On Basak's “Bounds On Factors Of Odd Perfect Numbers”
$begingroup$
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
asked Jan 8 at 8:03
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,26952045
$endgroup$
add a comment |
$begingroup$
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
asked Jan 8 at 8:03
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,26952045
$endgroup$
$begingroup$
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
$begingroup$
After taking to Basak: "... the same year a paper was published by Anirudh Prabhu in the United States which used different methods to obtain bounds for the reciprocal sums of prime factors. The constraints on OPNs produced are different, but seem slightly stronger." The link to Prabhu's paper is here.
$endgroup$
– Jose Arnaldo Bebita-Dris
Feb 25 at 12:55
add a comment |
$begingroup$
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
asked Jan 8 at 8:03
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,26952045
$endgroup$
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
asked Jan 8 at 8:03
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,26952045
asked Jan 8 at 8:03
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,26952045
edited Jan 10 at 11:21
Jose Arnaldo Bebita-Dris
asked Jan 8 at 8:03
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,26952045
asked Jan 8 at 8:03
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,26952045
asked Jan 8 at 8:03
Jose Arnaldo Bebita-DrisJose Arnaldo Bebita-Dris
5,26952045
5,26952045
$begingroup$
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
$begingroup$
After taking to Basak: "... the same year a paper was published by Anirudh Prabhu in the United States which used different methods to obtain bounds for the reciprocal sums of prime factors. The constraints on OPNs produced are different, but seem slightly stronger." The link to Prabhu's paper is here.
$endgroup$
– Jose Arnaldo Bebita-Dris
Feb 25 at 12:55
add a comment |
$begingroup$
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
$begingroup$
After taking to Basak: "... the same year a paper was published by Anirudh Prabhu in the United States which used different methods to obtain bounds for the reciprocal sums of prime factors. The constraints on OPNs produced are different, but seem slightly stronger." The link to Prabhu's paper is here.
$endgroup$
– Jose Arnaldo Bebita-Dris
Feb 25 at 12:55
$begingroup$
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
$begingroup$
After taking to Basak: "... the same year a paper was published by Anirudh Prabhu in the United States which used different methods to obtain bounds for the reciprocal sums of prime factors. The constraints on OPNs produced are different, but seem slightly stronger." The link to Prabhu's paper is here.
$endgroup$
– Jose Arnaldo Bebita-Dris
Feb 25 at 12:55
$begingroup$
After taking to Basak: "... the same year a paper was published by Anirudh Prabhu in the United States which used different methods to obtain bounds for the reciprocal sums of prime factors. The constraints on OPNs produced are different, but seem slightly stronger." The link to Prabhu's paper is here.
$endgroup$
– Jose Arnaldo Bebita-Dris
Feb 25 at 12:55
add a comment |
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$begingroup$
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
$begingroup$
After taking to Basak: "... the same year a paper was published by Anirudh Prabhu in the United States which used different methods to obtain bounds for the reciprocal sums of prime factors. The constraints on OPNs produced are different, but seem slightly stronger." The link to Prabhu's paper is here.
$endgroup$
– Jose Arnaldo Bebita-Dris
Feb 25 at 12:55