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.

Forum category: Ideas / Student Ideas

Forum thread: Goldbach's twin primes ]]>

…which is what we gave in class.

But yes, it does come from the Rhind Papyrus.

Forum category: Ideas / Student Ideas

Forum thread: Pi and the ancient Egyptians ]]>

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).

Forum category: Ideas / Student Ideas

Forum thread: Merten's Conjecture ]]>

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.

Forum category: Ideas / Student Ideas

Forum thread: Generating Primitive Pythagorean Triples ]]>

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

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

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

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

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

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

Forum category: Ideas / Student Ideas

Forum thread: Generating Primitive Pythagorean Triples ]]>

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!

Forum category: Ideas / Student Ideas

Forum thread: Generating Primitive Pythagorean Triples ]]>

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).

Forum category: Ideas / Student Ideas

Forum thread: Nontotient numbers ]]>

The smallest such number is 14.

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

Forum category: Ideas / Student Ideas

Forum thread: Nontotient numbers ]]>

Forum category: Ideas / Student Ideas

Forum thread: What's Special About This Number? ]]>

Forum category: Ideas / Student Ideas

Forum thread: Fermat's Last Theorem Documentary ]]>

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

Forum category: Ideas / Student Ideas

Forum thread: Fermat's last theorem for other n ]]>

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

http://www.lrz-muenchen.de/~hr/numb/pi-irr.html

http://www.mathpages.com/home/kmath313.htm

Forum category: Ideas / Student Ideas

Forum thread: Other proofs that Pi is irrational ]]>

Forum category: Ideas / Student Ideas

Forum thread: Pi and the ancient Egyptians ]]>

\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?

Forum category: Ideas / Student Ideas

Forum thread: Merten's Conjecture ]]>

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

Forum category: Ideas / Student Ideas

Forum thread: Reciprocal Fibonacci constant ]]>

$\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).

Forum category: Ideas / Student Ideas

Forum thread: Series Converging 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 $\pi$.

Forum category: Ideas / Student Ideas

Forum thread: Series Converging to Pi ]]>

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

Forum category: Ideas / Student Ideas

Forum thread: Riemann Zeta Function ]]>

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

Forum category: Ideas / Student Ideas

Forum thread: Riemann Zeta Function ]]>

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.

Forum category: Ideas / Student Ideas

Forum thread: Primality Testing ]]>

Forum category: Ideas / Student Ideas

Forum thread: A look at Carl Gustav Jacob Jacobi ]]>

\begin{align} \left(\frac{a}{n}\right) = \left(\frac{a}{p_1}\right)^{b_1} \left(\frac{a}{p_2}\right)^{b_2}... \left(\frac{a}{p_r}\right)^{b_r} \end{align}

where $\left(\frac{a}{p_1}\right),\left(\frac{a}{p_2}\right),...,\left(\frac{a}{p_r}\right)$ are Legendre symbols.

Now, that's cool and all, but did you know:

- There is a crater on the Moon named after him?
- The phrase "Invert, always invert," is associated with Jacobi because he thought that problems were best addressed backwards?
- He contributed to a great number of mathematical areas, including number theory, vector theory, and partial differential equations.

C.J. Scriba once compared Jacobi to Euler, saying,

"Jacobi and Euler were kindred spirits in the way they created their mathematics. Both were prolific writers and even more prolific calculators; both drew a great deal of insight from immense algorithmical work; both laboured in many fields of mathematics (Euler, in this respect, greatly surpassed Jacobi); and both at any moment could draw from the vast armoury of mathematical methods just those weapons which would promise the best results in the attack of a given problem."

Wikipedia also mentioned that Jacobi was "the most inspiring teacher of his time" and I was wondering if there were any other mathematicians out there who were similarly inspiring and how so. Does anyone out there have a mathematician who inspires them to tackle tough problems and plow through mathematics?

I've always found Paul Erdős to be a huge inspiration because of his undying dedication to mathematics and his treatment of it as a social discipline and one that is always thoroughly interesting. Check out Paul Hoffman's The Man Who Loved Only Numbers for a great read on Erdős.

