Wednesday, December 28, 2011

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!

Sunday, December 25, 2011

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.

Sunday, November 13, 2011

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.

Wednesday, November 09, 2011

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.

Wednesday, October 26, 2011

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!

Wednesday, October 12, 2011


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.

Thursday, September 01, 2011

Wednesday, August 31, 2011


David 'Honeyboy' Edwards died on Monday, so yesterday I listened to a bit of what I had of his in iTunes (mostly the Delta Bluesman album). One of the songs was Just a Spoonful, which brought me to list all of the 'Spoonful' covers that I had: about a dozen. Cream's short six-and-a-half-minute version from Fresh Cream played next and I was reminded that they had a long version on their Wheels of Fire double-album which, back in the day, I once owned. But the album was missing from my iTunes library, so I downloaded it — and am listening to it today. Catherine actually brought into our relationship the Fresh Cream album (along with The Incredible String Band's Wee Tam & The Big Huge), acquired back then from Kay Owen, her British cousin. Much appreciated music, even after all these years.

Tuesday, August 16, 2011

The largest base-24 left-truncatable prime


As previously aspired to, I have now replicated Martin Fuller's 2008 determination of all base-24 left-truncatable primes. I don't know if Martin thought to store the largest of these as part of his program, but I don't see it anywhere on the Net and more specifically not in the a-file for A103443, where it belongs. It's the 53-digit number {17, 22, 19, 14, 19, 15, 10, 3, 10, 7, 1, 13, 18, 13, 9, 22, 15, 22, 23, 15, 14, 3, 13, 3, 20, 19, 17, 10, 6, 1, 20, 17, 9, 2, 18, 15, 12, 10, 3, 21, 11, 8, 16, 15, 4, 4, 4, 15, 11, 11, 7, 10, 17}.


Back on May 7, I created a 9 times 48969-digits-each, type-2 Belgian-numbers html-file to elegantly display solutions to Eric Angelini's second type of Self Belgian Numbers. On May 17, I published a list of thirteen type-2 Belgian primes plus six more probable primes, the largest composed of 27867 digits. By May 24, I had determined for what number-length (after length 1) all nine type-2 Belgian-number templates again produced solutions: length 1899283.

All that needed to be done now was to find all solutions to length 1899283 so that these might be as elegantly displayed as my initial length-48969 set. The calculation took about a week for each of the nine templates and, on July 31, I html-colourized the result. Only problem was, the numbers would not display properly in any of the browsers that I tried. The far-right digits did not align, even though all nine numbers have the same number of digits and they are rendered in a monospace font.

Thankfully, Firefox 6 has just been released and it did display the numbers correctly! The length-1899283 solutions turned out to be terms 3594728-3594736 of A107070 and because these nine solutions along with the first nine solutions form "bookends" to all the other solutions within the nine templates, that's what I ended up naming the over-80-MB file.

Monday, July 25, 2011

Friday, July 08, 2011

Kaiser Wilhelm

Just now, I fixed the birth year and the 1921 stated age in the Wikipedia entry for baseballer Irvin K. Wilhelm — yet again! This time I added references.

Wednesday, June 29, 2011

Wednesday, June 15, 2011

A076623 decomposed

I've been busy with a number of ongoing projects. I'm quite proud of my "number of left-truncatable primes by digit length" which I web-published yesterday and extended today. I still want to get actuals for bases 73, 32, 28, 79, 55, 83, 34, 89, 51, 65, 38, and (finally) 24, but it'll take a couple of months.

August 2011: They're all done! I've put a summary of bases tackled (with specifics) here.

Tuesday, May 03, 2011

Aliquot sequence 46758

I have gotten back to pursuing the fate of aliquot sequence 46758. On April 7, I found the 53-digit factor of term #3237, and early this morning, the 44-digit factor of term #3238. Terms 3239 (this used to be my Beer Store employee number) to 3249 proved relatively easy to factorize, so now I am working on the 113-digit composite of term #3250.

Sunday, May 01, 2011

iMac interruptus

There was a power interruption this morning. This is the first time that my wife's iMac has rebooted since I started it running back in January. I make that to be 100 days.

Saturday, April 09, 2011


Three 3-digit numbers, a, b, and c, have no digits in common and a^b ± c are both prime.

