_{substack{k|r \ k leq n}} mu left({ {k}}right)$$



and then summing over to see how it compared to my result? Also is there any method of calculating it without using the Riemann Hypothesis?



Proof of my method



Consider the following sums:
$$ 1+1 + 1+ 1+1 +underbracedots_{n! text{ times}} = n! $$
$$ 0 + 1+ 0+1 +0 +underbracedots_{n! text{ times}} = frac{n!}{2} $$
$$ 0 + 0+ 1+0 +0 +underbracedots_{n! text{ times}} = frac{n!}{3} $$
We proceed to do so $n$ times:
$$ 0+ 0 +underbracedots_{n text{ times}} +1+underbracedots_{n! text{ times}} = frac{n!}{n} $$



Multiplying the $r$th row with $a_r$ which is an arbitrary variable:



$$ a_1+ a_1 + a_1+ a_1+a_1 +underbracedots_{n! text{ times}} = n! a_1 $$
$$ 0 + a_2+ 0+a_2 +0 +underbracedots_{n! text{ times}} =n! frac{a_2}{2} $$
$$ 0 + 0+ a_3+0 +0 +underbracedots_{n! text{ times}} = n! frac{a_3}{3} $$
We proceed to do so $n$ times:
$$ 0+ 0 +underbracedots_{n-1 text{ times}} +a_n+underbracedots_{n! text{ times}} = n!frac{a_n}{n} $$



Adding all the above equations vertically (as it is a finite sum):



$$ underbrace{b_1}_{a_1} + underbrace{b_2}_{a_1+a_2}+ underbrace{b_3}_{a_1+a_3} + dots+ b_n +sum_{r=n+1}^{n!} tilde b_r = n!sum_{r=1}^n frac{a_r}{r}$$



Writing the above properly:



$$ sum_{r=1}^n b_r + sum_{r=n+1}^{n!} tilde b_r = n! sum_{r=1}^n frac{a_r}{r} $$



where $b_r = sum_{r|k} a_k$ and $tilde b_r = sum_{substack{k|r \ kleq n}} a_k $. Using the mobius function we know $ a_{n}=sum_{dmid n}mu left({frac {n}{d}}right)b_{d}$. Re-expressing the expression the above in terms of $b_r$:



$$ sum_{r=1}^n b_r + sum_{r=n+1}^{n!} sum_{substack{k|r \ kleq n}} sum_{dmid k}mu left({frac {k}{d}}right)b_{d} = n! sum_{r=1}^n frac{sum _{dmid r}mu left({frac {r}{d}}right)b_{d}}{r} $$



Applying $frac{partial}{partial b_1} $ both sides:



$$ 1 + sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! sum_{r=1}^n frac{mu left({{r}}right)}{r} $$



Using RH:



$$ 1 + sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! O(frac{1}{sqrt n}) $$






combinatorics number-theory mobius-function







share|cite|improve this question
























  • You’ve put $rmid k$ but it should be $kmid r$ in all instances.
    – Jacob
    Nov 26 '18 at 1:39










  • Also, if we go straight from our formula for $n!frac{a_k}{k}$ to $sum_{k=1}^{n!}sum_{dmid k \ dleq n}a_d=n!sum_{k=1}^nfrac{a_k}{k}$, and then use $a_k=mu(k)$ we have $n!sum_{k=1}^nfrac{mu(k)}{k}=sum_{k=1}^{n!}sum_{dmid k \ dleq n}mu(d)=left(sum_{k=1}^nsum_{dmid k}mu(d)right)+left(sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)right)$. Since $$sum_{dmid k}mu(d)=begin{cases} 1 & n=1 \ 0 & text{otherwise},end{cases}$$ we end up with $n!sum_{k=1}^nfrac{mu(k)}{k}=1+sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)$.
    – Jacob
    Nov 26 '18 at 2:03










  • Made it $k|r$ hopefully everywhere
    – More Anonymous
    Nov 26 '18 at 11:05










  • The title still has $rmid k$.
    – Gerry Myerson
    Nov 27 '18 at 2:49










  • My bad ... fixed now
    – More Anonymous
    Nov 27 '18 at 2:53
















3














Background & Question



I recently thought of a combinatoric method to get an interesting result:



$$ sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! O(frac{1}{sqrt n}) -1 $$



I was wondering how other methods compared to mine in calculating the quantity:



$$sum_{substack{k|r \ k leq n}} mu left({ {k}}right)$$



and then summing over to see how it compared to my result? Also is there any method of calculating it without using the Riemann Hypothesis?