Forum category: Ideas / Student Ideas

Forum thread: A look at Carl Gustav Jacob Jacobi ]]>

Forum category: Ideas / Student Ideas

Forum thread: Project Euler ]]>

Here is the conjecture take from wikipedia article Artin's conjecture:

Let a be an integer which is not a perfect square and not -1. Denote by S(a) the set of prime numbers p such that a is a primitive root modulo p. Then

1. S(a) has a positive Schnirelmann density inside the set of primes. In particular, S(a) is infinite.

2. under the condition that a be squarefree, this density is independent of a and equals the Artin constant which can be expressed as an infinite product

I would guess that the Schnirelmann density obtain is greater when a particular *a* has a large set of primes for which *a* is a primitive root.

Forum category: Ideas / Student Ideas

Forum thread: Artin's conjecture on primitive roots ]]>

For example:

$13 \equiv 1 \mod{12}$, 13 is prime

$25 \equiv 1 \mod{12}$, $25=5*5, 5 \equiv 5 \mod{12}$

$37 \equiv 1 \mod{12}$, 37 is prime

$49 \equiv 1 \mod{12}$, $49=7*7, 7 \equiv 7 \mod{12}$

$61 \equiv 1 \mod{12}$, 61 is prime

$85 \equiv 1 \mod{12}$, $85=17*5, 17 \equiv 5 \mod{12}$

$109 \equiv 1 \mod{12}$, 109 is prime

$121 \equiv 1 \mod{12}$, $121=11*11, 11 \equiv 11 \mod{12}$

$133 \equiv 1 \mod{12}$, $133=19*7, 19 \equiv 7 \mod{12}$

$157 \equiv 1 \mod{12}$, 157 is prime

$169 \equiv 1 \mod{12}$, $169=13*13, 13 \equiv 13 \mod{12}$

I have no idea if what this means (if it means anything at all), but I haven't found anything through Google on it yet. I also haven't tried to prove it yet, so it might not even hold true for all *k*, but I still thought it was worth posting.

Forum category: Ideas / Student Ideas

Forum thread: An interesting find... ]]>

This Wednesday, one of UIUC's best grad students will be giving a talk entitled "The Birth of Analytic Number Theory" to the local undergrad math club (MATRIX). It will be held at 7pm in Altgeld 245, and should be great. The abstract is below. I **highly** encourage people to attend.

ABSTRACT: In the mid-1700's Euler made the first significant advance in the study of the distribution of prime numbers that had been made since Euclid's proof that there are infinitely many prime numbers. He established what is now known as the Euler product formula and used it to show that the sum of reciprocals of prime numbers diverges. This last result can be interpreted as a statement about how slow the prime numbers thin out. The influence of Euler's paper has been vast and can be felt on much contemporary research in number theory and related areas. In this talk I will derive Euler's product formula and show how Euler proved that the sum of reciprocals of primes diverges. If time permits I will show how he subsequently used his method to establish that the prime numbers are in some sense equally split between the arithmetic progressions 4k + 1 and 4k + 3.

Forum category: Ideas / Student Ideas

Forum thread: The Birth of Analytic Number Theory ]]>

http://www.ocf.berkeley.edu/~gagnanda/mathstuff/Touchard.pdf

Forum category: Ideas / Student Ideas

Forum thread: Odd Perfect Numbers ]]>

\begin{align} a_n = \left[ \frac{2^{n}+10}{6}} \right] = \left[ \frac{2^{n-1}+5}{3}} \right] \end{align}

Forum category: Ideas / Student Ideas

Forum thread: Quadratic Residues Mod 2^k ]]>

More importantly: good find! Any idea of the function which is behind the table that you posted?

Forum category: Ideas / Student Ideas

Forum thread: Quadratic Residues Mod 2^k ]]>

It is fairly easy to see that all integers are *quadratic* residues mod 2, and that 0 and 1 are both *quadratic* residues mod 2.

But what about for larger powers of 2?

For $k = 2$, that is, mod 4, the *quadratic* residues are also 0 and 1. ($0^2 \equiv 2^2 \equiv 0 \mod 4, 1^2 \equiv 3^2 = 9 \equiv 1 \mod 4$). For $k = 3$, the *quadratic* residues are 0, 1, 4; for $k = 4$, the *quadratic* residues are 0, 1, 4, 9. Do we have a pattern? Not quite.

