Wednesday, February 29, 2012

Happy trails to you

One of my current preoccupations is the extension of an idea that Eric Angelini mentioned to Sequence Fanatics eleven days ago. I wondered what would happen if we allowed the "digit trail" of a negative number to be the negatives of that number's digits. For starting numbers greater than ten, this had the resulting sequence oscillate between negative and positive territory, with each "run" taking another shot at hitting zero. The number of steps needed to get to the first zero is A208059. I subsequently extended the concept to other bases. Eric did a nice job of summarizing (in French) all of this.

Wednesday, February 15, 2012

Apple TV

No, not the television that Apple will introduce later this year but the 720p digital media receiver that I purchased, four weeks ago, along with a fifth-generation AirPort Extreme transmitter. The idea was to allow my new television access to the many movies and TV shows I'd amassed on my computer, without the painfully-slow process of putting them on those USB flash (thumb) drives that I acquired back then. (In retrospect, I shouldn't have bought those two extra 128GB ones.)

The problem was that the seven-and-a-half-year-old wireless card in my eight-and-a-half-year-old Power Mac G5 (Dual 2GHz) wasn't up to speed and, as a result, I was encountering fatal errors watching my movies. Thankfully, the Apple-TV receiver remembered where I left off, but it was annoying at best and unwatchable at worst.

So, yesterday, I purchased a 100-foot ethernet cable (75-foot was all I needed, but Tiger Direct didn't have 'em) to bridge the gap between the diametrically-opposite corners of our main floor. I then spent the better part of half a day (until the wee hours of this morning) configuring my Local Area Network so that the Apple TV and Catherine's iMac (upstairs) were getting internet while my domain was still pointing to my computer. It shouldn't have been that difficult.

Apple TV will likely get updated next month, which is fine: I'm all in. And I will surely complement my now-enhanced LAN with an iPad 3!

Tuesday, February 07, 2012

A061205 quintuples, anyone?

One month ago, Franklin T. Adams-Watters asked about numbers that occur more than twice in A061205. The result was A203924, to which, yesterday, I added a link to my augmented table of 21313 terms. The majority of these are the third and fourth terms of A061205 quadruples: 101556 being the smallest, appearing at positions 156, 273, 372, and 651. There are 10554 quadruples in my less-than-ten-million position search-range, so these account for 21108 of the entries. The remaining 205 are taken from 67 A061205 triples, 24 sextuples, and 7 octuples.

A solution may be said to be "trivial" if all of its position numbers end in zero. (A non-trivial solution allows for an infinite number of trivial ones by multiplying all of its position numbers by the same power of ten.) After looking at the 40 non-trivial A061205 triples in my table, I conjectured that an odd-tuple exists if the product of a number multiplied by its reversal is either the square of a palindrome or, less frequently, the square of ten times a palindrome.

This insight allowed me to create a table of A061205 triples well beyond my original search range and held out the hope of finding a quintuple, if ever the palindrome could be got at in more than one way. Alas, my program is still considerably brute-force-ish: It searches for a palindrome in the square root of a number multiplied by its reversal. It might well be more efficient doing this the other way 'round.