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Numbers 101

Some random thoughts on pseudo-random things

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Numbers 101 | arnienumbers.blogspot.com Reviews
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Some random thoughts on pseudo-random things
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1 search this blog
2 case 1
3 case 2
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5 then$
6 posted by
7 and also that
8 frac{2n 2}{q k}$
9 note that
10 we obtain
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Numbers 101 | arnienumbers.blogspot.com Reviews

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Some random thoughts on pseudo-random things

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1

Numbers 101: On a Conjecture of Dris Regarding Odd Perfect Numbers

http://arnienumbers.blogspot.com/2014/11/on-conjecture-of-dris-regarding-odd.html

Some random thoughts on pseudo-random things. On a Conjecture of Dris Regarding Odd Perfect Numbers. Redirects to an arXiv document. Here is the abstract:. On a Conjecture of Dris Regarding Odd Perfect Numbers. Arnie B. Dris. There was an error in this gadget. There Is No Place, Like Home. There Is No Place, Like Home. Arnie B. Dris. View my complete profile.

2

Numbers 101: The Abundancy Index of Divisors of Spoof Odd Perfect Numbers

http://arnienumbers.blogspot.com/2015/03/the-abundancy-index-of-divisors-of.html

Some random thoughts on pseudo-random things. The Abundancy Index of Divisors of Spoof Odd Perfect Numbers. I have a new paper out there (currently already in Scribd. To summarize: I extended the results that I have obtained in my previous papers on odd perfect numbers, to the case of spoof odd perfect numbers, also known as Descartes numbers in the literature. Arnie B. Dris. There was an error in this gadget. There Is No Place, Like Home. There Is No Place, Like Home. Arnie B. Dris.

3

Numbers 101: Improving the bound $q < n\sqrt{3}$ for an odd perfect number $N = {q^k}{n^2}$ given in Eulerian form

http://arnienumbers.blogspot.com/2014/12/improving-bound-q-nsqrt3-for-odd.html

Some random thoughts on pseudo-random things. Improving the bound $q. Http:/ math.stackexchange.com/questions/1009929. Arnie B. Dris. There was an error in this gadget. There Is No Place, Like Home. There Is No Place, Like Home. Arnie B. Dris. View my complete profile.

4

Numbers 101: Improving the lower bound for $I(n)$ where $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, with $n < q$

http://arnienumbers.blogspot.com/2015/01/improving-lower-bound-for-in-where-n.html

Some random thoughts on pseudo-random things. Improving the lower bound for $I(n)$ where $N = {q k}{n 2}$ is an odd perfect number given in Eulerian form, with $n. Per Will Jagy's answer (and a subsequent comment by Erick Wong) to the following MSE question. We have the (sharp? 1 leq frac{I(x 2)}{I(x)} leq prod {p}{ frac{p 2 p 1}{p 2 p} = frac{ zeta(2)}{ zeta(3)} approx 1.3684327776 ldots$. Now, let $N = {q k}{n 2}$ be an odd perfect number given in Eulerian form. We have the following biconditional:.

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danvk.org » Beal’s Conjecture

http://www.danvk.org/wp/beals-conjecture

Keepin’ static like wool fabric since 2006. If m, n, r ≥ 3 and x, y, z are integers, and x. Then x, y and z share a common factor. Several years ago, Peter Norvig. Wrote a computer program to search for counterexamples. Norvig’s program was written in Python and run on a 400 MHz processor. I have access to a recent 64-bit AMD CPU with lots of RAM, so I decided to update his program and see how far I could push the threshhold. And the exponents ( max pow. In this case, I used max base = max pow = 1000.

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Perfect numbers | Irrational Inquiries

https://polytopal.wordpress.com/2010/10/25/perfect-numbers

A continuous injection of pure mathematics. Equations, and undoing operations. The first infinite number →. One would be hard put to find a set of whole numbers with a more fascinating history and more elegant properties surrounded by greater depths of mystery and more totally useless than the perfect numbers. Martin Gardner, 1997. This entry is dedicated to my beloved wife, Jennifer—for supporting me in my goals, for holding me to my promises, and for putting up with my obsessions. 1 2 4 7 14 = 28.

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Numbers 101

Some random thoughts on pseudo-random things. If $ frac{σ(x)}{x}= frac{p 2}p$ where $p$ is an odd prime, does it follow that $x$ is an odd square? Note: The following proof was copied verbatim from the answer. Of MSE user Alex Francisco. First, note that for any coprime $a, b in mathbb{N} $, there is$. I(ab) = I(a) I(b). Suppose there is an even number $n = 2 k cdot l$, where $k geq 1$ and $l$ odd, such that$. I(n) = frac{p 2}{p}. K geq 2$. Then$. Which implies $4(p 2) geq 7p$, contradictory to $p geq 3$.

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