bio  website  

location  Seattle, WA  
age  71  
visits  member for  3 years, 4 months 
seen  6 hours ago  
stats  profile views  1,956 
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 worldclass 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.
18h

awarded  Popular Question 
Dec 4 
comment 
$\zeta(0)$ and the cotangent function
From Edwards, p 12, (1): for $n=0$,$$\zeta (2 n)=\frac{(1)^{n+1} 2^{2 n1} \pi ^{2 n} B_{2 n}}{(2 n)!}=\frac{(1)^{n+1} 2^{2 n1} \pi ^{2 n}}{(2 n)!}=\frac{1}{2},$$ with and without the Bernoulli number. 
Sep 11 
suggested  rejected edit on experimentalmathematics tag wiki excerpt 
Sep 2 
comment 
Recognize this strange expression from linear algebra?
+1 for indexspaghetti 
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 counterexample. See modified OP. $2193$ is a multiple of 17, inside the 17seg 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 subsequences. 
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 reinvented 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 