Proof of my method



Consider the following sums:
$$ 1+1 + 1+ 1+1 +underbracedots_{n! text{ times}} = n! $$
$$ 0 + 1+ 0+1 +0 +underbracedots_{n! text{ times}} = frac{n!}{2} $$
$$ 0 + 0+ 1+0 +0 +underbracedots_{n! text{ times}} = frac{n!}{3} $$
We proceed to do so $n$ times:
$$ 0+ 0 +underbracedots_{n text{ times}} +1+underbracedots_{n! text{ times}} = frac{n!}{n} $$



Multiplying the $r$th row with $a_r$ which is an arbitrary variable:



$$ a_1+ a_1 + a_1+ a_1+a_1 +underbracedots_{n! text{ times}} = n! a_1 $$
$$ 0 + a_2+ 0+a_2 +0 +underbracedots_{n! text{ times}} =n! frac{a_2}{2} $$
$$ 0 + 0+ a_3+0 +0 +underbracedots_{n! text{ times}} = n! frac{a_3}{3} $$
We proceed to do so $n$ times:
$$ 0+ 0 +underbracedots_{n-1 text{ times}} +a_n+underbracedots_{n! text{ times}} = n!frac{a_n}{n} $$



Adding all the above equations vertically (as it is a finite sum):



$$ underbrace{b_1}_{a_1} + underbrace{b_2}_{a_1+a_2}+ underbrace{b_3}_{a_1+a_3} + dots+ b_n +sum_{r=n+1}^{n!} tilde b_r = n!sum_{r=1}^n frac{a_r}{r}$$



Writing the above properly:



$$ sum_{r=1}^n b_r + sum_{r=n+1}^{n!} tilde b_r = n! sum_{r=1}^n frac{a_r}{r} $$



where $b_r = sum_{r|k} a_k$ and $tilde b_r = sum_{substack{k|r \ kleq n}} a_k $. Using the mobius function we know $ a_{n}=sum_{dmid n}mu left({frac {n}{d}}right)b_{d}$. Re-expressing the expression the above in terms of $b_r$:



$$ sum_{r=1}^n b_r + sum_{r=n+1}^{n!} sum_{substack{k|r \ kleq n}} sum_{dmid k}mu left({frac {k}{d}}right)b_{d} = n! sum_{r=1}^n frac{sum _{dmid r}mu left({frac {r}{d}}right)b_{d}}{r} $$



Applying $frac{partial}{partial b_1} $ both sides:



$$ 1 + sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! sum_{r=1}^n frac{mu left({{r}}right)}{r} $$



Using RH:



$$ 1 + sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! O(frac{1}{sqrt n}) $$






combinatorics number-theory mobius-function







share|cite|improve this question
























  • You’ve put $rmid k$ but it should be $kmid r$ in all instances.
    – Jacob
    Nov 26 '18 at 1:39










  • Also, if we go straight from our formula for $n!frac{a_k}{k}$ to $sum_{k=1}^{n!}sum_{dmid k \ dleq n}a_d=n!sum_{k=1}^nfrac{a_k}{k}$, and then use $a_k=mu(k)$ we have $n!sum_{k=1}^nfrac{mu(k)}{k}=sum_{k=1}^{n!}sum_{dmid k \ dleq n}mu(d)=left(sum_{k=1}^nsum_{dmid k}mu(d)right)+left(sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)right)$. Since $$sum_{dmid k}mu(d)=begin{cases} 1 & n=1 \ 0 & text{otherwise},end{cases}$$ we end up with $n!sum_{k=1}^nfrac{mu(k)}{k}=1+sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)$.
    – Jacob
    Nov 26 '18 at 2:03