For $k = 5$, the *quadratic* residues are 0, 1, 4, 9, 16, 25, and **17**.

For $k = 6$, the *quadratic* residues are 0, 1, 4, 9, 16, 25, 36, 49, and 17, 33, 41, 57.

For $k = 7$, we get 0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, and 17, 33, 41, 57, and 65, 73, 89, 97, 105, 113, and 68.

We can surely generate the *quadratic* residues of $2^k$ from the first $2^{k-1}$ elements. But $\forall k \geq 2$, we can generate the *quadratic* residues from the first $2^{k-2}$ elements. (I cannot figure out an easy proof of this.)

According to Wikipedia, a non-zero number will be a *quadratic* residue $\mod 2^k, k > 3$ if it is in the form $4^n \times (8m+1)$ with $0 \leq n \leq \frac{k}{2}, 0 \leq m < 2^{k-3}$.

A rough sketch of a proof was offered; but my observations can confrim it (i.e., $68 = 4 \times 17 = 4 \times (2 \times 8 + 1)$).

Including the trival *quadratic* residue 0, for the given $k$, there are # residues:

$k$ = | # quadratic residues |

1 | 2 |

2 | 2 |

3 | 3 |

4 | 4 |

5 | 7 |

6 | 12 |

7 | 23 |

8 | 44 |

*edit: added "quadratic" to every instance of the word "residue" for Andy's benefit.*

Forum category: Ideas / Student Ideas

Forum thread: Quadratic Residues Mod 2^k ]]>

Forum category: Ideas / Student Ideas

Forum thread: Legendre ]]>

Forum category: Ideas / Help!

Forum thread: Ch.4 24. c ]]>

and so that both *a* and *b* are squares modulo *p*.

Does that help?

Forum category: Ideas / Help!

Forum thread: Ch.4 24. c ]]>

Forum category: Ideas / Help!

Forum thread: Ch.4 24. c ]]>

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

Forum category: Ideas / Student Ideas

Forum thread: Legendre ]]>

Forum category: Ideas / Student Ideas

Forum thread: Credit Card Numbers and Very Large Primes ]]>

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Forum thread: proofs of quadratic reciprocity ]]>

Forum category: Ideas / Student Ideas

Forum thread: Legendre/Jacobi Symbol Extended ]]>

$\delta_{ij} = \begin{cases} 1 &\text{if } i = j \\ 0 &\text{if } i \neq j \end{cases}$

where the numbers $i$ and $j$ can be integers.

A version of this is used in digital signal processing (DSP) where signals are represented by a sequence of numbers. (Note: DSP includes subfields such as audio and speech signal, radar signal, statistical signal, and seismic data processing. Credit goes to Wikipedia.) It it is then displayed as

$\delta[n] = \begin{cases} 1 &\text{if } n = 0 \\ 0 &\text{if } n \neq 0 \end{cases}$

for $n \in \mathbb{Z}$.

This bit of information was taken from a Wikipedia article found while looking up the Kronecker Symbol.

Forum category: Ideas / Student Ideas

Forum thread: The Kronecker Symbol ]]>

The Kronecker symbol is an extension of the Jacobi Symbol to all integers.

For general n, we write n = $2^c m$, with m odd and define: (1)

\begin{align} \left( \frac {a}{n} \right) = \left( \frac {a}{2} \right) ^c \left( \frac {a}{m} \right) \end{align}

where $\left( \frac {a}{m} \right)$ is the Jacobi symbol.

And define:

\begin{align} \left( \frac {a}{2} \right) = \begin{cases} 0 &\text{if }a \equiv 0\pmod{4}\ & 1& \text{if } a \equiv 1\pmod{8}\ & -1& \text{if } a \equiv 5\pmod{8}\end{cases}\ \end{align}

Forum category: Ideas / Student Ideas

Forum thread: The Kronecker Symbol ]]>

Given that 9907 is prime calculate $\left(\frac{1001}{9907}\right)$.

