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15.3.16

Some Recent (Re-)Uploads

Email Thread Regarding OPNThesis1.PDF (as of March 6 2013)




4.2.16

If $N = qn^2$ is an odd perfect number, is it possible to have $q + 1 = \sigma(n)$?

(Note:  This blog post was pulled from this [MO link].)

The title says it all.

Question

If $N = qn^2$ is an odd perfect number, is it possible to have $q + 1 = \sigma(n)$?

Heuristic

From the Descartes spoof, with quasi-Euler prime $q_1$:
$$n_1 = 3003 < \sigma(n_1) = 5376 < q_1 = 22021$$

So it appears that it might be possible to prove that $q + 1 \neq \sigma(n)$.

Motivation

If $q + 1 \neq \sigma(n)$, then it follows that
$$I(q) + I(n) \neq \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q}$$
from which we obtain
$$I(q) + I(n) < \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q},$$
since the reverse inequality
$$\frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < I(q) + I(n)$$
will violate the inequality $I(q) < \sqrt[3]{2} < I(n)$ (see this [paper]).

But the inequality
$$I(q) + I(n) < \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q}$$
implies that the biconditional
$$q < n \Longleftrightarrow \sigma(q) < \sigma(n)$$
holds.

This biconditional is then a key ingredient in the proof of the main result in this [arXiv preprint].

The methods in that preprint are only sufficient to rule out $\sigma(q) = q + 1 = \sigma(n)$ if $3$ does not divide $n$, since we obtain
$$2.799 \approx 1 + 2^{\frac{\log(6/5)}{\log(31/25)}} \leftarrow \frac{q + 1}{q} + \left(\frac{2q}{q + 1}\right)^{\frac{\log(I(5))}{\log(I(5^2))}} \leq I(q) + \left(I(n^2)\right)^{\frac{\log(I(u))}{\log(I(u^2))}}$$
$$< \frac{\sigma(q)}{q} + \frac{\sigma(n)}{n} = \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < \sqrt{3}\left(1 + {10}^{-500}\right) + \left(1 + {10}^{-500}\right) \approx 2.732,$$
(where the smallest prime factor $u$ of $N$ satisfies $u \geq 5$), whence we arrive at a contradiction.

Further Considerations

If $\sigma(q) = q + 1 = \sigma(n)$ and $3 \mid n$, then the same method does not force a contradiction, because we then have
$$2.7199 \approx 1 + 2^{\frac{\log(4/3)}{\log(13/9)}} \leftarrow \frac{q + 1}{q} + \left(\frac{2q}{q + 1}\right)^{\frac{\log(I(3))}{\log(I(3^2))}} \leq I(q) + \left(I(n^2)\right)^{\frac{\log(I(u))}{\log(I(u^2))}}$$
$$< \frac{\sigma(q)}{q} + \frac{\sigma(n)}{n} = \frac{\sigma(q)}{n} + \frac{\sigma(n)}{q} < \sqrt{3}\left(1 + {10}^{-500}\right) + \left(1 + {10}^{-500}\right) \approx 2.732,$$
where $u$ is the smallest prime factor of $N$.

30.1.16

On even almost perfect numbers other than the powers of two, as compared to odd perfect numbers given in Eulerian form

(This question has been cross-posted from MSE to MO.)

Antalan and Tagle (in a 2004 preprint titled Revisiting forms of almost perfect numbers) show that, if $M \neq 2^t$ is an even almost perfect number, then $M$ takes the form $M = {2^r}{b^2}$ where $r \geq 1$ and $b$ is an odd composite.  We call $b^2$ the odd part of the even almost perfect number $M$.

