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So I was messing around with Tau in base 12 and Mathematica (very amateur), and at some point started converting 5 same digit long base 10 sequences to base 12, and inputing these sequences in a Notepad file with 1,250,000 Tau decimals in base 12. Weirdly, many of these (>1/2) pretty rare sequences tend to end up only ~5,000 digits from, and most of these even closer to 1,200,000th decimal. Here's the data:

X = 10
E = 11
sqs = sequences

("closest to..." is calculated from the first decimal the number makes appearance on)

  • $11111_{10} = 651E_{12}$ - 68 sqs found, closest to 1,200,000 on 1,201,556th decimal, (67th is 1,189,183rd),(this sequence is most likely not relevant - too common).
  • $22222_{10} = 10X3X_{12}$ - 4 sqs found, closest to 1,200,000 on 698,601st decimal.
  • $33333_{10} = 17359_{12}$ - 4 sqs found, closest to 1,200,000 on 1,197,845th decimal.
  • $44444_{10} = 21878_{12}$ - 5 sqs found, closest to 1,200,000 on 1,195,972nd decimal.
  • $55555_{10} = 28197_{12}$ - 4 sqs found, closest to 1,200,000 on 1,196,762nd decimal.
  • $66666_{10} = 326E6_{12}$ - 3 sqs found, closest to 1,200,000 on 1,188,745th decimal.
  • $77777_{10} = 39015_{12}$ - 3 sqs found, closest to 1,200,000 on 1,533,163rd decimal (679,798th).
  • $88888_{10} = 43534_{12}$ - 4 sqs found, closest to 1,200,000 on 1,066,571st decimal.
  • $99999_{10} = 49X53_{12}$ - 5 sqs found, closest to 1,200,000 on 1,199,871st decimal.

Of course then I wonder, what mathematical explanation does this have? It surely must have some explanation, the coincidence is just too big. (Now I will try to do the opposite, convert 5 of the same digit sequences from base 12 to base 10, and input it in a Tau base 10 file. There could be some similarities?)

edit: I did try what I suggested above, with little luck. There were some minor curiosities. Like 11111 = 22621 had 5 appearances within ~31,000 digits from 500,000, closest being 495,703, 496,433, 497,578 - and 495,703 was shared by XXXXX = 226210 (extra 0), but that must be 10% chance of happening, so, very minor curiousity.

Just another minor unrelated curiousity I noticed was the symmetrical number "06X999X60", where the middle "9" is decimal number 3,330.

To mods: Suggest and edit my tags, I'm really not sure what I should tag this (amateur).

Check this out, if you add these numbers together, almost the same pattern does emerge here as well (bold equations are the ones that were close to 1,200,000):

  • $11111_{10} = 651E_{12}$ | 6+5+1+E= 23
  • $22222_{10} = 10X3X_{12}$ | 1+0+X+3+X= 24
  • $33333_{10} = 17359_{12}$ | 1+7+3+5+9= 25
  • $44444_{10} = 21878_{12}$ | 2+1+8+7+8= 26
  • $55555_{10} = 28197_{12}$ | 2+8+1+9+7= 27
  • $66666_{10} = 326E6_{12}$ | 3+2+6+E+6= 28
  • $77777_{10} = 39015_{12}$ | 3+9+0+1+5= 18 (why not 29?)
  • $88888_{10} = 43534_{12}$ | 4+3+5+3+4= 19 (why not 30?)
  • $99999_{10} = 49X53_{12}$ | 4+9+X+5+3= 31

Do you see the pattern? 2 of the last numbers that were not close to the 1,200,000th decimal did not follow the logical steps. Only the 1st of them "10X3X" still followed the pattern.

$ 23, 24, 25, 26, 27, 28, 18, 19, 31 $

Isn't it a bit strange that the only number "24" that "does not follow" the "pattern", is the only number divisible by 12! :)

I now continue adding "11111", and the most exciting discovery so far is number $122221_{10} = 5X891_{12}$ makes its 13th appearance on decimal 2,222,222 !

PS: I did do many more +11111, but no apparent pattern emerged. The next 9 numbers weren't that close to 1,200,000 in general as the first 9 (tho still had some bias towards it). But nothing really suggests that the first 9 numbers, 0r the second were more than just slightly lucky.