Calculations using the Legendre symbol:

Calculations using the Jacobi symbol:

Example take from Jacobi symbol at wikipedia

Forum category: Ideas / Student Ideas

Forum thread: Legendre/Jacobi Symbol Extended ]]>

$\forall \;d \leq k, d \mid k!$

$\sum_{d \mid k!} \mu(d)d^{-s}& = \sum_{j=1}^k \mu(j)j^{-s}& + \sum_{j=1}^k \mu(\frac{k!}{j})(\frac{k!}{j})^{-s}$

Let's call the latter sum $\alpha_{k}$.

Now, the function $f(n) = n!$ increases extremely faster than $g(n) = n$.

($n!$ increases much faster than $e^{n}$ in fact.)

So as $k \rightarrow \infty,\; (\frac{k!}{j})^{-s} \rightarrow 0$

This implies $\alpha_{k} \rightarrow 0$ as $k \rightarrow \infty$

Therefore $\lim_{k \to \infty} \sum_{d \mid k!} \mu(d)d^{-s}& = \lim_{k \to \infty} \sum_{j=1}^k \mu(j)j^{-s}$

$\sum_{j=1}^\infty \frac{\mu(j)}{j^{s}}& = \prod_{p} (1-p^{-s})& = \frac{1}{\zeta(s)}$ as desired. □

Forum category: Ideas / Student Ideas

Forum thread: Reciprocal of the Riemann zeta function ]]>

The direct comparison test requires $|a_{n}| \leq |b_{n}|$ for sufficiently large n.

But that's ok because $|\mu(j)j^{-s}| \leq j^{-s}$, so the test still gives us the result we wanted.

Forum category: Ideas / Student Ideas

Forum thread: Reciprocal of the Riemann zeta function ]]>

There seems to be a problem with your argument. Namely, you claim that for every $d \leq k!$ we have $d \mid k!$. Unfortunately, though, this isn't quite true. For instance, $4! = 24$, but not all $j \leq 24$ are divisors of 24.

There might be a fix for this, though…can you see what it would be?

Forum category: Ideas / Student Ideas

Forum thread: Reciprocal of the Riemann zeta function ]]>

$\zeta(s) =\sum_{n=1}^\infty \frac{1}{n^s}$ where $s \in \mathbb{R}$ and $s>1$.

This function can be extended to all $s \in \mathbb{C}, s \neq 1$ which leads to much more such as the Riemann hypothesis, but I'm just going to focus on $s \in \mathbb{R}$. This is my favorite function and I've seen the formula for its inverse but never seen a derivation. But I think I was able to prove it myself thanks to our homework. So here it is:

One cool identity is called the Euler product formula:

$\zeta(s) = \prod_{p \text{ prime}} \frac{1}{1-p^{-s}}$

-Proof*:

$\zeta(s) = 1+\frac{1}{2^s}+\frac{1}{3^s}+\frac{1}{4^s}+\frac{1}{5^s}+ \cdots$

$\frac{1}{2^s}\zeta(s) = \frac{1}{2^s}+\frac{1}{4^s}+\frac{1}{6^s}+\frac{1}{8^s}+\frac{1}{10^s}+ \cdots$

Subtracting the second from the first we remove all elements that have a factor of 2:

$\left(1-\frac{1}{2^s}\right)\zeta(s) = 1+\frac{1}{3^s}+\frac{1}{5^s}+\frac{1}{7^s}+\frac{1}{9^s}+\frac{1}{11^s}+\frac{1}{13^s}+ \cdots$

Repeating for the next term:

$\frac{1}{3^s}\left(1-\frac{1}{2^s}\right)\zeta(s) = \frac{1}{3^s}+\frac{1}{9^s}+\frac{1}{15^s}+\frac{1}{21^s}+\frac{1}{27^s}+\frac{1}{33^s}+ \cdots$

Subtracting again we get:

$\left(1-\frac{1}{3^s}\right)\left(1-\frac{1}{2^s}\right)\zeta(s) = 1+\frac{1}{5^s}+\frac{1}{7^s}+\frac{1}{11^s}+\frac{1}{13^s}+\frac{1}{17^s}+ \cdots$

where all elements having a factor of 3 or 2 (or both) are removed.