Since $M$ is almost perfect, we have
$$\left(2^{r+1} - 1\right)\left(\sigma(b^2) - b^2\right) = \sigma(b^2) - 1$$
which further implies that

$$2^{r+1} = 1 + \frac{b^2 - 1}{\sigma(b^2) - b^2}.$$

Since $b$ is composite, then we have $\sigma(b^2) > b^2 + b + 1$, from which we obtain the inequality $2^{r + 1} < b$ using the last equation.  This last inequality implies that
$$2^r < 2^{r+1} < b < \sigma(b)$$
and
$$\sigma(2^r) = 2^{r+1} - 1 < b - 1 < b$$
so that we have
$$\frac{\sigma(2^r)}{b} < 1 < 2 < \frac{\sigma(b)}{2^r}.$$

Additionally, since $b^2$ is deficient, we can write $\sigma(b^2) = 2b^2 - c$, where we compute $c$ to be
$$c = b^2 - \frac{b^2 - 1}{\sigma(2^r)}$$
from which we obtain the upper bound
$$\frac{\sigma(b)}{b} < \frac{\sigma(b^2)}{b^2} < \frac{4}{3}.$$
Lastly, since $r \geq 1$ and $2 \mid 2^r$, then
$$\frac{3}{2} = \frac{\sigma(2)}{2} \leq \frac{\sigma(2^r)}{2^r},$$
so that
$$\frac{\sigma(2^r)}{b} < 1 < \frac{\sigma(b)}{b} < \frac{4}{3} < \frac{3}{2} \leq \frac{\sigma(2^r)}{2^r} < 2 < \frac{\sigma(b)}{2^r}.$$

Compare the results we have obtained for even almost perfect numbers other than powers of two with the conjectured inequalities for the divisors of odd perfect numbers $N = {q^k}{n^2}$ given in Eulerian form (see this [link1] and [link2]):

$$\frac{\sigma(q^k)}{n} < 1 < \frac{\sigma(q^k)}{q^k} < \frac{5}{4} < \sqrt{\frac{8}{5}} < \frac{\sigma(n)}{n} < \frac{\sigma(n)}{q^k}$$

Lastly, observe that, for the lone spoof odd perfect number $D = m{n_1}^2 = 198585576189$ that we know of (see this [link3]), we actually have

$$\frac{\sigma(n_1)}{m} = \frac{5376}{22021} < 1 < \frac{22022}{22021} = \frac{\sigma(m)}{m} < \frac{5376}{3003} = \frac{\sigma(n_1)}{n_1} < \frac{\sigma(m)}{n_1} = \frac{22022}{3003},$$

where $m$ denotes the quasi-Euler prime of $D$.


My question is:  Could there be a simple logical explanation for the discrepancies in the inequalities relating the divisors of even almost perfect numbers other than powers of two, odd perfect numbers, and spoof odd perfect numbers

28.1.16

On odd perfect numbers given in Eulerian form - Part 3

Let $N = q^k n^2$ be an odd perfect number given in Eulerian form.

Dr. Patrick Brown (PatrickBrown496 AT gmail.com) has communicated to me an attempt to prove my 2008 conjecture that $q^k < n$.  In particular, Dr. Brown appears to have completed a proof for the inequality $q < n$.  He accomplished this by proving the implication $k = 1 \Rightarrow q < n$. (Update (Feb 6 2016):  Dr. Brown's preprint has appeared in the arXiv.)

In this post, we will investigate the implications of a proof for $k = 1$, in addition to Dr. Brown's claim that $q < n$.

Hereinafter, we will assume the Descartes-Frenicle-Sorli conjecture that $k = 1$.

We then have
$$\frac{\sigma(n^2)}{q} = D(n^2) = 2n^2 - \sigma(n^2) = \gcd\left(n^2, \sigma(n^2)\right).$$

If 
$$\frac{\sigma(n^2)}{q} \leq q,$$
then $n^2 < \sigma(n^2) \leq q^2$, which would contradict $q < n$.

Hence
$$\frac{\sigma(n^2)}{q} > q.$$

We want to show that 
$$\frac{\sigma(n^2)}{q} \neq n^2.$$

Assume that
$$\frac{\sigma(n^2)}{q} = n^2.$$

Then
$$\frac{\sigma(n^2)}{q} = D(n^2) = 2n^2 - \sigma(n^2) = n^2$$
which implies that
$$\sigma(n^2) = n^2.$$
This contradicts the fact that $n > \sqrt[3]{N} > {10}^{500}$.

Observe that the inequality
$$\frac{\sigma(n^2)}{q} > n^2$$
cannot hold because this, together with $q \geq 5$, will imply that $n^2$ is abundant, contradicting $I(n^2) < 2$.

