Freeman Dyson and 1/19
I just read this article about Freeman Dyson and a math puzzle that asks you to find a number that is doubled when you tear off the rightmost digit and stick it on the left. For example, tearing the 2 off 12 and sticking it in front gets you 21, which isn’t 2×12, so 12 isn’t right. Moving the 1 in 7654321 gets you 1765432 which also dosn’t double it, so that’s not the number either.
Anyway, it turns out the number we’re looking for is 18 digits long. It’s 052,631,578,947,368,421 which doubles to 105,263,157,894,736,842. (You’re allowed to stick a zero in front.)
Well, my son loves this kind of base-10 number tomfoolery, so I asked him if he could figure it out. And he reeled off the answer without pause. All 18 digits.
“Uhh, how’d ya know?” I asked.
He said it’s because 1/19 is 0.052631578947368421… (repeated forever), and 2/19 is 0.105263157894736842… (just move the 1). And of course 2/19 is twice 1/19. So if you know your N/19 repeating decimals, this is apparently easy peasy. And all the N/19’s are buried in that infinite sequence. You just have to start at different places. 3/19 is 0.157894736842105263…, 4/19 is 0.210526315789473684…, 5/19 is 0.263157894736842105…, etc.
So the next time someone asks “what’s a number that’s quadrupled when you tear the last two digits off the right and stick them on the left” you can say “doy, it’s 052,631,578,947,368,421 of course” and then roll your eyes. Because you’ve seen 1/19 and 4/19 written out as decimals.
Anyway, my son also told me there’s something similar for 1/29, except you triple the number by tearing off the rightmost digit and putting it on the left. I think it’s 28 digits long.
And also for 1/39, except you quadruple (x4) the number. And with 1/49 you x5 the number, and 1/59 gets you x6. Even 1/99 (0.0101010101…) works (x10 gives you 0.1010101010…).
As for 1/109, I don’t know, I’ll bet there’s some kind of x11 trick there. My son’s in bed now, but I’ll ask him tomorrow when he gets up.
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One other note: The linked NYTimes article describes how to generate 052,631,578,947,368,421. Write down a 1, double it and put down the 2, double the 2 and put down the 4, and keep going until you cycle all the way back. The sequence looks like this:
1
21
421
8421
(16)8421
When you get to the 16 you just put down the 6 and carry the 1. Then you double the 6 and add the carry. You get:
(16)8421 -> (+1)68421 -> (6+6+1)68421 -> (13)68421 -> (+1)368421
At the next step you get
(3+3+1)368421 -> 7368421
Eventually you end up back at 1 (with no carry) and the whole thing starts again.
Another note: The method described in my comment above is the same as shifting and summing summing all the powers of 2. Like this:
If you keep doing this forever, with all the powers of 2 (to infinity), you end up with …052631578947368421 repeated over and over. Since you repeat on the left and the sequence is infinite, the first digit is undefined. If you start with 0.XXXX and glue the infinite sequence after the decimal point, there are 18 places in the sequence where you could start. They correspond to the decimal representations of 1/19 thru 18/19.
There is another way to generate this sequence, but left-to-right instead of right-to-left like we’ve done so far. It uses powers of 5 instead of powers of 2, and it shifts 2 by digits (not just 1) for each new term. Here’s how it looks:
You can also do this 2 digits at a time but you have to predict how much carrying you’re going to get. It’s not hard though.
And if you do generate this 2 digits at a time, you can start anywhere. You can start with 10 or 05 or 52 or 26. You don’t have to worry about even or odd alignment — it works both ways.
Since you can generate the 1/19 sequence right-to-left with powers of 2, or left-to-right with powers of 5, what about 1/49? The post mentions that you can generate the 1/49 sequence (42 digits before it repeats) right-to-left with powers of 5. So can you also generate 1/49 left-to-right with powers of 2?
Yes. 1/49 is 0.020408163265306122448979591836734693877551… You can travel left-to-right through this series 2 digits at a time by doubling the previous 2 digits mod 100. And if the result is 50 or above you add one (to carry from the next double). So from 02 you get 04, 08, 16, 32, and then 65 (you add 1 to 64 because it 64 >= 50). From 65 you get 130 which is 30 when you mod 100.
What’s really cool about this sequence is that you can start doubling from anywhere, even though this generates 2 digits at a time. You can start with 02 or with 20. Or with 08 or 81 or 16. It’s like we have two sequences of 2-digit numbers that zip together. The even-aligned sequence (02 04 08 16 32 65 30 etc) and the odd-aligned sequence (20 40 81 63 26 53 06 12 24 etc). Both follow the same doubling rule, and both are the same when viewed as a sequence of 1-digit numbers.
WOW! That is really cool… How do you think he figured that n/19 pattern?
Did he already know the pattern or did he figure a clever way to solve the problem and then seek the pattern out?
Hey SATT, nice to see you!
My son likes calculating stuff out and looking for patterns. He’s done it since he was 2. Squares, cubes, primes, Pascal triangle stuff, all kinds of integer sequences.
And you know how N/7 works:
1/7 0.142857…
2/7 0.285714…
3/7 0.428571…
4/7 0.571428…
5/7 0.714285…
9/7 0.857142…
Same 6 digits (repeated forever), just started at different places. And 14*2=28, 28*2=56, and you add to 56 to make 57 because 57*2=114 which is 3 digits so you have to take the 1 off and carry it back to the 56. Or you can start doubling at 42 instead of 14 — the even/odd patterns zip together.
So once my son saw how the 7ths worked he’s looked for the same thing elsewhere. The 19ths are almost the same.
WOW AGAIN!… Ill have to start taking a closer look at numbers… I never realized how interesting all these natural patterns are.