Wednesday, March 30

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

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

Friday, March 18

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

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 8

Thursday, March 3

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 1

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'.