# Glad Hobo Express

## Friday, December 2

### As it happens

When you hear/see a helicopter hovering overhead here in Toronto, it's an indication of a nearby news story.

## Friday, November 25

### A deep walk in base-two pi

After I managed to calculate 8 billion ternary digits of π, I figured that I could probably generate 10 billion binary digits thereof. My previous effort was 1 billion binary digits. I won't bother graphing the walk but I will report that my output of the indices of zero therein has jumped from 45915 to 147043.

**Addendum:**I've stepped through those 10 billion digits a couple more times. In the walk, I counted 5738590822 terms that are positive and 4261262135 terms that are negative. I've also checked record maxima and minima and submitted them as two new sequences: A278737 and A278738.### A deep walk in base-three pi

In my previous post I set up a graphical depiction of the ternary digits of π by "walking" them from the origin to (950,2678) in 25 million steps. Here is an alternate view — in orange — of the same walk (click on the picture to better see the detail):

This is a

*compressed*version of my previous graph as only one of every one thousand points is shown and connected. I have added an x=y line to indicate where along the walk the accumulated number of ones and twos might be equal. (950,2678) is shown as a black dot. Starting at that point — in cyan — I have*continued*the walk (a bit of the left and a lot of the right of it has been cropped). The*full*walk to 8 billion looks like this:
A hint of orange near the origin betrays the change in scale of the continuation. After step 8 billion we have reached (42488,102078), also marked by a black dot. Which is to say that we have a digit deficit at this point of 102078 zeros and 59590 twos! I counted 96619 times that the number of ones and twos were exactly equal. And we are left with 0, 15, and 18 initial ternary-π digits as the only known instances where the number of all three digits are exactly equal.

**Addendum:**I have upped the walk from 8 billion steps to 10 billion steps, where we have reached (35427,93367). I'm also reworking the graphics, which I will showcase in a future article.

## Wednesday, November 16

### A walk in base-three pi

Looking at the output of my previous post, it would be natural to believe that A039624 is infinite. What about a base-three analogue? For how many of the initial digits in the ternary representation of π are there solutions where the number of zeros, ones, and twos are all exactly equal? In base three, π = 10.010211012222010211... After a preliminary no-digits, the first 15 and 18 digits also clearly qualify. Then what?

I wanted to explore this graphically, so I converted π's ternary digits to vectors in a Cartesian coordinate system. I let the digit 1 be the vector (0,1); the digit 2, (1,0); and the digit 0, (-1,-1). If we start at (0,0), every time we revisit the origin we should have an exactly equal number of zeros, ones, and twos:

1

2 (0,1) + (-1,-1) = (-1,0)

3 (-1,0) + (-1,-1) = (-2,-1)

4 (-2,-1) + (0,1) = (-2,0)

5 (-2,0) + (-1,-1) = (-3,-1)

6 (-3,-1) + (1,0) = (-2,-1)

7 (-2,-1) + (0,1) = (-2,0)

8 (-2,0) + (0,1) = (-2,1)

9 (-2,1) + (-1,-1) = (-3,0)

10 (-3,0) + (0,1) = (-3,1)

11 (-3,1) + (1,0) = (-2,1)

12 (-2,1) + (1,0) = (-1,1)

13 (-1,1) + (1,0) = (0,1)

14 (0,1) + (1,0) = (1,1)

15 (1,1) + (-1,-1) =

16 (0,0) + (0,1) = (0,1)

17 (0,1) + (-1,-1) = (-1,0)

18 (-1,0) + (1,0) =

19 (0,0) + (0,1) = (0,1)

20 (0,1) + (0,1) = (0,2)

...

And indeed, we re-arrive at the origin at steps 15 and 18. Then what?

The graph shows 25 million steps, with the initial (0,0) and final (950,2678) points shown in red. Yes, we should get back to the origin eventually but it's not as easy to believe as it is for base-two. It should be more understandable then that providing an exactly-equal number of all base-b digits in some initial terms of base-b π with b > 3 will be difficult. Regardless, we have in base-four, the first 4 digits (3.021) and in base-five, the first 75 digits!