Wednesday, March 30, 2011

444085 Ruth-Aaron pairs

I contacted Donovan Johnson and he was kind enough to share his list of Ruth-Aaron numbers less than 10^12. So I added the additional terms to my compilation: 444085 factored pairs @ 66.5 MB.

Tuesday, March 29, 2011

150000 Ruth-Aaron pairs

I am done with my Ruth-Aaron pairs compilation for now. I finished around 3:30 a.m. this morning: 150000 factored pairs @ 22.5 MB. I discovered that the number of terms less than 10^11 was not 106770 as indicated here, but 106670 as indicated here (I had not been aware of this sequence before today). I also now know that the Herman Baer Ruth-Aaron number 174024968568 sits at index 149492. Mr. Baer estimates here that its index is ~2050, but reduces that guess to ~1123 just a few weeks later.

Sunday, March 20, 2011

Friday, March 18, 2011

28, 11, 12, 13, ... (reprise)

Back on January 4, I asked: What might be this sequence's final term? After asking Lars Blomberg this question two days ago, this morning he replied: 3008889. Additionally, he provided me with all 1124577 terms. My own calculation — after 73 days of computation — had only achieved 481185 terms. Why do I bother?

Tuesday, March 15, 2011

Ides of March

I had given today as the date when I might have finished my Ruth-Aaron pairs compilation. Well, I have another two weeks to go: Running the third Mathematica did take away from the iMac's available clock cycles. I've been keeping busy working with Eric Angelini's colorless green ideas. More specifically, I'm looking for minimal loops of a given length.

Tuesday, March 08, 2011

Thursday, March 03, 2011

Snow big deal (reprise)

I had submitted the photo of the squirrel that I shared here last month to the Toronto Star's "camera club" (topic: snow) and was delighted, yesterday, to see it online. Here it is in the context of the other nine winners. I looked for the promo in today's print edition but it wasn't there. :(

Tuesday, March 01, 2011

250 komets, mostly done

While Lars Blomberg went about extending the columns (up and down), I colored in the cells of our spreadsheet (on the basis of the ratio of a cell's value to that of its predecessor):

•     f(n)/f(n-1) < 1   pale orange
•     f(n)/f(n-1) = 1   yellow
• 1 < f(n)/f(n-1) < 2   cyan
•     f(n)/f(n-1) = 2   light green
• 2 < f(n)/f(n-1) < 3   blue

Hmm. I can't get the 'blue' color here to match the one that I used in the spreadsheet. The tops of the 250 komets remain colorless, pending calculations that show their predecessors to be larger than they are, in which case they'll end up 'pale orange'.

Sunday, February 27, 2011

A missed opportunity

I rarely go anywhere without my camera and even at home it is always close by. Last night I lent the Nikon to my daughter (for today) and this morning I woke up to winter wonderland: delicate heaps of snow on thin branches — already only a memory just a few hours later. I'll recall my maplepan captured 2 March 2003 (saved photo #52 of my previous Coolpix).

Saturday, February 26, 2011

Tuesday, February 15, 2011

Monday, February 14, 2011

Komets and planits, oh my

Lars Blomberg and I have been busy populating entries in a Google docs spreadsheet that extrapolates "komet" k2 (A038834 <-> A038807) to k6, k11, k13, ... k781: 250 kometary path fragments (and 5 planits) in all. I have had a (much smaller) version of this table as long ago as 1997 but it is certainly nice to see it updated, extended, and out in the open. Almost all of the really large numbers in the chart were calculated by Lars.

Monday, February 07, 2011

60000 Ruth-Aaron pairs

I have added another 10000 Ruth-Aaron pairs to my list: 60000 factored pairs @ 7.6 MB. I was tempted to run the third Mathematica on this effort — cutting my expected completion date by a couple of weeks — but, on Saturday night, I started Mathematica 8 on a different venture: duplicating the extension of A038807 sent me by Lars Blomberg last summer.

Saturday, February 05, 2011

Reality check

By this morning, Charles Greathouse and Donovan Johnson had calculated, respectively, the sixth and seventh terms of the sequence that I mentioned in my last entry: A185584. At that point, my own effort had only reached 2 billion. Ordinarily, I might have let things run — or started a new search beginning with the larger value — but I thought better of it.

A number of folk thought it was a good idea to prepend zero to the sequence — which, in my mind, lacked logic — and I objected*. Nevertheless, it was added to the submitted sequence — but later dropped, after I pointed out that the closely related A064510 lacked the zero.

