370 reputation
2525
bio website
location Seattle, WA
age 71
visits member for 3 years, 4 months
seen 6 hours ago

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.


18h
awarded  Popular Question
Dec
4
comment $\zeta(0)$ and the cotangent function
From Edwards, p 12, (1): for $n=0$,$$\zeta (2 n)=\frac{(-1)^{n+1} 2^{2 n-1} \pi ^{2 n} B_{2 n}}{(2 n)!}=\frac{(-1)^{n+1} 2^{2 n-1} \pi ^{2 n}}{(2 n)!}=-\frac{1}{2},$$ with and without the Bernoulli number.
Sep
11
suggested rejected 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