So…

Oddly enough, this sequence is in the OEIS… A054735 (Idon't know why I couldn't find it earlier!)

But it did mention that the formula to find any term is $p^q + q^p \mod pq$, where $p,q$ is the n^{th} twin prime pair.

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So…

Oddly enough, this sequence is in the OEIS… A054735 (Idon't know why I couldn't find it earlier!)

But it did mention that the formula to find any term is $p^q + q^p \mod pq$, where $p,q$ is the n^{th} twin prime pair.

Well, after a little math…

(1)\begin{align} \pi_{Egyptian} = 3 + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} = 3 + \frac{13}{81} = \frac{3*81+13}{81} = \frac{256}{81} = \frac {(4)^4}{(3)^4} = {\Big{(} \frac{4}{3} \Big{)}}^4 \end{align}

…which is what we gave in class.

But yes, it does come from the Rhind Papyrus.

This post moved me to re-visit Goldbach's conjecture…

I am guessing that the expected value of the difference between consecutive primes, $x = p_{k+1} - p_k$, has something to do with the proof (or lack thereof).

It also may have to do with the fact that we can partition the prime numbers into groups of equal magnitude (in particular, if $m = 2^k$, we get $2^{k-1}$ groups) such that $p_i \equiv r \equiv 2j-1 \mod m, \; \forall \; 1 \leq j \leq 2^{k-1}$.

Thus, for a given $p_1, p_2$ we get $p_1 + p_2 \equiv (r_a + r_b) \mod m$. Also note that $r_a + r_b \equiv 2i \mod m \forall \; 0 \leq i \leq 2^{k-1}-2$ for all possible choice of $r_a, r_b$. These sums are sets even numbers. (Note that these groups should more than span $2 \mathbb{Z}$, although it would be nice to know if there is a reasonably calculable amount of overlap between the groups).

eguzman2 07 Dec 2008 06:55

in discussion Ideas / Student Ideas » Generating Primitive Pythagorean Triples

in discussion Ideas / Student Ideas » Generating Primitive Pythagorean Triples

Here is something that I wanted to tell you about in class but didn't not had time:

Plato’s Formula:

Let m be contained the set of integers with m > 1. Then we can generate a Pythagorean triple by letting a = 2m, b = m^2 – 1, c = m^2 + 1.

Ex. Let m = 2, so we have that a = 2(2) = 4, b = 4 – 1 = 3, c = 4 + 1 = 5.

Then 4^2 + 3^2 = 5^2, which is true.

The formula generates finitely many triple but not all Pythagorean triples.

Formula for generating all Pythagorean triples

Consider the Pythagorean triple x – y – z where (x,y,z) = d. Let x = du,

y = dv, and z = dw where (u,v,w) = 1. So we have u^2 + v^2 = w^2, thus

u – v – w is a triple. Then we can say that every Pythagorean triple is a multiple of primitive triple.

Ex. Let x = 6, y = 8, z = 10. We have that 6^2 + 8^2 = 10^2. Then (6,8,10) = 2. We have that:

x = 2(3) = 6

y = 2(4) = 8

z = 2(5) = 10

This tells that 3 – 4 – 5 is a Pythagorean triple, which we know is true.

Using this idea in the equations for generating infinitely many primitive triples that Dan showed we can generate all triples by adding a parameter d. So we have:

a = d*(m^2 – n^2), b = d*(2mn), c = d*(m^2 + n^2) where m, n, d are contained in the set of integers with m>n>0 and d positive, n or m is odd and (m,n) = 1.

Conclusion:

The formula generates all Pythagorean triples but not uniquely.