Consequently, we have
$$\frac{\sigma(n^2)}{q} < n^2.$$

Claim:
$$\frac{\sigma(n^2)}{q} > n$$

Suppose that $\sigma(n^2)/q \leq n$.  Then $\sigma(n^2) \leq qn < n^2$, a contradiction.

18.1.16

On odd perfect numbers given in Eulerian form - Part 2

Let $N = q^k n^2$ be an odd perfect number given in Eulerian form.

The following equations can be easily derived:

$$N - (q^k + n^2) + 1 = \sigma(q^{k-1})(q-1)(n+1)(n-1)$$

$$\sigma(n^2) = {q^k}\cdot\left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)$$

$$n^2 = {\frac{\sigma(q^k)}{2}}\cdot\left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)$$

From the last two equations, it can be proved that
$$\gcd(n^2, \sigma(n^2)) = \frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})},$$
and
$$\sigma(n^2) - n^2 = \left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)\cdot\left(q^k - \frac{\sigma(q^k)}{2}\right)$$
$$= \left(\frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}\right)\cdot\left(\frac{q^k - \sigma(q^{k-1})}{2}\right).$$
This last equation expresses a relationship among the quantities
$$\sigma(n^2) - n^2,$$
$$2n^2 - \sigma(n^2),$$
$$\sigma(q^{k-1}),$$
and
$$q^k.$$

In particular, we know that
$$\frac{\sigma(n^2)}{q^k} = \frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}$$
and
$$\frac{\sigma(q^k)}{n^2} = 2\cdot\left(\frac{\sigma(q^{k-1})}{2n^2 - \sigma(n^2)}\right).$$

Notice that, since $\sigma(q^k)\sigma(n^2) = \sigma(N) = 2N = 2{q^k}{n^2}$ and $\gcd(q^k,\sigma(q^k)) = 1$, then$$\frac{\sigma(n^2)}{q^k} = \frac{2n^2 - \sigma(n^2)}{\sigma(q^{k-1})}$$
is an odd integer.

The following papers obtain (ever-increasing) lower bounds for $\sigma(n^2)/q^k$:

(1)  J. A. B. Dris, The Abundancy Index of Divisors of Odd Perfect Numbershttps://cs.uwaterloo.ca/journals/JIS/VOL15/Dris/dris8.html
(2)  J. A. B. Dris and F. Luca, A note on odd perfect numbershttp://arxiv.org/pdf/1103.1437v3.pdf
(3)  F. J. Chen and Y. G. Chen, On Odd Perfect Numbershttp://journals.cambridge.org/action/displayAbstract?fromPage=online&aid=8738171
(4)  K. A. Broughan, D. Delbourgo, and Q. Zhou, Improving the Chen and Chen result for odd perfect numbers, http://www.emis.de/journals/INTEGERS/papers/n39/n39.pdf
(5)  F. J. Chen and Y. G. Chen, On the index of an odd perfect numberhttp://zmath.sub.uni-goettingen.de/ZMATH/serials/en/search/zmath/?q=an:1301.11003&format=complete

From this M. Sc. thesis, we know that
$$\frac{11}{3} \leq \frac{\sigma(q^k)}{n^2} + \frac{\sigma(n^2)}{q^k}.$$

3.1.16

On odd perfect numbers given in Eulerian form - Part 1


Let $N = {q^k}{n^2}$ be an odd perfect number given in Eulerian form.

That is, $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q, n) = 1$.

Since prime powers are not perfect and $\gcd(q, n) = 1$, then $q \neq n$ and $q^k \neq n$ both hold.

We want to prove the following proposition.

PROPOSITION 1:  If $N = {q^k}{n^2}$ is an odd perfect number given in 
Eulerian form with $\sigma(q^k) \neq \sigma(n)$, then the biconditional

$$q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n) \Longleftrightarrow \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$

holds.

PROOF:

Note that the inequation
$$\frac{\sigma(q^k)}{n} \neq \frac{\sigma(n)}{q^k}$$
is trivial, and follows from the fact that $\gcd(q, n) = 1$ and $1 < I(q^k)I(n) < 2$.

