Limited persistence (part 2)

I lied about leaving the other bases to you.

In base 2, as I said, every number > 1 has persistence 1. (Either it contains a 0, so goes to 0, or it’s all 1s, so goes to 1.)

In base 3, only one number in each decade (of the form 222…2223) is worth looking at. From 222223 on most of the digit products seem to contain zeros so the persistence is 2. 223 and 22223 also have persistence 2. The only numbers up to 1000 digits with larger persistence seem to be 2223 and 2222222222222223 both with persistence 3.

For bases 3 through 16, as far as I’ve checked in each:

BaseMax persistMin example

One thing that seems to be happening is that you tend to get larger maximum persistences in prime bases, smaller ones in composite bases. Presumably that’s because of the analog to the 5-and-even situation in base 10: in a prime base, the digit product cannot be a multiple of the base, while in a composite base it can, in which case the digit product ends in a zero and terminates the sequence. Notice how every prime base from 5 on up has larger maximum persistence than the subsequent base. Also, recall 12 and 16 have respectively four and three proper divisors larger than 1, and then notice how small the maximum persistence is in bases 12 and 16 compared to bases 10, 11, 13, 14, and 15.


Limited persistence

There’s a numberphile video about the number 277,777,788,888,899:

But in written words, here it is: Take a number and multiply its digits (in base 10 unless otherwise noted) together. Then the product of the digits of that number. Then the product of the digits of that number. Keep going. Eventually you will reach a single digit number (the digit product for a multi-digit number is always less than that number, so it decreases at each step until a single digit is reached), and of course the digit product of a single digit number is itself, the end.

For instance: Starting from 28, we have 2×8 = 16, then 1×6 = 6 and you get to a single digit number in two steps. We say the multiplicative persistence of 28 is 2. From 88 it goes: 88, 64, 24, 8. Persistence is 3.

You might guess there’s no upper limit on persistence, that there can be numbers with persistence of 100 or 1000 or 1,000,000 or whatever; but actually the conjecture is that no number has persistence higher than 11. The smallest number with persistence 11 is 277,777,788,888,899:

277777788888899, 4996238671872, 438939648, 4478976, 338688, 27648, 2688, 768, 336, 54, 20, 0

Obviously the persistence doesn’t depend on the order in which the digits occur, so for instance 998,888,887,777,772 also has persistence 11. So does 27,777,772,228,888,899, where an 8 has been replaced by three 2s. Or by replacing all the 8s, and all the 9s by two 3s, you get numbers like 22,222,222,222,222,222,223,333,777,777, again with persistence 11. And of course adding a 1 digit doesn’t change anything, so 111,122,222,222,222,222,222,223,333,777,777 has persistence 11 too.

But no one has ever found a number with greater persistence. Why not?

Well, there’s a zero trap. Any number > 9 with a 0 digit in it has persistence 1. Any number that has no zero in it but does have a 5 and an even digit has persistence 2, because the product of its digits ends in 0. And even if a number has no zero and either no 5 or no 2, the same must be true of the product of its digits, and of the product of the digits of that product, and so on, for eleven steps in order for there to be a twelfth step. And when you’re dealing with numbers up around 277,777,788,888,899, the likelihood of that gets vanishingly small.

But how would you check if it’s really vanishing? The naive thing to do is to check all numbers up through, say, 1040, but that would take, ah, a while. (What’s your computer’s clock speed? The age of the Universe is 4.3 × 1026 nanoseconds…). Checking all the n-digit numbers would take around ten times longer than the n–1-digit ones, and that gets intractable pretty fast.

But as we just noticed, there’s no point in checking numbers containing a zero. And up around 277,777,788,888,899, a lot of numbers do. You can also skip any number containing a 5 and an even digit. Numbers containing a 1 are equivalent to a smaller number without a 1, so no point in bothering with them either. And of course, there’s no point in checking a number whose digits are a permutation of a number you’ve already checked.