It can be seen that the right side is being sieved. Repeating infinitely we get:

$\cdots \left(1-\frac{1}{11^s}\right)\left(1-\frac{1}{7^s}\right)\left(1-\frac{1}{5^s}\right)\left(1-\frac{1}{3^s}\right)\left(1-\frac{1}{2^s}\right)\zeta(s) = 1$

Dividing both sides by everything but the $\zeta(s)$ we obtain:

$\zeta(s) = \frac{1}{\left(1-\frac{1}{2^s}\right)\left(1-\frac{1}{3^s}\right)\left(1-\frac{1}{5^s}\right)\left(1-\frac{1}{7^s}\right)\left(1-\frac{1}{11^s}\right) \cdots }$

This can be written more concisely as an infinite product over all primes ''p'':

$\zeta(s)\;=\;\prod_{p} (1-p^{-s})^{-1}$

*Taken from Wikipedia

Now, we proved in our homework if $f(n)$ is multiplicative and $n = p_{1}^{a_{1}}... \;p_{k}^{a_{k}}$, then $\sum_{d \mid n} \mu(d)f(d)& = \prod_{i=1}^k (1-f(p_{i}))$ where $\mu$ is the Möbius function.

Now let $n = k!$ and $f(n) = n^{-s}$ (which is multiplicative).

So we obtain;

$\sum_{d \mid k!} \mu(d)d^{-s}& = \prod_{i=1}^r (1-p_{i}^{-s})$, but since all $d \leq k!$ divide $k!$

$\sum_{j=1}^{k!} \mu(j)j^{-s}& = \prod_{i=1}^r (1-p_{i}^{-s})$

$\lim_{k \to \infty}\sum_{j=1}^{k!} \mu(j)j^{-s}& = \lim_{r \to \infty}\prod_{i=1}^r (1-p_{i}^{-s})$, so

$\sum_{j=1}^{\infty} \mu(j)j^{-s}& = \prod_{p} (1-p^{-s}) = \frac{1}{\zeta(s)}$

Now we must test $\sum_{j=1}^{\infty} \mu(j)j^{-s}$ for convergence, but this is easy to see since

$\mu(j)j^{-s} \leq j^{-s} \;\forall\; j \in \mathbb{N}$

By the direct comparison test, $\sum_{j=1}^{\infty} \mu(j)j^{-s}$ converges if $\zeta(s)$ does as well.

SO…

$\frac{1}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^{s}}$

Forum category: Ideas / Student Ideas

Forum thread: Reciprocal of the Riemann zeta function ]]>

It's pretty interesting because the Jacobi Symbol have same properties as the Legendre Symbol, and then some.

Forum category: Ideas / Student Ideas

Forum thread: Legendre/Jacobi Symbol Extended ]]>

Here's the link

http://mathworld.wolfram.com/QuasiamicablePair.html

Forum category: Ideas / Student Ideas

Forum thread: Augmented Amicable Pairs ]]>

Some colossally abundant numbers are 2, 6, 12, … But six is also a perfect number. I didn't think that a number could be perfect and abundant. But apparently, a number can be perfect and colossally abundant.

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

Forum category: Ideas / Student Ideas

Forum thread: Abundant and Deficient numbers ]]>

A pair of numbers m and n such that

(1)\begin{align} \sigma (m) = \sigma (n) = m + n - 1 \end{align}

It said that only 11 pairs of augmented amicable pairs have been found.

I was wondering if there were any other types of augmented pairs, maybe such that

\begin{align} \sigma (m) = \sigma (n) = m + n + 1 or =m + n - 2 \end{align}

Forum category: Ideas / Student Ideas

Forum thread: Augmented Amicable Pairs ]]>

\begin{align} f(n) = \left\{\begin{array}{ll} 1 & n = 1 \\ f\left(\frac{n}{2}\right) & n\mbox{ is even} \\ f(3n+1) & n\mbox{ is odd and }n>1\end{array}\right. \end{align}

An example would be $f(11)$:

(2)\begin{align} 11\to 34 \to 17 \to 52 \to 26 \to 13 \to 40 \to 20 \to 10 \to 5 \to 16 \to 8 \to 4 \to 2 \to 1. \end{align}

So, 11 finally terminates at 1. The big (and unsolved) question is: does every positive integer terminate, i.e., does every input eventually reach 1?