*The crux of my objection was that the primary sense of the word "sum" is an aggregate of two or more things. Aggregating one thing is already a stretch but I'll buy it. Aggregating no things? { } != {0} but (apparently) Apply[Plus, { }] == 0.

Thursday, February 03, 2011

An unexpected bonus

Claudio Meller today asked Sequence Fanatics about an extension to the sequence 1, 130, 1860, 148480, 3039520, ... So I fired up Mathematica 8 on my wife's iMac and started a search for the next term. I had indicated here on Saturday my expectation that doing so would degrade the performance of Mathematica 6 and Mathematica 7, concurrently running their respective Ruth-Aaron numbers compilations, but when I consulted the computer's Activity Monitor, I found — to my surprise, and delight — all three Mathematica kernels running at ~100%. The effect — and I wonder if it is just the appearance of a processing-power benefit — may be due to hyper-threading.

Wednesday, February 02, 2011

Snow big deal

Here in Toronto, they certainly oversold the snowmageddon that was to befall us today. Nevertheless, our school boards — concerned that it had been over 12 years since the last one — decided to declare a snow day. The kids were happy, at least. My photo was taken 50 days ago.

Monday, January 31, 2011

50000 Ruth-Aaron pairs

This 5000 Ruth-Aaron-pairs addition to my compilation (50000 factored pairs @ 6.3 MB) arrived much faster than my previous updates, because two days of the calculation were done on the new, much faster iMac. From now on, therefore, I'll update only when I have collected another 10000 pairs. My current aim for the compilation is ~150000 pairs (up to ~175 billion) @ ~20 MB.

Sunday, January 30, 2011

A small premium

I have run that program now for 24 hours. Whereas Mathematica was checking 1.5 billion numbers per day in the greater-than-100-billion search-range, this search has managed to look at 1.8 billion numbers in its first day (it's at 27 billion) — because factoring smaller numbers takes less time. In spite of that small premium, it'll still take until mid-March to complete the run to 100 billion. This is, nevertheless, a significant improvement over my previous estimate of August-or-so, had I not had access to the new iMac.

Saturday, January 29, 2011

Double take

One of my pre-retirement, beer-store acquaintances emailed me yesterday: Are you using Mathematica 8 yet? The copies of Mathematica 4 and 5 on my old machine are Power-PC only, so I can't run them on my wife's Intel. Mathematica 6 is what is collecting the Ruth-Aaron numbers greater than 100 billion. This evening, I installed Mathematica 7 on her iMac and had it running the same program — but starting at 25.2 billion (which is where I was at). Each Mathematica runs at close to 100% CPU because the iMac's "i5" chip is dual-core. Running Mathematica 8 on top of that would only serve to degrade the performance. ;)

Thursday, January 27, 2011

How fast?

I've now had my wife's new iMac collecting, in Mathematica, Ruth-Aaron numbers (greater than 100 billion) for one full week. In that period of time, it has looked at some 10.5 billion candidates — roughly one million every minute. This is more than four times the rate that I have been getting on my old Mac — in a range where the numbers are five times the size. So, I could reach 100 billion (from my current position of 24.5 billion) by mid-March (at the latest) if I were to transfer the job to my wife's iMac today. Instead, I will let her computer continue its current compilation for another two weeks.

45000 Ruth-Aaron pairs

I've added another 5000 Ruth-Aaron pairs to my list: 45000 factored pairs @ 5.7 MB.

Tuesday, January 25, 2011


I have now collected more than 275000 terms of A179066 and have encountered the first few values thereof exceeding 1124577 (the assumed length of this sequence) — which, therefore, lack a symmetrical counterpoint in the x=y mirror. Term #261379 is the very first of these with a value of 4000555.

Thursday, January 20, 2011


It's been a long time coming but I finally got my wife to replace her "Bondi Blue" with a new "mid-2010" iMac. She insisted on the 21.5" screen but I maxed out every other spec. This will allow me to run Mathematica in the background and give me a handle on just how much faster this machine is compared to my old workhorse. I have started it collecting Ruth-Aaron numbers greater than 10^11.

Tuesday, January 18, 2011

My secret journal

When I started my "blahgodo" blog on 6 April 2008, I deliberately disabled its search and comment capabilities. It sat on my computer with a single link in my "chesswanks" portal, hidden in my avatar's glasses. I notified a handful of friends/acquaintances about the blog and eventually ended up with a handful of followers (who, by and large, were not the people that I had notified). The first Google mention of the word "blahgodo" did not appear until 14 August 2008.