In fact, suppose you don’t worry about the 5-and-even check. Then you can just look at numbers containing only digits > 1 in nondecreasing order. Like this:

2, 3, 4, 5, 6, 7, 8, 9, 22, 23, 24, 25, 26, 27, 28, 29, 33, 34, 35… , 77, 78, 79, 88, 89, 99, 222, 223, 224…

There are 8 valid single digit numbers. For 2-digit numbers there are 8 valid first digits, but the second digit must not be less than the first, so there are 8 numbers starting with 2, 7 starting with 3, 6 starting with 4, and so on: 8+7+6+5+4+3+2+1 = 36. That’s the 8th triangular number. For three digits, the number of valid numbers is similarly the 8th tetrahedral number, and generally for n digits, the 8th n-simplex number:

N(n) = \frac{(7+n)!}{7!n!}

That increases pretty slowly with n. You have to check 8 out of 10 1-digit numbers, but only 36 out of 90 2-digit numbers and 120 out of 900 3-digit numbers, and by the time you’re up to 40-digit numbers, you only have to check 62,891,499 of them! If you do skip over numbers with a 5 and an even digit, that drops even further to only 9,378,299 numbers. For 41 digits the number is 10,749,914, so instead of ten times as many numbers to check in that decade, there’s only about 15% more. A Python script can blow pretty easily through 50 or more digits in a fairly short time.

What you find is this: For valid (nondecreasing, all digits > 1) 15 digit numbers only one, 277,777,788,888,899, has persistence 11. For 16 digits, there’s only 2,247,777,778,888,899. For 17 digits there are two: 22,227,777,778,888,899 and 27,777,789,999,999,999. The first of these is a trivial modification of the 15-digit number, with the same digit product; the second has a different digit product, but the digit product of its digit product is the same as for the 15-digit number; it goes 27777789999999999, 937638166841712, 438939648, 4478976, 338688, 27648, 2688, 768, 336, 54, 20, 0.

Up to 29 digits there are only two valid numbers per decade with persistence 11, based on those two 17 digit numbers. The longest valid variant of 277,777,788,888,899 is 22,222,222,222,222,222,223,333,777,777 as noted above, and the other also has variants up to 29 digits. From 30 digits on up the maximum persistence is 10, with variants of a single number that peter out at 36 digits. For 37 through 42 digits maximum persistence is 8; for 43 through 44, 6: for 45 through 58, 5, and that’s all I’ve got so far.

The sequence of the persistences of the counting numbers is A031346 in OEIS, and the smallest number with persistence n is A003001. Per comments there, Martin Gardner wrote about the subject (of course), and there are results implying there are no numbers with persistence greater than 11 through 1020585.

That’s in base 10. In binary, every number > 1 has persistence 1. I’ll leave the other bases to you.

Crafting grafting, part 4

(I’ve gone back and revised my notation. a becomes j and b becomes 10^m.)

Check out this grafting number, the second biggest I’ve got at the moment:


There’s some structure to understand here. This is generated using j = 9261, m = 8, and n = 7. You can read those values off the numbers above, though. There are 22 digits in the integer on the left, and that’s 2n+m. On the right, those 22 digits recur starting 7 digits left of the decimal point; that’s n. Before them are the digits 9261, which is j.

Likewise, examine this one:

\sqrt{4110105646} = 64110.105646457...

You can see 2n+m = 10 and n = 4, so m = 2, and j = 6.

Though here’s an evil one:

\sqrt{576276646} = 24005.76276646922...

Here 2n+m = 9 and n = 1, so m = 7, and j = 2400. Or… wait, is it? If you write the integer on the left as 0576276646 then 2n+m = 10 and n = 2 so m = 6 and j = 240. Or should you write it as 00576276646 and conclude 2n+m = 11 and n = 3 so m = 5 and j = 24? Turns out all three interpretations are valid; they lead to three different equations yielding the same grafting number.

What about this miserable attempt at a grafting number?

\sqrt{100} = 10.000...

Well, 2n+m = 3 and n = 2, so m = , um, -1? And j = 0? Those give x = 0.1000... (with the + sign in the quadratic formula, unlike all normal grafting numbers, because the - sign gives x=0) and the equation 10^2(0+0.1) = \sqrt{10^4(10^{-1}(0.1))} or 10 = \sqrt{100}. Yes, that sort of fits the profile, but in an ugly way.