I wanted to explore this graphically, so I converted π's ternary digits to vectors in a Cartesian coordinate system. I let the digit 1 be the vector (0,1); the digit 2, (1,0); and the digit 0, (-1,-1). If we start at (0,0), every time we revisit the origin we should have an exactly equal number of zeros, ones, and twos:

1

**(0,0)**+ (0,1) = (0,1)2 (0,1) + (-1,-1) = (-1,0)

3 (-1,0) + (-1,-1) = (-2,-1)

4 (-2,-1) + (0,1) = (-2,0)

5 (-2,0) + (-1,-1) = (-3,-1)

6 (-3,-1) + (1,0) = (-2,-1)

7 (-2,-1) + (0,1) = (-2,0)

8 (-2,0) + (0,1) = (-2,1)

9 (-2,1) + (-1,-1) = (-3,0)

10 (-3,0) + (0,1) = (-3,1)

11 (-3,1) + (1,0) = (-2,1)

12 (-2,1) + (1,0) = (-1,1)

13 (-1,1) + (1,0) = (0,1)

14 (0,1) + (1,0) = (1,1)

15 (1,1) + (-1,-1) =

**(0,0)**16 (0,0) + (0,1) = (0,1)

17 (0,1) + (-1,-1) = (-1,0)

18 (-1,0) + (1,0) =

**(0,0)**19 (0,0) + (0,1) = (0,1)

20 (0,1) + (0,1) = (0,2)

...

And indeed, we re-arrive at the origin at steps 15 and 18. Then what?

The graph shows 25 million steps, with the initial (0,0) and final (950,2678) points shown in red. Yes, we should get back to the origin eventually but it's not as easy to believe as it is for base-two. It should be more understandable then that providing an exactly-equal number of all base-b digits in some initial terms of base-b π with b > 3 will be difficult. Regardless, we have in base-four, the first 4 digits (3.021) and in base-five, the first 75 digits!

### A walk in base-two pi

In base two, π = 11.001001000011111101101010100010001... Ignoring the decimal point and replacing every zero with minus one, we get: 1, 1, -1, -1, 1, -1, -1, 1, -1, -1, -1, -1, 1, 1, 1, 1, 1, 1, -1, 1, 1, -1, 1, -1, 1, -1, 1, -1, -1, -1, 1, -1, -1, -1, 1, ... We can now treat these numbers as a one-dimensional walk on the y-axis. Accumulating: 1, 2, 1, 0, 1, 0, -1, 0, -1, -2, -3, -4, -3, -2, -1, 0, 1, 2, 1, 2, 3, 2, 3, 2, 3, 2, 3, 2, 1, 0, 1, 0, -1, -2, -1, ... Finally, we graph this sequence:

The picture is very much compressed, showing only

The picture is very much compressed, showing only

*one*of every one hundred points. What looks like a single crossing from positive to negative territory at 10^8, plotting the values from 100000000 to 101400000 details to this:
Even this detail is lacking. What appear to be a couple of dozen y = 0 values are actually 2409 such, ranging from 100023378 to 101375384. The list of zeros constitute the terms of A039624 and I've just calculated 45915 of them.

## Monday, November 14

### Fear of height

The Humber river retaining wall is approaching its northern limit. In the photo (above), the house at the top-right is actually at the end of the street on which I live. Yesterday, I climbed to the top of the wall at its southern end — five meters above the (shallow) water. Here are two pictures taken from that vantage — looking across to the other side of the river, and towards the north:

It didn't feel the least bit safe and I was glad to climb down again.

## Saturday, November 5

### More 3-balanced factorization integers

Four weeks ago I discovered what I now know to be the

1045675984884 = 2^2 * 3 * 17 * 67 * 103 * 359 * 2069

1046959786860 = 2^2 * 3 * 5 * 13 * 43 * 2087 * 14957

1047697856460 = 2^2 * 3^2 * 5 * 109 * 347 * 153889

My ill-fated, eight digits

*smallest*base-ten 3-balanced factorization integer. By the end of October I had completed my base-ten k-balanced list to 10^12 and decided to extend the search to 1.5 * 10^12, thus including (when finished) the first of the four 13-digit 2-balanced factorization integers. I was of course also hoping for more (small)*3-balanced*examples. Yesterday and today one of my processors found*these*(which thusly become the second-, third-, and fourth-smallest):1045675984884 = 2^2 * 3 * 17 * 67 * 103 * 359 * 2069

1046959786860 = 2^2 * 3 * 5 * 13 * 43 * 2087 * 14957

1047697856460 = 2^2 * 3^2 * 5 * 109 * 347 * 153889

My ill-fated, eight digits

*reverse*search is now back on track, though I realize that it will take a good while to complete. It will rediscover the*twenty-one*12-digit 2-balanced factorization integers and — more excitingly — has already come up with a k-balanced solution for k > 2.## Sunday, October 30

### Weston's John Street pedestrian bridge

On Friday, the John Street UP/GO rail crossover was, at long last, opened to the public. Historically, this location may be considered to be the heart of Weston because it is where its train stations (CNR, CPR) used to be. Architecturally, I thought the bridge to be quite nice.