By contrast, Glad Hobo is both searchable and commentable, but I have yet to tell a single soul about it. And although it is tempting, I will not, for I am curious just how long it will take to get that out-of-the-blue first comment.

Monday, January 17, 2011

A large Ruth-Aaron pair

A large (3109-digit) Ruth-Aaron pair from Jens Kruse Andersen in 2006:

and the adjacent

Saturday, January 15, 2011

Friday, January 14, 2011

Automobile number

An article describing Thomas A. Edison's Latest Invention appeared in Scientific American (the weekly journal of practical information) one hundred years ago today. Right beside it is a piece called The Modern Pleasure Electric Vehicle. (William Hudson writes: "The electric pleasure car is coming into its own. For certain kinds of service it is ideal. Its simple and responsive control will always be its most remarkable feature. The gasoline car has perhaps obscured the development which the electric has been undergoing in recent years.") Several other car-related articles follow in this special automobile number: Some Remarkable Mechanical Road Guides, Why You Can Buy a Good Car for Little Money, The Commercial Motor Truck vs. the Horse, A Few Shop Jobs on an Old Car, and Repainting the Old Car at Home. There follows a nice two-page spread showing twenty-five motor cars, complete with vital statistics (except for some damaged entries at the bottom of my pages). There's an article titled The Small, Inexpensive Garage. Then one about Automobile Cylinder Lubricating Oils and, on the page opposite, an advertisement for that new Edison storage battery (pictured) made by the Edison Storage Battery Company.

Thursday, January 13, 2011

The big picture

I had Mathematica create a very large (4834 x 4766 pixels) graph of A179066. Most browsers will shrink the image to fit window size — which is how it looks best — but then one can expand it to actual size to better appreciate detail. The highest-up points in this part of the graph — there are higher-ups (>200000) that are not shown — are these: {97860,121125}, {98065,121114}, {98067,121134}, {98076,121143}, {98470,121123}, {98506,121132}, {98560,121141}, {98605,121177}, {98607,121152}, {98650,121213}, {98670,121215}, {98704,121222}, {98706,121224}, {98740,121231}, {98760,121233}. Their x=y mirror images are on the far right.

I had not really intended for Glad Hobo to be only recreational mathematics (as it has been so far) but rather a less-constrained, different-emphasis successor to my "blahgodo" blog. Tomorrow I will publish something different.

Tuesday, January 11, 2011


= 11 + 22 - 33/44*55 + 66/77 - 88/99 - 110 + 121

This is a pretty good approximation to the number e.


= 3*7/15 + 1/292/1 + 1/1 + 2/1/3 + 1/14

The fraction in the title approximates π, and the 3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14 are the first thirteen terms of π's simple continued fraction (= 80143857/25510582).

Monday, January 10, 2011

Fractals everywhere

Perhaps it should not come as a surprise that sequence A179066 has fractal properties. I have put here two images: First, of values less than 7500, then of values less than 75000. The document has shrunk the pictures for easier comparison: Click on each one to look at the graphs full-size.

Saturday, January 08, 2011

Progress and prognosis

Last night I reached term #35000 in my evaluation of A179066. I had been trying a few program improvements on yet a fourth copy of Mathematica to see if I could speed things up a little. Most of these didn't really pan out — the overhead of the new code canceling any hoped-for benefits — but one obvious implementation did shave some time off the calculations, so I aborted my run and started the new routine from where I left off. Now I am about to pass term #70000 and — putting one of my two Ruth-Aaron-pairs tabulations on hold in order to devote more CPU muscle to this — tomorrow morning I should be passing term #100000.

My expectation is that — any day now — someone else will have calculated all 1124577 terms and that will be the end of my effort. In particular, Maximilian Hasler — who co-authored A179066 with Eric Angelini — thinks that "it should not take that long".

Thursday, January 06, 2011

Computing power

How difficult might it be to calculate term #1124577 of A179066? For me, very. I've been running my program now for a couple of days and have only managed some 25000 terms. I can probably see my way to 100000 terms but, short of a significantly more optimized version of my search routine (and even with that, the search beyond would prove taxing), there it will end. Remember also that I am running two other copies of Mathematica in the background (my ongoing Ruth-Aaron pairs tabulation) and these will eat up available computer-clock cycles, reducing the Mathematica kernel's CPU usage to some 60% of the available 2 MHz. Have I mentioned yet that I could really use a modern replacement for my seven-and-a-half-year-old machine?