Likewise, \sqrt{98} = 9.8995... and \sqrt{99} = 9.9498... have similarly pathological analyses. Those I think really do need to be counted as grafting numbers, but not normal ones.

Two things I have not figured out yet is why it’s always the ceiling, not the floor, that gives the grafting number, and why j = 2 and m = 1 gives grafting numbers for, apparently, all values of n while other values of j and m don’t. I suspect these are related. Observe

\sqrt{\lceil y \rceil}  = \sqrt{y-\{y\}+1} = \sqrt{y}\sqrt{1-\frac{\{y\}-1}{y}}

(with braces denoting fractional part) which is, to first order

\sqrt{y}\left (1-\frac{\{y\}-1}{2y} \right ) = \sqrt{y} - \frac{\{y\}-1}{2\sqrt{y}}

and \{y\} is whatever it is, but it’s in 0<\{y\}<1 and for an approximate result we can take \{y\}\approx1/2 so

\sqrt{\lceil y \rceil}  \approx \sqrt{y} + \frac{1}{4\sqrt{y}}.

By taking the square root of the ceiling of a number y instead of y itself, we add approximately 1/(4\sqrt{y}) to the square root. By the same token, using floor subtracts about 1/(4\sqrt{y}).

That’s sort of an average, though, and in the case of grafting numbers, it’s often an overestimate. For instance, for n = 3, j = 8079, we get x=49991903-\sqrt{2499190300000000}=6557202816420999992418.... Then with m = 8 we end up with grafting number \lceil 65572028164209.99992418... \rceil. Clearly the ceiling in this case changes the number by several orders of magnitude less than \approx 1/2. But why don’t numbers looking hypothetically like \lfloor 65572028164209.00002418... \rfloor work out?

Crafting grafting, part 3

In the first two parts we found some grafting numbers using the equation (j+x)^2=10x with j=1~{\mathrm or~}2. Real solutions 0 < x < 1 can be turned into integer grafting number candidates \lceil 10^{2n+1}x\rceil. Or potentially \lfloor 10^{2n+1}x\rfloor but we haven’t found a case where that works.

We could use other values for the coefficient of x, though. Any power of 10 in fact. (For grafting numbers in base 10. If you want grafting numbers in another base, use powers of that base.) We have (j+x)^2=10^mx. Solutions are x = ((10^m-2j)\pm\sqrt{10^{2m}-4\times 10^mj})/2. Real solutions are obtained up to j=10^m/4 but we want 0 < x < 1. You can figure out this means j < \sqrt{10^m}-1.

For instance, with m=2, we can use 0 < j < 9. With j=1, x = 0.0102051443.... Now \lceil 10^{2n}(100x)\rceil = 2 (n=0), 103, 10206, 1020515, 102051444… not one of which, sorry to report, is a grafting number. With j=8, though, and n = 0, we get \sqrt{77} = 8.77496... and that is a grafting number, though not very impressive. j=6 and n = 4 gives us \sqrt{4110105646} = 64110.105646457.... Yes! That’s what I’m talking about! And j=7 with n = 1, 3, 5 gives \sqrt{5736} = 75.73638..., \sqrt{57359313} = 7573.5931366..., and \sqrt{57359312880715} = 7573593.1288071582....

And on it goes. Those two I started off the first post with, 60,755,907 and 63,826,090,875, arise from j=7794, m=8, and n=0 and j=252, m=5, and n=3, respectively. Here’s another: \sqrt{44144658239614} = 6644144.658239614216..., and that’s the only one I’ve found so far using floor instead of ceil. Edit: This does not use floor; I must have been fooled by a rounding error. This comes from j=66, m=4, and n=5. Then x = 0.4414465823961399708..., and 10^{14}x = 44144658239614 when rounded up. In fact I’ve found no cases where floor gives the grafting number, and exactly one case where both ceil and 1+ceil work: j=2, m=1, and n=1 giving grafting numbers 764 and 765. Otherwise it’s always ceil. Which suggests other questions to ask. Sometime.

