bio | website | |
---|---|---|
location | Seattle, WA | |
age | 70 | |
visits | member for | 3 years, 3 months |
seen | 17 mins ago | |
stats | profile views | 1,935 |
Contact: rudytoody.AT.comcast.DOT.net
I retired in 2006 and bought Mathematica and a stack of math books with the goal to teach myself to become a world-class mathematician. I am on pace to achieve that goal sometime shortly after the next Ice Age.
I consider myself a Mathematical Mutt (no papers) who occasionally ventures off the back porch to play in the yard with the big dogs.
I donate regularly to the The OEIS Foundation.
When I look at the patterns, I can hear the wheels turning. When I look at the math, I find out the hamsters have died.
Sep 11 |
suggested | suggested edit on experimental-mathematics tag wiki excerpt |
Sep 2 |
comment |
Recognize this strange expression from linear algebra?
+1 for index-spaghetti |
Aug 11 |
awarded | Notable Question |
Aug 11 |
revised |
$\prod_{n=1}^{\infty} n^{\mu(n)}=\frac{1}{4 \pi ^2}$
added note that the bug has been corrected in Mathematica V.10 |
Aug 10 |
comment |
Palindromic Patterns of Greatest Divisors $\leq k$
@PerAlexandersson, math.stackexchange.com/q/886041/28555 shows another identity (conjecture) based on the same palindromic divisor sequence. |
Aug 10 |
comment |
Palindromic Patterns of Greatest Divisors $\leq k$
@PerAlexandersson, math.stackexchange.com/q/867135/28555 points to a post that explains how I found it. mathematica.stackexchange.com/q/48452/973 points to the Mathematica code for the recursive routine. |
Aug 10 |
comment |
Palindromic Patterns of Greatest Divisors $\leq k$
Actually, it is not a counter-example. See modified OP. $2193$ is a multiple of 17, inside the 17-seg which shows it is a derangement. |
Aug 10 |
revised |
Palindromic Patterns of Greatest Divisors $\leq k$
added one assumption to constructing the sequence and examining sub-sequences. |
Jul 4 |
revised |
$\prod_{n=1}^{\infty} n^{\mu(n)}=\frac{1}{4 \pi ^2}$
expanded link description |
Jul 4 |
revised |
$\prod_{n=1}^{\infty} n^{\mu(n)}=\frac{1}{4 \pi ^2}$
removed two previous edits which no longer apply to the problem |
Jul 4 |
comment |
$\prod_{n=1}^{\infty} n^{\mu(n)}=\frac{1}{4 \pi ^2}$
@MichaelHardy, thanks for the edit. Thanks for the notation tip. |
Jul 4 |
revised |
$\prod_{n=1}^{\infty} n^{\mu(n)}=\frac{1}{4 \pi ^2}$
added a link to the reciprocal |
May 22 |
accepted | New identity for lcm of the first n integers and the second Chebyshev function |
May 22 |
comment |
New identity for lcm of the first n integers and the second Chebyshev function
@joro, thanks. Somehow, I missed that. |
May 22 |
comment |
New identity for lcm of the first n integers and the second Chebyshev function
Thanks. The RHS is my middle Mathematica statement. So, it seems I have re-invented the wheel and called it "Fire." |
May 22 |
awarded | Nice Answer |
May 22 |
revised |
New identity for lcm of the first n integers and the second Chebyshev function
added a link to function definition |
May 22 |
asked | New identity for lcm of the first n integers and the second Chebyshev function |
May 22 |
comment |
Correct spelling of names, Chebychev and Cholesky
@Marco, it's the third word of the shaded area. |
Apr 23 |
awarded | Benefactor |