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.


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