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:

$\sqrt{8578201027979935320056}=92618578201.027979935320056863...$

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.

$\sqrt{100} = 10.000...$

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?