Greg Gifford 05 Dec 2008 04:41

in discussion Ideas / Student Ideas » Generating Primitive Pythagorean Triples

in discussion Ideas / Student Ideas » Generating Primitive Pythagorean Triples

There's also another way to generate primitive triples that I don't think was discussed in class, although it isn't linear algebra. If you let $a = 2k+1$ (any odd number), then $b = (a^2-1)/2$ and $c = (a^2+1)/2$. To prove it:

$a^2+b^2 = c^2$

$(2k+1)^2+(((2k+1)^2-1)/2)^2 = (((2k+1)^2+1)/2)^2$

$(2k+1)^2+((4k^2+4k+1-1)/2)^2 = ((4k^2+4k+1+1)/2)^2$

$(2k+1)^2+(2k^2+2k)^2 = (2k^2+2k+1)^2$

$4k^2+4k+1+4k^4+8k^3+4k^2 = 4k^4+8k^3+8k^2+4k+1$

$4k^4+8k^3+8k^2+4k+1 = 4k^4+8k^3+8k^2+4k+1$

jonas2 02 Dec 2008 17:34

in discussion Ideas / Student Ideas » Generating Primitive Pythagorean Triples

in discussion Ideas / Student Ideas » Generating Primitive Pythagorean Triples

Taken from the Wolfram website. I am pretty sure this method wasn't covered but I can't always be sure…

If $<a_0, b_0, c_0>$ is a Primitive Pythagorean Triple, then $<a_0, b_0, c_0>U_i$ generates a new primitive triple $<a_i, b_i, c_i>$ where

$U_1 = \left| \begin{array}{ccc} 1 & 2 & 2 \\ -2 & -1 & -2 \\ 2 & 2 & 3 \end{array} \right| ,$ $U_2 = \left| \begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 3 \end{array} \right| ,$ $U_3 = \left| \begin{array}{ccc} -1 & -2 & -2 \\ 2 & 1 & 2 \\ 2 & 2 & 3 \end{array} \right|$.

Let's exam a simple example. We were told that $<3, 4, 5>$ is a primitive triple. So…

$<3 ,4, 5>U_1 = <5, 12, 13> ,$ $<3, 4, 5>U_2 = <21, 20, 29> ,$ $<3, 4, 5>U_3 = <15, 8, 17> ,$

…which can easily be verified to be valid!

So remembering that $\phi(x)$ is multiplicative, we note the following: suppose $n = p_1^{a_1} * p_2^{a_2}$, with $p_1, p_2 \neq 2$ (there could be a 2 in there, but not more than one, since $\phi(2) = 1$ but that 4|$\phi(2^k)$ and 14 isn't divisible by 4). Then $\phi(n) = p_1^{a_1 -1}(p_1 -1) + p_2^{a_2 -1}(p_2 -1)$. But 2|$p_1 -1$ and 2|$p_2 -1$ and thus that implies that 4|$\phi(n)$ which would be impossible. And thus n cannot be a product of odd primes. So the only case left is that $n=p^{a}$ (again possibly times 2), and since p is odd let p = 2k+1, $k \in \mathbb{Z}$. But then $\phi(n) = p^{a-1}(p-1)$ which implies that $\frac{14}{2} = 7 = (2k+1)^{a-1}*k$. And then there's no k that makes that work (that would need to be more rigorous, but it's certainly a start).

A nontotient number $n \in \mathbb{N}$ is a number such that there is no $x \in \mathbb{N}$ where $\phi(x) = n$.

The smallest such number is 14.

My question is how would one prove for example that 14 is nontotient?

Dan Bergren 28 Nov 2008 23:55

in discussion Ideas / Student Ideas » What's Special About This Number?

in discussion Ideas / Student Ideas » What's Special About This Number?

I was using StumbleUpon and I stumbled across a website that listed something special about a lot of the numbers between 0 and 9999. If anything, it seems like a great jumping off point for exploring unfamiliar topics in number theory.

here's an amazing link for a bunch of proofs of Fermat's last theorem for other n, such as 3 and 5. There's even a proof for n being a prime number.

http://fermatslasttheorem.blogspot.com/2005/05/fermats-last-theorem-proof-for-n3.html

If you found yourself thinking what other proofs are out there take a look at these:

http://www.mcs.csuhayward.edu/~malek/Mathlinks/Pi.html

Define

(1)\begin{align} M(n) = \sum_{1\le k \le n} \mu(k) \end{align}

where $\mu$ is the Moebius function

Merten's conjecture says that

(2)\begin{align} \left| M(n) \right| < \sqrt {n} \end{align}

Now in 1985, this conjecture was proved false and also that a counter example exists somewhere between $10^{14}$ and $e^{1.59 \cdot 10^{40}}$.

