On Basak's “Bounds On Factors Of Odd Perfect Numbers”












2












$begingroup$


Let $N = q^k n^2$ be an odd perfect number given in Eulerian form, i.e. $q$ is the special/Euler prime satisfying $q equiv k equiv 1 pmod 4$ and $gcd(q,n)=1$. In what follows, we denote the abundancy index of $x in mathbb{N}$ by $I(x)=sigma(x)/x$, where $sigma(x)$ is the sum of the divisors of $x$.



In Case 1 under Remark 3.1 on page 4 of Basak's Bounds On Factors Of Odd Perfect Numbers, it is proven that



$$frac{16}{7zeta(3)} < frac{16q^3}{7zeta(3)(q^3 - 1)} < bigg(1 + frac{1}{q}bigg)prod_{p mid n, hspace{0.05in} p neq q}bigg(1 + frac{1}{p} + frac{1}{p^2}bigg).$$



But we also have
$$bigg(1 + frac{1}{q}bigg)prod_{p mid n, hspace{0.05in} p neq q}bigg(1 + frac{1}{p} + frac{1}{p^2}bigg) < I(q)I(n^2) leq frac{6I(n^2)}{5},$$
since $q$ is prime with $q equiv 1 pmod 4$ implies that $q geq 5$.



(Note that this last inequality is unconditional on the truth of the Descartes-Frenicle-Sorli Conjecture that $k=1$.)



This implies that
$$frac{16}{7zeta(3)} < frac{6I(n^2)}{5}$$
from which it follows that
$$I(n^2) > frac{5}{6}cdotfrac{16}{7zeta(3)} = frac{40}{21zeta(3)} approx 1.58458547158229994034881195966.$$



But then this resulting numerical lower bound for $I(n^2)$ is trivial, as it is known that
$$I(q^k) < frac{q}{q - 1} leq frac{5}{4} < frac{8}{5} leq frac{2(q - 1)}{q} < I(n^2),$$
so that we already know, unconditionally, that $I(n^2) > 1.6$.



Here is my question:




Would it be possible to tweak Basak's argument in order to come up with an improved lower bound for
$$bigg(1 + frac{1}{q}bigg)prod_{p mid n, hspace{0.05in} p neq q}bigg(1 + frac{1}{p} + frac{1}{p^2}bigg)?$$







elementary-number-theory upper-lower-bounds divisor-sum arithmetic-functions perfect-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    Note that, when $k=1$, then we have the lower bound $$I(n^2) geq frac{5}{3} = 1.overline{666}$$ where equality holds if and only if $q=5$.
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 11 at 7:39


















2












$begingroup$


Let $N = q^k n^2$ be an odd perfect number given in Eulerian form, i.e. $q$ is the special/Euler prime satisfying $q equiv k equiv 1 pmod 4$ and $gcd(q,n)=1$. In what follows, we denote the abundancy index of $x in mathbb{N}$ by $I(x)=sigma(x)/x$, where $sigma(x)$ is the sum of the divisors of $x$.



In Case 1 under Remark 3.1 on page 4 of Basak's Bounds On Factors Of Odd Perfect Numbers, it is proven that



$$frac{16}{7zeta(3)} < frac{16q^3}{7zeta(3)(q^3 - 1)} < bigg(1 + frac{1}{q}bigg)prod_{p mid n, hspace{0.05in} p neq q}bigg(1 + frac{1}{p} + frac{1}{p^2}bigg).$$



But we also have
$$bigg(1 + frac{1}{q}bigg)prod_{p mid n, hspace{0.05in} p neq q}bigg(1 + frac{1}{p} + frac{1}{p^2}bigg) < I(q)I(n^2) leq frac{6I(n^2)}{5},$$
since $q$ is prime with $q equiv 1 pmod 4$ implies that $q geq 5$.



(Note that this last inequality is unconditional on the truth of the Descartes-Frenicle-Sorli Conjecture that $k=1$.)



This implies that
$$frac{16}{7zeta(3)} < frac{6I(n^2)}{5}$$
from which it follows that
$$I(n^2) > frac{5}{6}cdotfrac{16}{7zeta(3)} = frac{40}{21zeta(3)} approx 1.58458547158229994034881195966.$$



But then this resulting numerical lower bound for $I(n^2)$ is trivial, as it is known that
$$I(q^k) < frac{q}{q - 1} leq frac{5}{4} < frac{8}{5} leq frac{2(q - 1)}{q} < I(n^2),$$
so that we already know, unconditionally, that $I(n^2) > 1.6$.



Here is my question:




Would it be possible to tweak Basak's argument in order to come up with an improved lower bound for
$$bigg(1 + frac{1}{q}bigg)prod_{p mid n, hspace{0.05in} p neq q}bigg(1 + frac{1}{p} + frac{1}{p^2}bigg)?$$







elementary-number-theory upper-lower-bounds divisor-sum arithmetic-functions perfect-numbers







share|cite|improve this question











$endgroup$












  • $begingroup$
    Note that, when $k=1$, then we have the lower bound $$I(n^2) geq frac{5}{3} = 1.overline{666}$$ where equality holds if and only if $q=5$.
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 11 at 7:39
















2












2








2


1



$begingroup$


Let $N = q^k n^2$ be an odd perfect number given in Eulerian form, i.e. $q$ is the special/Euler prime satisfying $q equiv k equiv 1 pmod 4$ and $gcd(q,n)=1$. In what follows, we denote the abundancy index of $x in mathbb{N}$ by $I(x)=sigma(x)/x$, where $sigma(x)$ is the sum of the divisors of $x$.