Maybe I will update this post occasionally with more interesting coincidences, if I find them, as with 122221 (mentioned above), the symmetrical number 06X999X60, in which the middle 9 is placed at 3,330th decimal, and the three XXX's (101010) where the middle X is placed on top of decimal 6,666.

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    $\begingroup$ Please forgive my ignorance, but who exactly is $\tau$ and why should we care? $\endgroup$
    – Alex M.
    Nov 17 '16 at 13:58
  • $\begingroup$ @AlexM. I did find another curiousity in Tau base 12 today. There are three "XXX" (101010) forked by the decimal number 6,666. So I would be careful talking about Tau in base 12 in such terms if I were you. $\endgroup$
    – someone
    Nov 17 '16 at 14:09
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    $\begingroup$ @AlexM.: I am curious, too. However, I believe they mean $\tau=2\pi$. There is a movement about to use $\tau$ instead of $\pi$. $\endgroup$
    – robjohn
    Nov 17 '16 at 15:26
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    $\begingroup$ Topics of interest: spigot algorithms for $\pi$ and normality of $\pi$. $\endgroup$
    – Sean Lake
    Nov 17 '16 at 17:08
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This is an expected behaviour. You found many (extremely) close occurrences but also several incomparably distant ones. The standard deviation of the offsets you list is roughly 207300 and counting the mean square deviation from our artificial reference point of 1200000 we get something even a bit higher, roughly 244943.

These numbers compare well with the inverse of the expected probability of occurrence of a pattern of particular 5-digit sequence in a random decimal string, i.e., $10^5$. Nothing would be out of normal still if they were twice as close each.

For $\pi$ (and consequently, $τ$), it is believed that the digits in any base should be distributed just like completely random numbers, and in the light of that any clusters are a mere coincidence that could happen, comparably sparingly, if you generated digits from a coin toss. The magnitude of 12 million relative to $10^5$ seems well within “sparingly”, albeit I did not calculate that in more detail.

While this may sound disappointing, please don't let that put you off looking for inexplicable phenomena in different combinations of bases! You're doing a great job and maybe you'll help disprove the conjecture. Many a big discovery started with a that's funny....

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    $\begingroup$ That's funny... $\endgroup$
    – qwr
    Nov 17 '16 at 21:31
  • $\begingroup$ @qwr That's the spirit! I did not know BBP was discovered using experimental mathematics! (Although that should not surprise me having seen some other Simon Plouffe's works.) $\endgroup$
    – The Vee
    Nov 18 '16 at 0:03
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About you second observation

=== newer answer after some thought =====

Consider in base 10, you have the number $1432= 1*10^3 + 4*10^2 + 3*10 + 2$ and you add up the digits $1 + 4+3 + 2 = 10$.

Then you do $2*1432 = 1432 + 1432 = 2*10^3 + 8*10^2 + 6*10 + 2 = 2864$ and you add up the digits $2+8+6+4 = 2(1+4+3+2) = 2*10 = 20$.

Now do $3*1432 = 1432 + 1432 + 1432 = 3*10^3 + 12*10^2 + 9*10 + 6$ and you add up the quasi-digits $3+12 + 9 + 6 = 3(1 + 4 + 3 + 2) = 3*10 = 30$.

But those are the quasi-digits the actual digits require carrying.

$3*1432 = 1432 + 1432 + 1432 = 3*10^3 + 12*10^2 + 9*10 + 6=$

$(3 + 1)*10^3 + (12 -10)*10^2 + 9*10 + 6$ and if we add up these psuedo-digits we get $(3 + 1) + (12 - 10)10^2 + 9 + 6 = (3 + 12 + 9 + 6) - 9 = 3(1+4+3+2) - 9 = 3*10 - 9 = 21$.

And indeed $3*1432 = 1432 + 1432 + 1432 = $

$3*10^3 + 12*10^2 + 9*10 + 6=(3 + 1)*10^3 + (12 -10)*10^2 + 9*10 + 6=$

$4*10^3 + 2*10^2 + 9*10 + 6 = 4296$ and the sum of the digits is $4+2+9+6 = 12$.

So our pattern went from $10,20$ and then when we expected $30$ it dropped by $9$ to $21$. i.e. instead of adding $10$, it added $1$.

For $4*1432$ we will get $4(16)(12)8$ and if we add $4+16+12+8 = 40$ we get the next in the expected pattern of adding $10$ each time.

