718 reputation
2526
bio website
location Seattle, WA
age 70
visits member for 2 years
seen yesterday

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.


Aug
3
revised Collatz Tool Box — any speed ups possible?
replaced uniqueRank to eliminate the `Floor`
Aug
3
revised Collatz Tool Box — any speed ups possible?
removed `|| 0 == w` from the `Break[]` which was a left-over from testing
Aug
2
revised Collatz Tool Box — any speed ups possible?
add final version of countOrbit
Aug
1
revised Collatz Tool Box — any speed ups possible?
replaced countOrbit with faster version
Aug
1
revised Collatz Tool Box — any speed ups possible?
added enhanced orbit count to example
Jul
31
revised Collatz Tool Box — any speed ups possible?
added one more function to replicate normal orbit counts
Jul
31
revised Collatz Tool Box — any speed ups possible?
added links to conjecture info
Jul
31
comment Collatz Tool Box — any speed ups possible?
@OleksandrR., also the $10^{477119}$ number is calculated in omegaSubOrbit[x] with x being the uniqueRank number. Both calculations are very quick--under 2 seconds.
Jul
31
comment Collatz Tool Box — any speed ups possible?
@OleksandrR.,no, I use my short-cut, which is what this is all about. Notice these numbers from the example: $4805005, 1903828$. The first is the count of multiplies required to get down to $1$ and the second is the number of sub orbits processed to get to that number. Each sub orbit contains its own length, so the first sub orbit adds $1000000$ to the count and moves on to the next sub orbit. So, the whole thing is done with a series of additions. The key is: IntegerExponent[x+1,2] applied to the first number of the sub orbit provides that count.
Jul
31
comment Collatz Tool Box — any speed ups possible?
@cormullion, and I had to show all the code because if you tweak one function, most likely you would have to tweak another to keep everything working together.
Jul
31
comment Collatz Tool Box — any speed ups possible?
@cormullion, Because I'm an old procedural programmer, I feel most comfortable using For and While and I don't usually see how to eliminate them. This post is two-fold---One, to let others grab this stuff and play with it and Two, to find out how to speed up the countOrbit function to maybe get it below 10 minutes duration on my pc.
Jul
31
revised Collatz Tool Box — any speed ups possible?
tweaked code alignment and questioned the down-votes
Jul
31
asked Collatz Tool Box — any speed ups possible?
Jul
24
awarded  Fanatic
Jul
12
comment Building graph based on the cities connection?
@rm-rf, I assumed that since he is selecting cities by population, and if he wanted one point in each state, then the resultant graph would have the most populus city of each state as the location of its vertex. I had used this approach on the travelling salesman problem 30-years ago (without any more success than the other methods.)
Jul
12
comment Building graph based on the cities connection?
Imagine a light-house with its revolving light at each city. When the light hits another city, draw the line, unless that line crosses another. The bottom graph has no crossing lines. (i.e., it's planar.) That's the best I can do to help.
Jul
12
comment Building graph based on the cities connection?
Bottom graph is most populus city in each state. Then there is a pruning algorithm to reduce the number of edges.
Jun
25
revised How do we solve N-Rooks variation using primes?
added another link
Jun
25
revised How do we solve N-Rooks variation using primes?
Cleaned up per suggestion
Jun
22
comment Trace of FullSimplify
I put a ( before the 2b and a ) before the / on the second line, which made only one fraction.