I did a more comprehensive search [3] for solutions to the Diophantine[1] equation R*S^2 = B^E-2 causing a square root of an integer to have repeating digits, with the digits resembling the coefficients of Series[K*sqrt(1/(1-2x))]. We want to have at least 2 repeated digits, because there are just too many examples if you allow just a single repeated digit (but see [2]). Some of the digit patterns are pretty darn pretty. sqrt(1820) in base 2 repeats 13 digits : 101010.1010100 1010101010100 1111111111111 1010101010100 ... sqrt(167772) in base 2 does a similar thing: 110011001.100110011000 11001100110011001100 100110011001100110010 111111111111111111111 ... This comes from 2^21-2 = 5^2*83886, and the fact that 5^2*2^2 = 100 gives us the base-10 similarity between 167772 and 2^24=16777216. Also in base 2: sqrt(28), sqrt(116508), and sqrt(85598) are similar. There are lots of 2^N-2 that are divisible by a square. sqrt(3) in base 3 repeats 3 digits: 1.20 120 212 222... This comes from 5^2+2=3^3 and is the only one I found where you're taking the square root of the base itself. Bases 4, 8 and 9 inherit all the cool patterns seen in bases 2 or 3. I gave a base 5 example earlier, another cool one is sqrt(1735): 131.31 13131 22222 13131 21211 44444 ... These go up to high bases too, and some are stunning. sqrt(16448427) in base 23 is: 7F7.F7 7F7F7 BBBBB 7F7F7 ACACA 5H5H5 where most pairs of digits add to M (the highest digit in base 23). sqrt(2434958270224475) in base 23 is a similar marvel with seemingly endless 7F7F7F7's I also want to mention that there is nothing arbitrary about the repeated digits themselves. In the case of sqrt(2), we have 2*7^2+2=100, so the repeated "14" is 2*7. This is also close to 100/7. In the case of sqrt(62), we have 62*127^2=999998, so the repeated digits "7874" are 62*127, close to 1000000/127. - Robert [1] Michael Reid mentions the Diophantine (2 a^2 - 1)(2 b^2 - 1)^2 = (2 c^2 - 1). I haven't tried to figure out whether or not the solutions to my Diophantine are in one-to-one correspondance with his. [2] A single repeated digit happens a lot: for any base B=K*N^2+2, sqrt(K*B) starts with a repeated digit. They all come from the same series expansion. In base 10, sqrt(20) shows the pattern: it is 4+0.4+0.06+0.010+... which is 4 times the series expansion of sqrt(1/(1-2x)) with x=0.1. [3] A longer list (bases up to 36 not including perfect powers like base 8 or base 9; R*S^2 up to 10^17, and R or R*B up to 999999): sqrt(28) in base 2 repeats 7 digits, because 14*3^2 = 2^7-2 = 126 sqrt(1820) in base 2 repeats 13 digits, because 910*3^2 = 2^13-2 = 8190 sqrt(116508) in base 2 repeats 19 digits, because 58254*3^2 = 2^19-2 = 524286 sqrt(167772) in base 2 repeats 21 digits, because 83886*5^2 = 2^21-2 = 2097150 sqrt(85598) in base 2 repeats 22 digits, because 85598*7^2 = 2^22-2 = 4194302 sqrt(3) in base 3 repeats 3 digits, because 1*5^2 = 3^3-2 = 25 sqrt(1735) in base 5 repeats 5 digits, because 347*3^2 = 5^5-2 = 3123 sqrt(276) in base 6 repeats 5 digits, because 46*13^2 = 6^5-2 = 7774 sqrt(977423) in base 7 repeats 10 digits, because 977423*17^2 = 7^10-2 = 282475247 sqrt(2) in base 10 repeats 2 digits, because 2*7^2 = 10^2-2 = 98 sqrt(62) in base 10 repeats 6 digits, because 62*127^2 = 10^6-2 = 999998 sqrt(836612) in base 14 repeats 5 digits, because 59758*3^2 = 14^5-2 = 537822 sqrt(392318) in base 22 repeats 6 digits, because 392318*17^2 = 22^6-2 = 113379902 sqrt(5711) in base 23 repeats 4 digits, because 5711*7^2 = 23^4-2 = 279839 sqrt(9326) in base 26 repeats 4 digits, because 9326*7^2 = 26^4-2 = 456974 sqrt(24584) in base 28 repeats 3 digits, because 878*5^2 = 28^3-2 = 21950 Allowing the residue to be a lot larger reveals some like this: sqrt(182044444444444440) in base 20 repeats 13 digits, because 9102222222222222*3^2 = 20^13-2 = 81919999999999998 This might seem like a freak or even an error, but it checks out. In fact, the statement "62*127^2 = 10^6-2 = 999998", written in base 5, becomes "222*1002^2 = 20^11-2 = 223444443". On Sun, Feb 5, 2012 at 21:13, Robert Munafo <mrob27@gmail.com> wrote:

Well, of course there's 2*7^2 = 98, which gives us the pretty digit pattern:

sqrt(2) = 1.4 14 ...

More on this later, plus a mnemonic for the coefficients of Series[sqrt(1/(1-2x))]

In base 5, because 111^2*3963+2 = 5^11, the square root of 3963*5 = 19815_10 = 1113230_5 = 1030.34033 00103034033 00132103324... where the digit-groups are 11 digits long.

In base 6, because 46*13^2+2 = 6^5, the square root of 46*6 = 276_10 = 1140_6 = 24.34 02434 04053... where the digit-groups are 5 digits long.

In base 7, because 977423*17^2+2 = 7^10, the square root of 977423_10 = 11210426_7 = 2611.43464 0261143464 0421565362... where the digit-groups are 10 digits long.

In each of these examples, the third and 4th pretty digit-groups are 3/2 and 5/2 times the first two digit groups. For example in the familiar square root of 2, the first 8 digits (in groups of 2) are 14, 14, 21, 35 and 14*1.5 = 21, 14*5/2=35.

(You can check the other pretty digit examples above, but remember I showed each in its native base and in odd bases 1/2 is an infinite "repeating decimal". So 261143464 times 3/2 in base 7 is 421565362.3333... where the "0.3333..." is 1/2 in base 7. These 3's overrun the following digit-group which hides the 5/2 term.)

So there are 4 examples (-:

The digits of the square root of 2, taken in pairs and divided by 14, are one way of remembering the coefficients of the series expansion of sqrt(1/(1-2x)), if you ever find yourself in the need and are stuck with just a pocket calculator. (-: The "62" ought to be 61.25, but it's overrun by the leading "1" of the following term which is 14*63/8=110.25) [...]

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