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 chesswanks.com 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!

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.

Saving grace

I (finally) got my big-screen TV last week. I had hoped to have an over-the-air antenna attached to it so that I could watch HD without having to engage my evil Rogers cable. But it was not to be. I tried attaching one of my hard-drive docks to the USB port on the TV but it didn't recognize it. So today I bought a 64GB Lexar flash drive from FutureShop and it saw that just fine: The saving grace! It worked so well that I just ordered two more (128GB each) flash drives from Amazon. The second movie I watched was Lola rennt (1998), at a paltry 368 lines of resolution — but at least the subtitle file had been finely honed by yours truly. The picture shows a scene from the first movie that I watched.

10*9*8+7+6-5+4*321

Happy New Year!

The one-billionth Kaprekar number

When I published the first 11 million Kaprekar numbers on September 1, I asked what the one-billionth Kaprekar number was. Today I fashioned my answer. This was a troublesome compute that included a six-week run calculating nothing: I had used in my program the old Mathematica NthSubset directive but had neglected to first load the Combinatorica package in which it resides. Ouch!

Digit pick-up

Consider the multiplication table for the number 125. For each product, we will pick up any digits thereof we have not already collected and add them to a growing list. When will we have collected all ten digits?

 1  125  +1 +2 +5 => {1,2,5}

 2  250        +0 => {0,1,2,5}

 3  375     +3 +7 => {0,1,2,3,5,7}

 4  500

 5  625        +6 => {0,1,2,3,5,6,7}

 6  750

 7  875        +8 => {0,1,2,3,5,6,7,8}

 8 1000

 9 1125

10 1250

11 1375

12 1500

13 1625

14 1750

15 1875

16 2000

17 2125

18 2250

19 2375

20 2500

21 2625

22 2750

23 2875

24 3000

25 3125

26 3250

27 3375

28 3500

29 3625

30 3750

31 3875

32 4000        +4 => {0,1,2,3,4,5,6,7,8}

33 4125

34 4250

35 4375

36 4500

37 4625

38 4750

39 4875

40 5000

41 5125

42 5250

43 5375

44 5500

45 5625

46 5750

47 5875

48 6000

49 6125

50 6250

51 6375

52 6500

53 6625

54 6750

55 6875

56 7000

57 7125

58 7250

59 7375

60 7500

61 7625

62 7750

63 7875

64 8000

65 8125

66 8250

67 8375

68 8500

69 8625

70 8750

71 8875

72 9000        +9 => {0,1,2,3,4,5,6,7,8,9}

So, it takes 72 multiplications before we've bagged all ten digits. In base ten, it turns out that 72 is the largest number of multiplications required for any given starting number.

What is the largest number of multiplications required in base b? Tomas Rokicki has worked them out for thousands of bases: It's the second number on each line in his list. I felt that these maxima (k) deserved to be graphed. Noting that in composite bases k is divisible by (b-1), I decided to do another graph of k/(b-1). Then, to explicitly list the k/(b-1) straight-line points, I did the families, with a question at the end.

Meloe yellow

Oops. Just over four minutes after taking a picture of an oil beetle at my usual spot above the river, I accidently stepped on it. At the time I thought that the exuded bright-yellow mass was liquid, but an examination of the photo revealed it to be eggs. Also visible in the image is the cantharidin-containing ochre (generally described as 'yellow') fluid exuding from the joints (detail above). Samuel Maunder's 1848 description of Meloe is here and a more modern treatment, here. I noticed my first oil beetle in the local, private cemetery three years ago. A large grassy section of this cemetery sports an extensive covering of ground-bee dwellings and the beetles have taken full advantage, being this fall every bit as bountiful as were the bees in the spring.

Scientific American, weakly

BoingBoing alerted me, last Friday, to the free accessing (this month only) of Scientific American weeklies from 28 August 1845 to 25 December 1909. A weeks-between lookup gives 3357 potential issues, but I think eight of those did not see publication. Of the remaining 3349 (that's four more than claimed by Scientific American), I managed to find 3246 complete, readable issues at the site, some hidden behind missing or misdirecting links.

I have been an aficionado (and eventual collector) of the magazine since discovering this issue near the end of 1967. Having a substantial, freely accessible pdf-archive of the old weeklies was, for me, a little like finding money on the street. I spent all day Saturday downloading, as a result of which I woke up Sunday morning with a debilitating lower-back issue (it became extremely painful to maintain an upright position) from which I did not recover until yesterday! I spent all of last night (until about 4:30 this morning, more carefully minding my posture) completing the heist.

Felix Aloysius Sullivan

Some three years ago I mentioned in my old blog, Felix, a relative of my wife. Today, I received a surprise email from a David Wilhelm about his relative, Kaiser Wilhelm. Wanting to share with David information about Felix, I did a search looking for my blog mention... and found instead this wonderful, previously unknown short biography!

Zillions

In response to the perhaps-little-known fact that the last prime in alphabetical order is two vigintillion, two undecillion, two trillion, two thousand, two hundred, ninety-three [Donald E. Knuth, Allan A. Miller: A Programming and Problem-Solving Seminar, page 12 (page 20 of the ftp-available pdf), June 1981 (the solution dating to October 1980)], I recently provided MathFun (28 September 2011) with:
____________________________________________________________
Bill Gosper:
"Yeah, what about 2*10^4679+3, (two zillion and three)-?"

Phil Carmody:
"It seems to precede 2*10^4772+2003 (two zillion two thousand and three)"

It's instructive here to look at the original (without using 'zillion') answer: 2*10^63 + 2*10^36 + 2*10^12 + 2*10^3 + 2*10^2 + 93 (two vigintillion, two undecillion, two trillion, two thousand, two hundred, ninety, three). The 93 is forced because it is the only two-digit number that will make the entire quantity prime, but 23 (twenty, three) would have been a more desirable choice to force the written number to the back of the dictionary. Just 2 (two) would have been even better but a number ending in 2 cannot be prime.

So, let's prepend the 'two zillion' (2*10^n) and force the presumably-optimal 23 ending.

 2*10^n   (two zillion)
+2*10^63  (two vigintillion)
+2*10^36  (two undecillion)
+2*10^12  (two trillion)
+2000     (two thousand)
+200      (two hundred)
+20       (twenty)
+3        (three)

Of course, n >=66 because we have perfectly good, widely-dictionaried names for numbers less than this. A quick run has n = {77, 113, 116, 158, 342, 464, 468, 565, 2171, 2274, 2340, 3347, 5724, ...}. That might have been the end of it, but I noticed that there is nothing to prevent our indefinite 'zillion' from being repeated. And every repetition forces the written number further back in the dictionary. So one begins to look at the divisors of n to see which combination results in the most number of greater-than-10^66 (zillion) amounts. For example, 10^5724 = 10^1431 * 10^1431 * 10^1431 * 10^1431, so, taking a zillion to be 10^1431, our large prime starts out with "two zillion zillion zillion zillion, two vigintillion, ..." Our n = 2340 solution allows for 30 written zillions (of 10^78 each).
____________________________________________________________

The term after n = 5724 is n = 39960. The resulting 39961-digit probable prime is now on Henri & Renaud Lifchitz's Probable Prime Records site.