We consider three cases:

Case 1.  $\frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k} < I(q^k) + I(n)$

Under this case, we have
$$\left(\sigma(q^k) - \sigma(n)\right)\left(q^k - n\right) < 0.$$

Consequently, we have the biconditional
$$q^k < n \Longleftrightarrow \sigma(n) < \sigma(q^k).$$

However, this biconditional contradicts $I(q^k) < I(n)$ (which can be clearly seen when written in the following form):
$$1 < \frac{\sigma(q^k)}{\sigma(n)} < \frac{q^k}{n} < 1.$$

Hence, in general, the inequality
$$I(q^k) + I(n) \leq \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}$$
holds.


Case 2.  $\frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k} = I(q^k) + I(n)$

Under this case, we have
$$\left(\sigma(q^k) - \sigma(n)\right)\left(q^k - n\right) = 0.$$

Since $q^k \neq n$, this means that
$$\sigma(q^k) = \sigma(n)$$
which is not true by assumption.

Case 3.  $I(q^k) + I(n) < \frac{\sigma(q^k)}{n} + \frac{\sigma(n)}{q^k}$

Under this case, we have
$$\left(\sigma(q^k) - \sigma(n)\right)\left(n - q^k\right) < 0.$$

Consequently, we have the biconditional
$$q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n).$$

This then implies that we have the biconditionals
$$q^k < n \Longleftrightarrow \sigma(q^k) < \sigma(n) \Longleftrightarrow \frac{\sigma(q^k)}{n} < \frac{\sigma(n)}{q^k}$$
under the assumption $\sigma(q^k) \neq \sigma(n)$.

QED.


24.12.15

From MO post http://mathoverflow.net/questions/226841

The title says it all.

What is wrong with this proof that $q < n$, if $N = qn^2$ is an odd perfect number with Euler prime $q$ and $\gcd(q, n) = 1$?

Acquaah and Konyagin showed that $q < (3N)^{1/3}$.  The following proof (communicated to me by Dr. Patrick Brown) is a modification of theirs to strengthen the result to show $q < n$.  (Update (Feb 6 2016):  Patrick's preprint containing a proof for $q < n$ (for all odd perfect numbers) as well as for $q^k < n$ (under some mild conditions) has appeared in the arXiv.)

For the three cases of the proof we write
$$N = q{p^{2b}}{{r_1}^{2\beta_1}}{{r_2}^{2\beta_2}}\cdots{{r_k}^{2\beta_k}}$$
where $p$ is the unique prime whereby $q \mid \sigma(p^{2b})$.  When convenient we will let $w^2 = {{r_2}^{2\beta_2}}\cdots{{r_k}^{2\beta_k}}$, and write $N = q{p^{2b}}{{r_1}^{2\beta_1}}{w^2}$.

Case 1
$$q = \sigma(p^{2b})$$

Note that the assumption $q = \sigma(p^{2b})$ means $p \not{\mid} \sigma(q)$ since $q + 1 \equiv 2 \pmod p$.  So we let $p^{c_i} || \sigma({r_i}^{2\beta_i})$ for $1 \leq i \leq k$.  It is possible that $p^{c_i} = \sigma({r_i}^{2\beta_i})$ for any particular $i$, but since we know $N$ has at least ten components, at least one of the $\sigma({r_i}^{2\beta_i})$ has to have factors other than $p$.  Thus we may rewrite subscripts and assume:
$$\sigma({r_1}^{2\beta_1}) = p^{c_1}{r_2}v$$
for $c_1 \geq 0$, where $v$ is equal to any other primes dividing $\sigma({r_1}^{2\beta_1})$, including multiplicities of $r_2$ should they appear.

We now have what we need to prove this case.  Observe,
$$2N = \sigma(N) = \sigma(q)\sigma(p^{2b})\sigma({r_1}^{2\beta_1})\sigma(w^2).$$
Since $p \not{\mid} \sigma(q)$, then $p^{2b - c_1} \mid \sigma(w^2)$, thus
$$2N > (q + 1)q({p^{c_1}}{r_2})(p^{2b - c_1})$$
$$2N > {q^2}{r_2}{p^{2b}}$$
Now, $r_2$ being an odd prime means $r_2 \geq 3$.  We also note that $p^{2b} > (2/3)\sigma(p^{2b})$.
Consequently,
$$2N > {q^2}(3)\frac{2}{3}\sigma(p^{2b})$$
$$N > q^3$$
from which it easily follows that $q < n$.