  • Made it $k|r$ hopefully everywhere
    – More Anonymous
    Nov 26 '18 at 11:05










  • The title still has $rmid k$.
    – Gerry Myerson
    Nov 27 '18 at 2:49










  • My bad ... fixed now
    – More Anonymous
    Nov 27 '18 at 2:53














3












3








3







Background & Question



I recently thought of a combinatoric method to get an interesting result:



$$ sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! O(frac{1}{sqrt n}) -1 $$



I was wondering how other methods compared to mine in calculating the quantity:



$$sum_{substack{k|r \ k leq n}} mu left({ {k}}right)$$



and then summing over to see how it compared to my result? Also is there any method of calculating it without using the Riemann Hypothesis?



Proof of my method



Consider the following sums:
$$ 1+1 + 1+ 1+1 +underbracedots_{n! text{ times}} = n! $$
$$ 0 + 1+ 0+1 +0 +underbracedots_{n! text{ times}} = frac{n!}{2} $$
$$ 0 + 0+ 1+0 +0 +underbracedots_{n! text{ times}} = frac{n!}{3} $$
We proceed to do so $n$ times:
$$ 0+ 0 +underbracedots_{n text{ times}} +1+underbracedots_{n! text{ times}} = frac{n!}{n} $$



Multiplying the $r$th row with $a_r$ which is an arbitrary variable:



$$ a_1+ a_1 + a_1+ a_1+a_1 +underbracedots_{n! text{ times}} = n! a_1 $$
$$ 0 + a_2+ 0+a_2 +0 +underbracedots_{n! text{ times}} =n! frac{a_2}{2} $$
$$ 0 + 0+ a_3+0 +0 +underbracedots_{n! text{ times}} = n! frac{a_3}{3} $$
We proceed to do so $n$ times:
$$ 0+ 0 +underbracedots_{n-1 text{ times}} +a_n+underbracedots_{n! text{ times}} = n!frac{a_n}{n} $$



Adding all the above equations vertically (as it is a finite sum):



$$ underbrace{b_1}_{a_1} + underbrace{b_2}_{a_1+a_2}+ underbrace{b_3}_{a_1+a_3} + dots+ b_n +sum_{r=n+1}^{n!} tilde b_r = n!sum_{r=1}^n frac{a_r}{r}$$



Writing the above properly:



$$ sum_{r=1}^n b_r + sum_{r=n+1}^{n!} tilde b_r = n! sum_{r=1}^n frac{a_r}{r} $$



where $b_r = sum_{r|k} a_k$ and $tilde b_r = sum_{substack{k|r \ kleq n}} a_k $. Using the mobius function we know $ a_{n}=sum_{dmid n}mu left({frac {n}{d}}right)b_{d}$. Re-expressing the expression the above in terms of $b_r$:



$$ sum_{r=1}^n b_r + sum_{r=n+1}^{n!} sum_{substack{k|r \ kleq n}} sum_{dmid k}mu left({frac {k}{d}}right)b_{d} = n! sum_{r=1}^n frac{sum _{dmid r}mu left({frac {r}{d}}right)b_{d}}{r} $$



Applying $frac{partial}{partial b_1} $ both sides:



$$ 1 + sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! sum_{r=1}^n frac{mu left({{r}}right)}{r} $$



Using RH:



$$ 1 + sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! O(frac{1}{sqrt n}) $$






combinatorics number-theory mobius-function







share|cite|improve this question















Background & Question



I recently thought of a combinatoric method to get an interesting result:



$$ sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! O(frac{1}{sqrt n}) -1 $$



I was wondering how other methods compared to mine in calculating the quantity:



$$sum_{substack{k|r \ k leq n}} mu left({ {k}}right)$$



and then summing over to see how it compared to my result? Also is there any method of calculating it without using the Riemann Hypothesis?