It is easy to see that every $n = 2^a$ reaches 1 eventually, so in some sense, this is some measure of how close numbers are to powers of 2. As for determining whether the "$3x+1$ conjecture" is true, one method is to do some sort of strong induction argument: if $f(n) = f(k)$ for some $k < n$, then we say $f(n)$ terminates by our inductive hypothesis.

If we use this method, we don't need to look at any even numbers since $f(2m) = f(m)$. In addition, we don't need to examine numbers of the form $4m + 1$ since $3(4m + 1) + 1 = 12m + 4 = 4(3m + 1)$ and hence $f(4m + 1) = f(3m + 1)$ for $m > 1$. Thus, we've narrowed it down to numbers of the form $4x + 3$. However, this is still a rather harrowing task since there doesn't seem to be much of a pattern in the number of iterations (at least to me): $f(27)$ and $f(31)$ take over 90 iterations to dip below their starting points, but $f(35)$ takes less than 10.

Incidentally, there's a group in Portugal attempting to "prove" the $3x+1$ conjecture via computing "all" the possibilities. In September, they verified the conjecture for numbers through $19\cdot 2^{58} \approx 5.5 \cdot 10^{18}$. They also have done work on the similar $5x+1$ and $7x+1$ problems.

Forum category: Ideas / Student Ideas

Forum thread: 3x+1 Problem ]]>

Forum category: Ideas / Student Ideas

Forum thread: Sociable Numbers and the Collatz Conjecture ]]>

I noticed, though, when I made a step-counting function to calculate the total number of steps (both $3x+1$ and $\frac{x}{2}$ steps) it takes $N$ to arrive at 1, I saw that multiple numbers can share the same step count (this is different from the dropping time, which is the number of steps it takes to reach an $N' < N$). I rapidly deduced that for a given odd number $n = (2x+1)$, it will share the same total number of steps to reach 1 as the number $6n + 2 = 2 \times (3n+1) = (12x+8)$. But this does not account for the behavior I describe below.

Both 14 and 15 take 17 steps; 3, 20, 21 and 128 each take 7 steps. This phenomenon occurs extensively (I made a MSExcel sheet checking values up of $N$ up to $16384 = 2^{14}$, 10(+) $N$ shared $s$ steps $\forall s: 14 < s \leq 184$. For $s \leq 14$, I know (strictly, as $2^m$ will be the largest value to take $s = m$ steps), that eighteen numbers take 14 steps, fourteen numbers take 13 steps, ten numbers take 12 steps, eight numbers take 11 steps, six numbers take 10 steps, six numbers take 9 steps, four numbers take 8 steps, four numbers take 7 steps, two numbers take 6 steps, two numbers take 5 steps, and one number takes each of 4, 3, 2, 1, and 0 steps (these last five numbers are the first five powers of 2: 16, 8, 4, 2, 1). The longest step count so far is 275 (for $N = 13255 = 5 \times 11 \times 241$).

*Edits: increased amount of calculations done in Excell and updated the relevant info. And fixed typos. Again.*

Forum category: Ideas / Student Ideas

Forum thread: Sociable Numbers and the Collatz Conjecture ]]>

Forum category: Ideas / Student Ideas

Forum thread: Euler: The Master of Us All ]]>

Forum category: Ideas / Student Ideas

Forum thread: Euler: The Master of Us All ]]>

Forum category: Ideas / Student Ideas

Forum thread: Euler: The Master of Us All ]]>

IF $m \mid n$

THEN $\phi(mn) = m\phi(n)$.

This means that the converse says

IF $\phi(mn) = m\phi(n)$

THEN $m\mid n$

Your job is to determine whether this statement is true (in which case, prove it) or false (in which case, give a counterexample).

Hope that helps!

Forum category: Ideas / Help!

Forum thread: Ch 3, problem 21b) unclear ]]>

I was reading the wiki on Sociable numbers and I found it interesting that there are no known social numbers of order 3 (i.e. there are three numbers in the cycle). Do you think there's a way to prove that social numbers do or do not exist for any known order?