So this result kind of kills peoples argument that "Goldbach's Conjecture is true because we've seen it to be true for any number tested".

But the fact that it's true for so many numbers makes me think it's more than just a coincidence. What do you guys think?

Let $F_k$ be the $k$th Fibonacci number.

Define

(1)\begin{align} \psi = \sum_{k=1}^{\infty} \frac{1}{F_k} \approx 3.359885666243177553172011302918927179688905133731 \ldots \end{align}

the sum of the reciprocals of every Fibonacci number

It has been proved to be irrational which looks like a daunting task to me

heres a link for more info:

http://en.wikipedia.org/wiki/Reciprocal_Fibonacci_constant

Similarly, I have found other series converging to various multiples of $\pi$ in one of my other classes. Sadly, this of course requires methods outside the scope of this class but it's still pretty interesting that a large sum of rationals add to such a peculiar irrational.

$\frac{\pi^2}{8} = \[\sum_{n=0}^{\infty} \frac{1}{(2n+1)^2} \]$

This sum can be derived with the Fourier sine series of $f(x)=1$ on $(0,\pi)$.

$\frac{\pi^2}{6} = \[\sum_{n=1}^{\infty} \frac{1}{n^2} \]$

This sum can be derived with the Fourier sine series of $f(x)=x$ on $(0,l)$.

$\frac{\pi^4}{90} = \[\sum_{n=1}^{\infty} \frac{1}{n^4} \]$

This sum can be derived with the Fourier cosine series of $f(x)=x^2$ on $(0,l)$.

(Additionally, this required the use of Parseval's Equality).

Chip and Josh in class today gave the series $\sum_{i=1}^{k} (-1)^{i-1}\frac{4} {2 i - 1} = 4 - \frac{4}{3} + \frac{4}{5} - \frac{4}{7} ...$ as an example of a series that converges to $\pi$.

After doing some further research, I stumbled upon two other series that have convergence related to $\pi$:

**Ramanujan's Formula**

\begin{align} \pi = 2 \sqrt{3} \sum_{n=0}^{\infty}{\frac{(-1)^{n}}{(2n+1)3^{n}}} \end{align}

**Chudnovsky Algorithm**

\begin{align} \frac{1}{\pi} = 12\sum_{k=0}^{\infty}{\frac{(-1)^{k}(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}640320^{3k+\frac{3}{2}}} \end{align}

The Chudnovsky Algorithm is based on a rapidly converging hypergeometric series. It was used to generate over a billion digits of $\pi$! Mathematica uses it today to calculate $\pi$.

"We have known since the 1920s that the first two numbers are 1 and 2, but it wasn't until a few years ago that mathematicians conjectured that the third number in the sequence may be 42—a figure greatly significant to those well-versed in The Hitchhiker's Guide to the Galaxy."

Haha, that would be amazing if it really was 42. Then it really could be the answer to life, the universe, and everything!

"In their search for patterns, mathematicians have uncovered unlikely connections between prime numbers and quantum physics. Will the subatomic world help reveal the elusive nature of the primes?"

My boyfriend sent me this article a couple of weeks ago, and I had been meaning to post it to the forum. It seems only appropriate now! I don't know how much has changed since this was published in 2006, but it's still interesting anyway.

Here you go: Riemann

So after lecture last week I was interested in the difference between primality tests that require a factorization and tests that don't. We saw one or two tests last week that didn't require a factorization, So I decided to look for more and have also included some that do require a factorization.

Miller-Rabin-http://en.wikipedia.org/wiki/Miller-Rabin_primality_test

Solovay-Strassen:-http://en.wikipedia.org/wiki/Solovay–Strassen_primality_test

AKS:http://en.wikipedia.org/wiki/AKS_primality_test

there are many many more if you're interested and wikipedia has many links.

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