Case 2
$$q{r_1} \mid \sigma(p^{2b}), p \not{\mid} \sigma(q)$$

In this case, since $q < \sigma(p^{2b})$, then there is another prime dividing $\sigma(p^{2b})$.  We assume without loss of generality that $r_1 \mid \sigma(p^{2b})$.  This time, we set $w^2 = {{r_1}^{2\beta_1}}\cdots{{r_k}^{2\beta_k}}$.  Observe that $p \not{\mid} \sigma(q)$ implies $p^{2b} || \sigma(w^2)$.  This is all the machinery we need to prove $q < n$ for this case.
$$2N = \sigma(N) = \sigma(q)\sigma(p^{2b})\sigma(w^2)$$
$$2N > (q + 1)q(r_1)(p^{2b})$$
$$2N > {q^2}(r_1)\frac{2}{3}\sigma(p^{2b})$$
$$2N > {q^2}(r_1)\frac{2}{3}{r_1}q$$
$$2N > \frac{2}{3}{r_1}^2{q^3}$$
As above, $r_1 \geq 3$.  Therefore,
$$N > 3q^3$$
and again, $q < n$ easily follows.

Case 3
$$q{r_1} \mid \sigma(p^{2b}), p \mid \sigma(q)$$

We borrow the same proof method Acquaah and Konyagin borrowed from Luca and Pomerance.  We also would not really utilize the hypothesis that $r_1 \mid \sigma(p^{2b})$ as it would not buy us the extra factor we need.  For that, we look back to case 1, and let $p^{c_i} || \sigma({r_1}^{2\beta_i})$, for $1 \leq i \leq k$ and $p^{c_q} || \sigma(q)$.  Again, we assume without loss of generality that $\sigma({r_1}^{2\beta_1}) = p^{c_1}{r_2}v$ for $c_1 \geq 0$ as we did in case 1.

Let $u = \sigma(p^{2b})/q$.  Since
$$\sigma(p^{2b}) \equiv 1 \pmod p, q \equiv -1 \pmod p$$
we know $u \equiv -1 \pmod p$.  Since $u$ is odd we know $u \neq p - 1$ and thus $u \geq 2p - 1$.

By assumption, $c_q \geq 1$.  For $w^2 = {{r_2}^{2\beta_2}}\cdots{{r_k}^{2\beta_k}}$, we have $p^{2b - c_q - c_1} || \sigma(w^2)$, which implies
$$\sigma(w^2) \geq p^{2b - c_q - c_1}.$$
Observe now,
$$p^{2b+1} - 1 = (p - 1)\sigma(p^{2b}) = (p - 1)uq = (p - 1)u\sigma(q) - (p - 1)u.$$
Therefore, $(p - 1)u \equiv 1 \pmod{p^{c_q}}$, which implies that $(p - 1)u > p^{c_q}$.

Combining inequalities yields,
$$\sigma(w^2)(p - 1)u > p^{2b - c_1} \Longrightarrow \sigma(w^2)u > \frac{p^{2b - c_1}}{p - 1}.$$
This should be all we need (for $w$ as defined in case 1):
$$2N = \sigma(N) = \sigma(q)\sigma(p^{2b})\sigma({r_1}^{2\beta_1})\sigma(w^2)$$
$$2N = (q + 1)uq(p^{c_1}{r_2}v)\sigma(w^2)$$
$$2N > {q^2}\frac{p^{2b - c_1}}{p - 1}{p^{c_1}{r_2}}$$
$$2N > {q^2}{r_2}\frac{p^{2b}}{p - 1}$$
$$2N > {q^2}{r_2}\frac{2\sigma(p^{2b})}{3(p - 1)}$$
$$2N > {q^2}{r_2}\frac{2uq}{3(p - 1)}$$
Recall that $u \geq 2p - 1$ and again $r_2$ being an odd prime means $r_2 \geq 3$.
$$2N > {q^3}(3)\frac{2}{3}\frac{2p - 1}{p - 1}$$
$$2N > {q^3}(3)\frac{2}{3}(2)$$
$$N > 2q^3$$
And again, we get $q < n$.