Crafting grafting, part 2

Before going on, let me address an omission. I gave some examples of grafting numbers but I never actually specified what a grafting number is.

A grafting number is a counting number whose digits appear (consecutively, and in order) in its square root, starting at or to the left of the decimal point.

Matt Parker excludes cases where the digits appear further to the right of the decimal point, which may seem an arbitrary restriction, but it makes sense if you think about it. We don’t know that square roots of nonsquare counting numbers are normal numbers, but it does seem plausible. A normal number is an irrational number in which every finite string of digits occurs with equal density to every other digit string of the same length. If square roots are normal then including numbers whose digits appear anywhere in their square root as grafting numbers would mean every nonsquare number is a grafting number! For instance:

\sqrt{744} = 27.27636339397171178551719510385918447020016888957751385958655644587745126\mathbf{744}79307476086...

That would be a bore, so we just allow numbers whose digits appear beginning at or to the left of the decimal point in their square roots.

The above definition allows \sqrt{1} = 1 and \sqrt{100} = 10.0, for instance, and it probably shouldn’t, but let’s leave it as is for now.

We came up with a grafting number by first finding \sqrt {10(3-\sqrt{5})} = \sqrt{7.639320...}=2.7639320..., then multiplying both sides by 100, and then rounding up the number under the radical sign to an integer.

100 was an arbitrary choice, though. What if we used 10^n? That is, candidate grafting numbers would be \sqrt{\lceil 10^{2n}\times 7.639320... \rceil}.

Let’s check:

  • \sqrt{8} = 2.828...
  • \sqrt{764} = 27.6405...
  • \sqrt{76394} = 276.39464...
  • \sqrt{7639321} = 2763.932162...
  • \sqrt{763932023} = 27639.3202340...

Seems to work every time. Notice the floor function doesn’t work, though: \sqrt{763932022} = 27639.3202159... doesn’t match that last digit, for instance. It’s not obvious why the ceiling function works and the floor function doesn’t. One might think both could work, or floor but not ceiling, or neither. For that matter there’s the… superceil? The second integer higher? That works in one case: \sqrt{765} = 27.6586.... There’s stuff to dig into here. But let’s set it aside.

We have here a series of grafting numbers, none of which is 60755907 or 63826090875 so clearly there’s more exploration to be done.

We got the above by considering (j+x)^2=10x, where j=2. What about other values of j, though? Expanding gives x^2-(b-2j)x+a^2 = 0 and so x = ((b-2j)\pm\sqrt{b^2-4jb})/2. Then when j=1 one solution is 4-\sqrt{15}=0.1270166538.... (We want solutions in 0 < x < 1, so the other solution is discarded.)

So \sqrt{1.270166538...} = 1.1270166538... which looks promising. Let’s check:

  • \sqrt{2} = 1.414... — nope.
  • \sqrt{128} = 11.3237... — nope.
  • \sqrt{12702} = 112.70314... — nope.
  • \sqrt{1270167} = 1127.016858.... — nope.
  • \sqrt{127016654} = 11270.1665471... — finally!

And no, the floor function doesn’t work where the ceiling fails here (or where it succeeds). Looks like we have a way of generating candidate grafting numbers, but not necessarily good ones.

That was j=1 and j=2. For j=3, 10^2-40j < 0 and there are no real solutions. So that’s that.

Except we can vary the right hand side of (j+x)^2=10x. The coefficient of x can be any power of 10: (j+x)^2=10^mx. We’ll do that next time.

The grafting numbers less than a million (excluding the dubious 1, 100, and 10000) are:


Crafting grafting

Here’s a cool integer: 60,755,907. What’s cool about it? Take the square root. You’ll need more significant figures than usual; the calculator that came with my Android phone will do it:

\displaystyle \sqrt{60755907} = 7794.6075590756973...

Check out what’s to the right of the decimal point. Crazy, right? Here’s another similar number:

\displaystyle \sqrt{63826090875} = 252638.2609087546739...