Although the safety fencing gets in the way of on-the-bridge photography, I was able to find a lower platform that provided an unobstructed view of the new train station to the south:

## Sunday, October 9

### A small 3-balanced factorization integer

My eight digits reverse search has run into unexpected and unwelcome difficulties. I have put it on a back burner for now. The forward search to 10^12 continues on several fronts and, today, snagged on one of those fronts a 12-digit 3-balanced factorization integer! In my mind I had conjectured a 13-digit 3-balanced factorization integer but a 12-digit find came as a total surprise. This answers my query as to the existence of a 3-balanced integer smaller than the largest 2-balanced integer:

785984586660 = 2^2 * 3^2 * 5 * 7 * 13 * 149 * 307 * 1049

785984586660 = 2^2 * 3^2 * 5 * 7 * 13 * 149 * 307 * 1049

## Wednesday, September 28

### So long, Teksavvy

On 17 September I experienced a significant outage on my Teksavvy internet feed. When it finally came back, I rebooted the modem, noticed that my domain no longer pointed to my computer, and made the change to the new IP address at DynDNS. I then sent an email message to an acquaintance. I have an automatic BCC to myself to let me know that messages are in fact being sent — and this one was not. So I tried some more test messages, none of which were being delivered!

Two days later, the situation had not resolved itself and I Gmailed Teksavvy support for help. After the usual security clearance, they sent me a long questionnaire — meant to rule out some obvious causes of failure. Some of the directions in that questionnaire made it obvious that they had no more than a passing acquaintance with Apple OS X and, along with my answers, I expressed: "I'm a little concerned about your lack of Mac experience as evidenced by several mistakes in using-the-terminal instructions. Other OS X references are stale at best. The current OS X is 10.12 and the location of the 'Network Utility' hasn't been where you say it is for some two years!"

On 26 September, they finally opened a ticket with their "Tier 2" department. This meant, I suppose, that this was not an obvious failure and they were prepared to take my issue seriously. What followed was an attempt to have me send a message in something other than Apple Mail because

Well, not so fast. On 25 September I had created an email account with Bell. The setup was straightforward and my first message — using Apple Mail — arrived at its destination immediately. Nothing wrong with Apple Mail! From Teksavvy's perspective, all I had to do to fix my problem was ditch Apple Mail. From my perspective, that was never going to happen. I've been using Apple Mail for some fourteen years. I've been using Teksavvy for five years. So long.

Two days later, the situation had not resolved itself and I Gmailed Teksavvy support for help. After the usual security clearance, they sent me a long questionnaire — meant to rule out some obvious causes of failure. Some of the directions in that questionnaire made it obvious that they had no more than a passing acquaintance with Apple OS X and, along with my answers, I expressed: "I'm a little concerned about your lack of Mac experience as evidenced by several mistakes in using-the-terminal instructions. Other OS X references are stale at best. The current OS X is 10.12 and the location of the 'Network Utility' hasn't been where you say it is for some two years!"

On 26 September, they finally opened a ticket with their "Tier 2" department. This meant, I suppose, that this was not an obvious failure and they were prepared to take my issue seriously. What followed was an attempt to have me send a message in something other than Apple Mail because

*they*were able to send something (using my account) by way of Microsoft Outlook. And I did manage eventually to send something using Mozilla Thunderbird, so it was (obviously) only a problem with the Apple Mail program.Well, not so fast. On 25 September I had created an email account with Bell. The setup was straightforward and my first message — using Apple Mail — arrived at its destination immediately. Nothing wrong with Apple Mail! From Teksavvy's perspective, all I had to do to fix my problem was ditch Apple Mail. From my perspective, that was never going to happen. I've been using Apple Mail for some fourteen years. I've been using Teksavvy for five years. So long.

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