Proof of my method



Consider the following sums:
$$ 1+1 + 1+ 1+1 +underbracedots_{n! text{ times}} = n! $$
$$ 0 + 1+ 0+1 +0 +underbracedots_{n! text{ times}} = frac{n!}{2} $$
$$ 0 + 0+ 1+0 +0 +underbracedots_{n! text{ times}} = frac{n!}{3} $$
We proceed to do so $n$ times:
$$ 0+ 0 +underbracedots_{n text{ times}} +1+underbracedots_{n! text{ times}} = frac{n!}{n} $$



Multiplying the $r$th row with $a_r$ which is an arbitrary variable:



$$ a_1+ a_1 + a_1+ a_1+a_1 +underbracedots_{n! text{ times}} = n! a_1 $$
$$ 0 + a_2+ 0+a_2 +0 +underbracedots_{n! text{ times}} =n! frac{a_2}{2} $$
$$ 0 + 0+ a_3+0 +0 +underbracedots_{n! text{ times}} = n! frac{a_3}{3} $$
We proceed to do so $n$ times:
$$ 0+ 0 +underbracedots_{n-1 text{ times}} +a_n+underbracedots_{n! text{ times}} = n!frac{a_n}{n} $$



Adding all the above equations vertically (as it is a finite sum):



$$ underbrace{b_1}_{a_1} + underbrace{b_2}_{a_1+a_2}+ underbrace{b_3}_{a_1+a_3} + dots+ b_n +sum_{r=n+1}^{n!} tilde b_r = n!sum_{r=1}^n frac{a_r}{r}$$



Writing the above properly:



$$ sum_{r=1}^n b_r + sum_{r=n+1}^{n!} tilde b_r = n! sum_{r=1}^n frac{a_r}{r} $$



where $b_r = sum_{r|k} a_k$ and $tilde b_r = sum_{substack{k|r \ kleq n}} a_k $. Using the mobius function we know $ a_{n}=sum_{dmid n}mu left({frac {n}{d}}right)b_{d}$. Re-expressing the expression the above in terms of $b_r$:



$$ sum_{r=1}^n b_r + sum_{r=n+1}^{n!} sum_{substack{k|r \ kleq n}} sum_{dmid k}mu left({frac {k}{d}}right)b_{d} = n! sum_{r=1}^n frac{sum _{dmid r}mu left({frac {r}{d}}right)b_{d}}{r} $$



Applying $frac{partial}{partial b_1} $ both sides:



$$ 1 + sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! sum_{r=1}^n frac{mu left({{r}}right)}{r} $$



Using RH:



$$ 1 + sum_{r=n+1}^{n!} sum_{substack{k|r \ k leq n}} mu left({ {k}}right) = n! O(frac{1}{sqrt n}) $$






combinatorics number-theory mobius-function




combinatorics number-theory mobius-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 27 '18 at 2:51

























asked Nov 25 '18 at 23:07









More Anonymous

34419




34419












  • You’ve put $rmid k$ but it should be $kmid r$ in all instances.
    – Jacob
    Nov 26 '18 at 1:39










  • Also, if we go straight from our formula for $n!frac{a_k}{k}$ to $sum_{k=1}^{n!}sum_{dmid k \ dleq n}a_d=n!sum_{k=1}^nfrac{a_k}{k}$, and then use $a_k=mu(k)$ we have $n!sum_{k=1}^nfrac{mu(k)}{k}=sum_{k=1}^{n!}sum_{dmid k \ dleq n}mu(d)=left(sum_{k=1}^nsum_{dmid k}mu(d)right)+left(sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)right)$. Since $$sum_{dmid k}mu(d)=begin{cases} 1 & n=1 \ 0 & text{otherwise},end{cases}$$ we end up with $n!sum_{k=1}^nfrac{mu(k)}{k}=1+sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)$.
    – Jacob
    Nov 26 '18 at 2:03










  • Made it $k|r$ hopefully everywhere
    – More Anonymous
    Nov 26 '18 at 11:05