Here the digits of the integer on the left appear in the real number on the right starting in the 100s place. Matt Parker calls these things “grafting numbers“. What’s going on with them? This isn’t just weird coincidence, is it?

It isn’t.

Consider good old \phi = (1+\sqrt{5})/2. It’s a solution of the equation x^2=x+1. So \phi = 1.6180339... and \phi^2 = 2.6180339.... There a number and its square (or, looked at the other way, a number and its square root) have digits in common; an infinite number of them, in fact. There’s a hint here.

Now let’s look at solutions to (j+x)^2 = 10x. Suppose 0 < x < 1. Now if j is an integer, then j+x in decimal form is just the concatenation of j and x and 10x is just the digits of x shifted one to the left.

For instance, j = 2; then (2+x)^2 = 10x. One solution is 3-\sqrt{5} = 0.7639320..., and 2.7639320...^2 = 7.639320.... Or looked at another way, \sqrt{7.639320...}=2.7639320....

This starts to look like the grafting numbers idea, but grafting numbers are integers. But hang on. Multiply both sides by, say, 100: \sqrt{76393.20...} = 276.39320.... Round the number on the left up and you find

\displaystyle \sqrt{76394} = 276.394645...

There you go, a grafting number. How about more? We’ll see…


There’s nothing at all new about this, in fact it’s ancient, but maybe it’s new to you? So here goes.

Here’s a square inscribed in a unit circle:

Or to keep down the clutter, here’s just one quadrant of that:

One side of the square along with the radii at each endpoint forms a triangle. The length of one side of the square we’ll call s_4 and it is of course \sqrt 2. That means the perimeter of the square is p_4 = 4\sqrt 2\approx 5.656854249.

Now consider an inscribed octagon. Again, here’s just one quadrant.

One side of the square is labeled s_8; what’s its length? Well, drop a perpendicular from the vertex in the middle:

And now notice the right triangle with hypotenuse 1 and sides x_8 and h_8 is just half of a quadrant of an inscribed square, which means h_8=s_4/2. Then x_8^2 = 1-h_8^2 = 1-s_4^2/4. From that you can get y_8=1-x_8 and from that, s_8^2=h_8^2+y_8^2. The perimeter of the octagon is then p_8 = 8s_8\approx 6.122934918.

Well, that was so much fun, let’s do it again. Here’s a quadrant of a 16-gon:

The right triangle with hypotenuse 1 and sides x_{16} and h_{16} is just half of an eighth of an inscribed octagon, which means h_{16}=s_8/2. Then x_{16}^2 = 1-h_{16}^2 = 1-s_8^2/4, y_{16}=1-x_{16}, and s_{16}^2=h_{16}^2+y_{16}^2. The perimeter of the 16-gon is then p_{16} = 16s_{16}\approx 6.242890305.

You know the rest, right? From here you can do the 32-gon, 64-gon, 128-gon, and so on, getting x_{2n}^2 = 1-h_{2n}^2 = 1-s_n^2/4, y_{2n}=1-x_{2n}, s_{2n}^2=h_{2n}^2+y_{2n}^2, and perimeter = p_{2n} = 2ns_{2n}. For n = 1024, the perimeter is p_{1024}\approx 6.283175451 = 2\times 3.1415877255. As n increases, the perimeter gets closer and closer to the circumference of the circle, so this gives a way to calculate \pi.

What about circumscribed polygons? If you look at the figures above, you can see the ratio of the size of a circumscribed square to an inscribed square is 1/x_8. Likewise the ratio of the size of a circumscribed octagon to an inscribed octagon is 1/x_{16}, and so on. So the perimeter of a circumscribed n-gon is P_n = p_n/x_{2n}. As n increases, P_n approaches 2\pi from above. And the average of p_n and P_n is a better approximation to 2\pi than either, though not by a lot.


This is something like the way Archimedes calculated \pi around 250 BC (I told you this was ancient), although he started with a hexagon rather than a square and only went up to a 96-gon. He didn’t have Google Sheets, though.