Also, the wiki talked about the theorizing the possibility that all numbers (positive integers) are either part of a social cycle or terminate at 1 by summing up proper divisors to get numbers in the next sequence. While this seemed kind of mind blowing — that perhaps there might be a counterexample to this of a sequence of composite numbers that never reach a prime, since all primes determine the next number in the sequence to be 1 — it also reminded me of a similar conjecture that i enjoy: the Collatz Conjecture.

The Collatz Conjecture makes a statement about a certain sequence. Take any positive integer to be the first term of a sequence and consider the following operations to determine the next number in the sequence. If the number is even, divide it by two, but if it is odd, multiply it by three and add one. The Collatz Conjecture says that for any sequence starting at any positive integer, this sequence will always reach 1.

Pretty crazy, huh? I think so, anyway. Paul Erdős said of the problem, "Mathematics is not yet ready for such problems." He offered $500 for its solution, which if you know Paul (or know of him), is quite the sum.

Forum category: Ideas / Student Ideas

Forum thread: Sociable Numbers and the Collatz Conjecture ]]>

While reading up on the unlikelyhood of odd perfect numbers, I ran across the primorial function. Combine *prime* and *factorial*: **primorial**! That is:

\begin{align} J = \prod_{i=1}^{k}(p_i) \end{align}

For N primorial, we write N#. To calculate N#, we take $J$ such that $p_k \leq$ N.

Thus, 10# = 9# = 8# = 7# $= 2 \times 3 \times 5 \times 7$.

And of course, Wikipedia's article on primorials (O beloved reference!).

Summing up Pomerance's Heuristic that Odd Perfect Numbers are Unlikely (link above): We can calculate the probability that a large perfect number ($n: n = pm^2 > 10^{300}$) ($10^{300}$ chosen because no odd perfects have been found in the interval $[1,10^{300}]$) exists. For a large, arbitrary $m$, hence, $m^2$, the probability is really small. It so happens that for even perfect numbers we can get a much better estimate on $m$ than simply arbitrary (thus, increasing the probability we can find large even perfect numbers, which do exist. But, at the same time, we really have not gotten a good handle on $m$ for odd primes (see the other threads on Perfect numbers).

Primorials come in to play in getting an upper bound for $\sigma(m^2)$ (for $N$ to be perfect, $p| \sigma(m^2)$). The upper bound is then used as an estimate in the nasty probability integral done in the calculations.

$p_n$# +/- 1 is sometimes a prime.

$p_n$# $\pm 1$ are twin primes for $n = 3, 5$ (the OEIS lists were too short to determine if there were any more overlaps (for $n > 1829$, there are none $5 < n < 1829$)).

Highly composite numbers (in the sense that the HCN is the smallest integer such that $\nu(HCN)$ increases to a record) are products of primorials.

Many HCNs are formed by multiplying together smaller HCNs. New primes are introduced on some HCNs, such as 2 (2), 6 (3), 60 (5), and 840 (7). For some HCNs, a prime is "re-introduced", such as 1260 (7*180).

Forum category: Ideas / Student Ideas

Forum thread: Primorials (and Odd Perfect Numbers) ]]>

The smallest prime factor of N is less than (2k + 8) / 3

The largest prime factor of N is greater than 10^8

The second largest prime factor is greater than 10^4, and the third largest prime factor is greater than 100

N has at least 75 prime factors and at least 9 distinct prime factors. If 3 is not one of the factors of N, then N has at least 12 distinct prime factors

An odd perfect number is not divisible by 105

Every odd perfect number is of the form 12m + 1 or 324m + 81 or 468m + 117

This website also has a few links regarding unsolved problems. In the interest of the topic, this includes odd perfect numbers.

Forum category: Ideas / Student Ideas

Forum thread: Odd Perfect Numbers ]]>

"Let $m$ and $n$ be positive integers with $m \nmid n$. Prove that $\phi(mn) = m\phi(n)$."?

Or could/should it be:

"Let $m$ and $n$ be non-positive integers with $m | n$. Prove that $\phi(mn) = m\phi(n)$."?

Forum category: Ideas / Help!

Forum thread: Ch 3, problem 21b) unclear ]]>