In Case 1 under Remark 3.1 on page 4 of Basak's Bounds On Factors Of Odd Perfect Numbers, it is proven that



$$frac{16}{7zeta(3)} < frac{16q^3}{7zeta(3)(q^3 - 1)} < bigg(1 + frac{1}{q}bigg)prod_{p mid n, hspace{0.05in} p neq q}bigg(1 + frac{1}{p} + frac{1}{p^2}bigg).$$



But we also have
$$bigg(1 + frac{1}{q}bigg)prod_{p mid n, hspace{0.05in} p neq q}bigg(1 + frac{1}{p} + frac{1}{p^2}bigg) < I(q)I(n^2) leq frac{6I(n^2)}{5},$$
since $q$ is prime with $q equiv 1 pmod 4$ implies that $q geq 5$.



(Note that this last inequality is unconditional on the truth of the Descartes-Frenicle-Sorli Conjecture that $k=1$.)



This implies that
$$frac{16}{7zeta(3)} < frac{6I(n^2)}{5}$$
from which it follows that
$$I(n^2) > frac{5}{6}cdotfrac{16}{7zeta(3)} = frac{40}{21zeta(3)} approx 1.58458547158229994034881195966.$$



But then this resulting numerical lower bound for $I(n^2)$ is trivial, as it is known that
$$I(q^k) < frac{q}{q - 1} leq frac{5}{4} < frac{8}{5} leq frac{2(q - 1)}{q} < I(n^2),$$
so that we already know, unconditionally, that $I(n^2) > 1.6$.



Here is my question:




Would it be possible to tweak Basak's argument in order to come up with an improved lower bound for
$$bigg(1 + frac{1}{q}bigg)prod_{p mid n, hspace{0.05in} p neq q}bigg(1 + frac{1}{p} + frac{1}{p^2}bigg)?$$







elementary-number-theory upper-lower-bounds divisor-sum arithmetic-functions perfect-numbers







share|cite|improve this question











$endgroup$




Let $N = q^k n^2$ be an odd perfect number given in Eulerian form, i.e. $q$ is the special/Euler prime satisfying $q equiv k equiv 1 pmod 4$ and $gcd(q,n)=1$. In what follows, we denote the abundancy index of $x in mathbb{N}$ by $I(x)=sigma(x)/x$, where $sigma(x)$ is the sum of the divisors of $x$.



In Case 1 under Remark 3.1 on page 4 of Basak's Bounds On Factors Of Odd Perfect Numbers, it is proven that



$$frac{16}{7zeta(3)} < frac{16q^3}{7zeta(3)(q^3 - 1)} < bigg(1 + frac{1}{q}bigg)prod_{p mid n, hspace{0.05in} p neq q}bigg(1 + frac{1}{p} + frac{1}{p^2}bigg).$$



But we also have
$$bigg(1 + frac{1}{q}bigg)prod_{p mid n, hspace{0.05in} p neq q}bigg(1 + frac{1}{p} + frac{1}{p^2}bigg) < I(q)I(n^2) leq frac{6I(n^2)}{5},$$
since $q$ is prime with $q equiv 1 pmod 4$ implies that $q geq 5$.



(Note that this last inequality is unconditional on the truth of the Descartes-Frenicle-Sorli Conjecture that $k=1$.)



This implies that
$$frac{16}{7zeta(3)} < frac{6I(n^2)}{5}$$
from which it follows that
$$I(n^2) > frac{5}{6}cdotfrac{16}{7zeta(3)} = frac{40}{21zeta(3)} approx 1.58458547158229994034881195966.$$



But then this resulting numerical lower bound for $I(n^2)$ is trivial, as it is known that
$$I(q^k) < frac{q}{q - 1} leq frac{5}{4} < frac{8}{5} leq frac{2(q - 1)}{q} < I(n^2),$$
so that we already know, unconditionally, that $I(n^2) > 1.6$.



Here is my question:




Would it be possible to tweak Basak's argument in order to come up with an improved lower bound for
$$bigg(1 + frac{1}{q}bigg)prod_{p mid n, hspace{0.05in} p neq q}bigg(1 + frac{1}{p} + frac{1}{p^2}bigg)?$$







elementary-number-theory upper-lower-bounds divisor-sum arithmetic-functions perfect-numbers




elementary-number-theory upper-lower-bounds divisor-sum arithmetic-functions perfect-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 10 at 11:21







Jose Arnaldo Bebita Dris

















asked Jan 8 at 8:03









Jose Arnaldo Bebita DrisJose Arnaldo Bebita Dris

5,43641944




5,43641944












  • $begingroup$
    Note that, when $k=1$, then we have the lower bound $$I(n^2) geq frac{5}{3} = 1.overline{666}$$ where equality holds if and only if $q=5$.
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 11 at 7:39




















  • $begingroup$
    Note that, when $k=1$, then we have the lower bound $$I(n^2) geq frac{5}{3} = 1.overline{666}$$ where equality holds if and only if $q=5$.
    $endgroup$
    – Jose Arnaldo Bebita Dris
    Jan 11 at 7:39


















$begingroup$
Note that, when $k=1$, then we have the lower bound $$I(n^2) geq frac{5}{3} = 1.overline{666}$$ where equality holds if and only if $q=5$.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 11 at 7:39






$begingroup$
Note that, when $k=1$, then we have the lower bound $$I(n^2) geq frac{5}{3} = 1.overline{666}$$ where equality holds if and only if $q=5$.
$endgroup$
– Jose Arnaldo Bebita Dris
Jan 11 at 7:39












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