  • The title still has $rmid k$.
    – Gerry Myerson
    Nov 27 '18 at 2:49










  • My bad ... fixed now
    – More Anonymous
    Nov 27 '18 at 2:53


















  • You’ve put $rmid k$ but it should be $kmid r$ in all instances.
    – Jacob
    Nov 26 '18 at 1:39










  • Also, if we go straight from our formula for $n!frac{a_k}{k}$ to $sum_{k=1}^{n!}sum_{dmid k \ dleq n}a_d=n!sum_{k=1}^nfrac{a_k}{k}$, and then use $a_k=mu(k)$ we have $n!sum_{k=1}^nfrac{mu(k)}{k}=sum_{k=1}^{n!}sum_{dmid k \ dleq n}mu(d)=left(sum_{k=1}^nsum_{dmid k}mu(d)right)+left(sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)right)$. Since $$sum_{dmid k}mu(d)=begin{cases} 1 & n=1 \ 0 & text{otherwise},end{cases}$$ we end up with $n!sum_{k=1}^nfrac{mu(k)}{k}=1+sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)$.
    – Jacob
    Nov 26 '18 at 2:03










  • Made it $k|r$ hopefully everywhere
    – More Anonymous
    Nov 26 '18 at 11:05










  • The title still has $rmid k$.
    – Gerry Myerson
    Nov 27 '18 at 2:49










  • My bad ... fixed now
    – More Anonymous
    Nov 27 '18 at 2:53
















You’ve put $rmid k$ but it should be $kmid r$ in all instances.
– Jacob
Nov 26 '18 at 1:39




You’ve put $rmid k$ but it should be $kmid r$ in all instances.
– Jacob
Nov 26 '18 at 1:39












Also, if we go straight from our formula for $n!frac{a_k}{k}$ to $sum_{k=1}^{n!}sum_{dmid k \ dleq n}a_d=n!sum_{k=1}^nfrac{a_k}{k}$, and then use $a_k=mu(k)$ we have $n!sum_{k=1}^nfrac{mu(k)}{k}=sum_{k=1}^{n!}sum_{dmid k \ dleq n}mu(d)=left(sum_{k=1}^nsum_{dmid k}mu(d)right)+left(sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)right)$. Since $$sum_{dmid k}mu(d)=begin{cases} 1 & n=1 \ 0 & text{otherwise},end{cases}$$ we end up with $n!sum_{k=1}^nfrac{mu(k)}{k}=1+sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)$.
– Jacob
Nov 26 '18 at 2:03




Also, if we go straight from our formula for $n!frac{a_k}{k}$ to $sum_{k=1}^{n!}sum_{dmid k \ dleq n}a_d=n!sum_{k=1}^nfrac{a_k}{k}$, and then use $a_k=mu(k)$ we have $n!sum_{k=1}^nfrac{mu(k)}{k}=sum_{k=1}^{n!}sum_{dmid k \ dleq n}mu(d)=left(sum_{k=1}^nsum_{dmid k}mu(d)right)+left(sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)right)$. Since $$sum_{dmid k}mu(d)=begin{cases} 1 & n=1 \ 0 & text{otherwise},end{cases}$$ we end up with $n!sum_{k=1}^nfrac{mu(k)}{k}=1+sum_{k=n+1}^{n!}sum_{dmid k \ dleq n}mu(d)$.
– Jacob
Nov 26 '18 at 2:03












Made it $k|r$ hopefully everywhere
– More Anonymous
Nov 26 '18 at 11:05




Made it $k|r$ hopefully everywhere
– More Anonymous
Nov 26 '18 at 11:05












The title still has $rmid k$.
– Gerry Myerson
Nov 27 '18 at 2:49




The title still has $rmid k$.
– Gerry Myerson
Nov 27 '18 at 2:49












My bad ... fixed now
– More Anonymous
Nov 27 '18 at 2:53




My bad ... fixed now
– More Anonymous
Nov 27 '18 at 2:53










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Puebla de Zaragoza

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Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
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Musa

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Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
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