But we have to carry so we get $(4+1)(16-10+1)(12 - 10)8 = 5728$ and $5+7+2+8 = 22 = 4 + 16 + 12 + 8 - 9 - 9 = 40 - 18$. Instead of adding $10$, we only added $1$ again.

So in general, if the sum of the digits of $X$ is $k$ then the sum of the digits of $vX$ will be $vk - 9m$ for however $m$ many times we had to "carry" in calculating the digits.

It's the same thing for base 12. Except instead of carrying "$-10$" and add in "$1$" to get a net of $-9$. We carry $-12$ to get a net of $-11$.

$11111_{10} = 651E_{12}$ has the sum of the digits $6+5+1+11 = 23$.

So the sum of the digits of $xxxxx_{10} = x*651E_{12}$ will have the sum of $23x - 11m$ if you carry $m$ times.

$222222_{10} = 2*651E_{12} = (2*6)12^3 + (2*5)12^2 + (2*1)12 + 2*11$

$= 1*12^4 + (12 -12)12^3 + 10*12^2 + (2 + 1)12 + (22 - 12)11$ will have the sum of the digits be $2*23 - 2*11 = 24$ because we carried twice.

Continuing we get the patter $23, 24, 25,26,27,28, 18 (= 29 + 1 - 12), 19, 31 (=19 +1 - (1-12))$, etc. The pattern is basically $ + 23 - 2*11 = + 1$ as you will typically need to carry twice in each step. Sometimes you we need to carry more or less so you may drop by $+1 - 11 = -10$ or sometimes you will jump by $+1 + 11 = + 12$. You may even jump by $23$ if you don't carry at all. (Maybe, it'll be rare.)

=== older confusing, stating the answer as I think of it and observations as the ome to me, answer ======

As to your second observation:

$11111_{10} = 651E_{12}$; this can be thought of as more or less arbitrary.

The sum of the digits is $6 + 5 + 1 + E = 23_{10} = 1E_{12}$

If it weren't for "carrying", the sum of the digits of $xxxxx_{10}$ would be $x$ times the sum of the digits of $11111$. But because of the carrying and because we are adding the digits of a base 12, when we carry we are decreasing an expected value by 12 and increasing the next expected value by 1 for a total decrease by a multiple by 11.

So we expect $22222 = (12)X2(2*E)$ and $12 + X + 2 + (2*E) = 46 = 2*23$ but the $2*E = 1X$. So we have (12)X3X The $X$ is $12$ less than $1X_{12}=22_{10}$ but "$3$" is $1$ more than "$2$ so we have $12 + X + 3 + X = 35$ which is $11$ less than what we expected. But the the (12) becomes "$12_{10}=10_{12}->1+0=0$" so that decreases what we expected by another $11$ so we get $22222 = 10X3X$ and $1+0+X+3+X = 24$

By coincidence, our first sum was $23=2*11 + 1$ so we expected the next sum to yield $23+23$ but it actually yielded $23 + 23 - 11 - 11 = 23 + 1 = 24$

We will expect our third sum to yeild $23 + 23 + 23$ but $3*651E$ will have us carry some more. $3*E=29$ the that carries twice. $3*5 = 13$ so that carries once. and $3*6 = 16$ so that carries once so the sum is $23 +23 + 23 - 4(11)= 25$.

The rest of the numbers behave nicely. We carry $6$ times for $44444$ to get a value of $4*23 - 6*11 = 23 + 3$. And from $55555$ and $66666$ we carry $8$ and $10$ times respectively. But for $77777$ we carry $13$ no $12$ times so we get $7*23 - 13*11 = 7(2*11 + 1) - 13*11 = 11 + 7 = 18$ which is $11$ less than we were coming to expect. (You made an error $3+9 + 1 + 5 = 18$; not $17$)

$88888$ carries $15$ times so we return to our pattern of "adding one".

But $99999$ carries $16$, not $17$, times so we "pop back" to our first sequence now in the thirties.

The pattern will continue of usually adding one, but occasionally instead of adding $1$ it will drop $10$ or jump $12$. It may even occasionally drop $21$ or jump $23$.

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  • $\begingroup$ This was a bit too much over my head, hehe. I will try to understand it as well as I can. Thank you for correcting my mistake and writing this answer. Shame, one prime less. $\endgroup$
    – someone
    Nov 17 '16 at 18:15

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