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Sunday, June 23, 2013. Another number theory problem. This is the second post dedicated to the problems set posted on "Math, Math Education, Math Culture" LinkedIn group. Here. Is the original LinkedId discussion, again . if you happen to have a LinkedIn account. Here is the problem:. Prove that the sequence a. 1111, contains an infinite sub-sequence whose terms are pairwise relatively prime. So if $a {n}$ is divisible by $t in mathbb{N}, t 1$ then $a {m cdot n}$ is also divisible by $t$ (1). Both $r,s$ ...

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Sunday, June 23, 2013. Another number theory problem. This is the second post dedicated to the problems set posted on Math, Math Education, Math Culture LinkedIn group. Here. Is the original LinkedId discussion, again . if you happen to have a LinkedIn account. Here is the problem:. Prove that the sequence a. 1111, contains an infinite sub-sequence whose terms are pairwise relatively prime. So if $a {n}$ is divisible by $t in mathbb{N}, t 1$ then $a {m cdot n}$ is also divisible by $t$ (1). Both $r,s$ ...

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rtybase | rtybase.blogspot.com Reviews

https://rtybase.blogspot.com

Sunday, June 23, 2013. Another number theory problem. This is the second post dedicated to the problems set posted on "Math, Math Education, Math Culture" LinkedIn group. Here. Is the original LinkedId discussion, again . if you happen to have a LinkedIn account. Here is the problem:. Prove that the sequence a. 1111, contains an infinite sub-sequence whose terms are pairwise relatively prime. So if $a {n}$ is divisible by $t in mathbb{N}, t 1$ then $a {m cdot n}$ is also divisible by $t$ (1). Both $r,s$ ...

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rtybase: May 2012

http://rtybase.blogspot.com/2012_05_01_archive.html

Tuesday, May 22, 2012. Factoring big numbers and RSA. Last week I came across this article. It's an interesting report indeed and with this post I will explain what is the implication. But first of all, let's quickly recall what RSA encryption. There is a message m. Which needs to be encrypted. There is a public key (e, n). And a private key (d, n). Message is encrypted in the following way c = m. Cipher text is decrypted in the following way m = c. Puting all these together m. Equiv; m (mod n). Now, let...

2

rtybase: February 2012

http://rtybase.blogspot.com/2012_02_01_archive.html

Wednesday, February 29, 2012. The Game Of Life. Did I manage to drag your attention with the title? Good Let's proceed to the lyrical part of this post then. Few years ago I watched this film. So, what is the game about? The player with greater "power" wins. If the players have the same power, the richest one wins. If the players have the same power and wealth, the winner is picked randomly. The game continues for 10000 iterations. Here is the code (slightly adjusted with my help):. Now, let's analyse th...

3

rtybase: Tackling Andrica's conjecture. Part 2

http://rtybase.blogspot.com/2012/11/tackling-andricas-conjecture-part-2.html

Sunday, November 11, 2012. Tackling Andrica's conjecture. Part 2. With the previous post. I stopped at the argument that $f {4}(x)= pi (x) - sqrt{x}$ seems to be ascending for prime arguments. Apparently, this is a necessary and sufficient condition, i.e.:. Andrica's conjecture is true iff function $f {4}(x)= pi (x) - sqrt{x}$ is strictly ascending ($x y Rightarrow f(x) f(y)$) for prime arguments. The proof is given in the following discussion. With the math.stackexchange.com. Now, if $x {0} 0$, where we...

4

rtybase: June 2013

http://rtybase.blogspot.com/2013_06_01_archive.html

Sunday, June 23, 2013. Another number theory problem. This is the second post dedicated to the problems set posted on "Math, Math Education, Math Culture" LinkedIn group. Here. Is the original LinkedId discussion, again . if you happen to have a LinkedIn account. Here is the problem:. Prove that the sequence a. 1111, contains an infinite sub-sequence whose terms are pairwise relatively prime. So if $a {n}$ is divisible by $t in mathbb{N}, t 1$ then $a {m cdot n}$ is also divisible by $t$ (1). Both $r,s$ ...

5

rtybase: May 2013

http://rtybase.blogspot.com/2013_05_01_archive.html

Sunday, May 26, 2013. Tackling Andrica's conjecture. Part 3. Here is an interesting, but slightly detached from the previous two articles. Result . Let's look at the following two sequences: $ a {n}= sqrt{p {n 1} - sqrt{p {n} $ $ b {n}=ln left ( frac{1 sqrt{p {n 1} }{1 sqrt{p {n} } right ) {1 sqrt{p {n 1} }$. Here is a short Python code to visualise the sequences:. And here is how both sequences look like ($a {n}$ the first and $b {n}$ the second):. Indeed they are . According to Mean Value Theorem.

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rtybase

Sunday, June 23, 2013. Another number theory problem. This is the second post dedicated to the problems set posted on "Math, Math Education, Math Culture" LinkedIn group. Here. Is the original LinkedId discussion, again . if you happen to have a LinkedIn account. Here is the problem:. Prove that the sequence a. 1111, contains an infinite sub-sequence whose terms are pairwise relatively prime. So if $a {n}$ is divisible by $t in mathbb{N}, t 1$ then $a {m cdot n}$ is also divisible by $t$ (1). Both $r,s$ ...

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