Wednesday, January 05, 2011

Proof by contradiction

Yesterday I asked for the "final term" of a sequence. It may not be immediately obvious that there is for this sequence such a thing, but consider this: A179066 is defined in such a way that term #1124578 must have a digital root of 1 (the digital root of 1124578) and yet be composed of only the digits 0, 3, 6, and 9 (those digits that are not used in 1124578). But numbers made up of only 0s, 3s, 6s, and 9s will be evenly divisible by 3 or, alternatively, leave a remainder of 0, 3, or 6 when divided by 9 (making their digital roots 3, 6, or 9), thus contradicting our already established fact that the number that we are looking for has a digital root of 1. As defined, term #1124578 cannot exist and term #1124577, therefore, is likely the final term of this sequence.

Tuesday, January 04, 2011

28, 11, 12, 13, ...

Eric Angelini came up with this: Each successive term is the smallest positive integer not yet used, where both the term and its index have the same digital root (remainder when divided by 9) but do not have in common any of each other's base-ten digits. What might be this sequence's final term?

Monday, January 03, 2011

35000 Ruth-Aaron pairs

I've just added another 5000 Ruth-Aaron pairs to my ongoing tabulation thereof (bringing the total now to 35000 pairs). The project will likely consume the better part of this year. I realize that this would go much, much faster using a modern computer (my dual 2 GHz PowerPC G5 is now well over seven years old) but, in the absence of anyone else taking this on, at least I've got a decent start on it.

Sunday, January 02, 2011


Using the digits 1, 2, 3, 4, 5, 6, 7, 8, 9 in order, replace each comma with one of +, -, *, /, or a concatenation of adjacent numbers. Parentheses or other punctuation are not allowed and, as usual, multiplications and divisions are to be evaluated before additions and subtractions. My title today evaluates to a quantity fairly close to π. Can you come up with an expression that is even closer?

Saturday, January 01, 2011


Just how likely was my pre-midnight countdown? Consider the ordering of 10, 9, 8, 7, 6, 5, 4, 3, 2, 1 and allow the commas to be replaced by one of +, -, *, /, or a concatenation of adjacent numbers (with the added proviso of no concatenation after the 10 to force the generally accepted start of the countdown). We exclude parentheses, so the mathematical statements must be evaluated in the standard order of operations (i.e., multiplications and divisions evaluated before additions and subtractions). Of the 1562500 (4*5^8) possible numerical outcomes, 486501 are unique. Of these, only 51249 are integers (31373 positive), the largest being 10*987654321. The smallest positive integers that do not appear are 1992, 2102, 2133, etc., so every year from 1993 to 2101 will have at least one solution: The largest number of solutions in this interval for a given year is 13 and the average, 616/109. Only two of the solutions start with 10/ (1997 = 10/9*87*6*5 - 43*21 and 2021 = 10/9*8*765/4 + 321). While 1992 is missing from our integer solutions, it may be approximated by the nearby 10 + 9 + 8 + 76*543/21 (as well, if we allow concatenation after the 10, it's 1098 + 765 + 4*32 + 1). You might like to look for one of the four solutions that evaluate to 377/120, the closest that we can get to the irrational number π.

An interesting bonus in last night's countdown exemplar is the symmetry of the operations: +, *, *, /, *, *, +. How common might that be? Out of the 616 solutions in the 1993-2101 interval, I count 12 that qualify:

1994 = 10*9 - 8*7 + 654*3 - 2*1
2001 = 10 - 9*87 + 65*43 - 21
2011 = 10 + 9*8*7/6*5*4 + 321
2020 = 10 - 9 + 8*7 + 654*3 + 2 - 1
2023 = 10*987/6 + 54/3*21
2024 = 10 + 9 + 8*7*6*5 + 4 + 321
2040 = 10 + 9 + 8*7 + 654*3 + 2 + 1
2072 = 10 + 9*8 + 7 + 654*3 + 21
2088 = 10*9*87/6/5/4*32*1
2093 = 10*98 + 7*6*5 + 43*21
2098 = 10 - 98*7 + 65*43 - 21 or 10 - 9*8 + 7 - 6 + 5*432 - 1