The Trivial Troot

Here is the square root of 2:

√2 = 1·414213562373095048801688724209698078569671875376948073176679738...

Here is the square root of 20:

√20 = 4·472135954999579392818347337462552470881236719223051448541794491...

And here are the first few triangular numbers:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105, 120, 136, 153, 171, 190, 210, 231, 253, 276, 300, 325, 351, 378, 406, 435, 465, 496, 528, 561, 595, 630, 666, 703, 741, 780, 820, 861, 903, 946, 990, 1035...

What links √2 and √20 strongly with the triangular numbers? At first glance, nothing does. The square roots of 2 and 20 are very different from the triangular numbers. Square roots like those are irrational, that is, they can’t be represented as a fraction or ratio of integers. This means that their digits go on for ever, never falling into a regular pattern. So the digits are hard to calculate. The sequence of triangular numbers also goes on for ever, but it’s very easy to calculate. The triangular numbers get their name from the way they can be arranged into simple triangles, like this:

* = 1

*
** = 3

*
**
*** = 6

*
**
***
**** = 10

*
**
***
****
***** = 15

The 1st triangular number is 1, the 2nd is 3 = 1+2, the 3rd is 6 = 1+2+3, the 4th is 10 = 1+2+3+4, and so on. The n-th triangular number = 1+2+3…+n, so the formula for the n-th triangular number is n*(n+1)/2 = (n^2+n)/2. So what’s the 123456789th triangular number? Easy: it’s 7620789436823655 (see A077694 at the OEIS). But what’s the 123456789th digit of √2 or √20? That’s not easy to answer. But here’s something else that is easy to answer. If tri(n) is the n-th triangular number, what are the values of n when tri(n) is one digit longer than tri(n-1)? That is, what are the values of n when tri(n) increases in length by one digit? If you look at the beginning of the sequence, you can see the first three answers:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, 66, 78, 91, 105...

1 is one digit longer than nothing, as it were, and 1 = tri(1); 10 is one digit longer than 6 and 10 = tri(4); 105 is one digit longer than 91 and 105 = tri(14). Here are some more answers, giving triangular numbers on the left, as they increase in length by one digit, and the n of tri(n) on the right:

1 ← 1
10 ← 4
105 ← 14
1035 ← 45
10011 ← 141
100128 ← 447
1000405 ← 1414
10001628 ← 4472
100005153 ← 14142
1000006281 ← 44721
10000020331 ← 141421
100000404505 ← 447214
1000001326005 ← 1414214
10000002437316 ← 4472136
100000012392316 ← 14142136
1000000042485480 ← 44721360
10000000037150046 ← 141421356
100000000000018810 ← 447213595
1000000000179470703 ← 1414213562
10000000002237948990 ← 4472135955
100000000010876002500 ← 14142135624
1000000000022548781025 ← 44721359550
10000000000026940078203 ← 141421356237
100000000000242416922750 ← 447213595500
1000000000000572687476751 ← 1414213562373
10000000000004117080477500 ← 4472135955000
100000000000007771272992046 ← 14142135623731
1000000000000031576491575006 ← 44721359549996
10000000000000140731196136705 ← 141421356237310
100000000000000250760786750861 ← 447213595499958
1000000000000000638090771126060 ← 1414213562373095
10000000000000000479330922588410 ← 4472135954999579
100000000000000000169466805816725 ← 14142135623730950
1000000000000000025572412483843115 ← 44721359549995794
10000000000000000087657358700327265 ← 141421356237309505
100000000000000000097566473134542830 ← 447213595499957939
1000000000000000000987561276980703725 ← 1414213562373095049
10000000000000000003048443380954913921 ← 4472135954999579393
100000000000000000006832246143819194316 ← 14142135623730950488
1000000000000000000014155501020518731556 ← 44721359549995793928

Can you spot the patterns? When tri(n) has an odd number of digits, n approximates the digits of √2; when tri(n) has an even number of digits, n approximates the digits of √20. And what can you call the approximations? Well, in a way they’re triangular roots so I’m calling them troots. Here are the troots for tri(n) with an odd number of digits:

1 → 1
14 → 105
141 → 10011
1414 → 1000405
14142 → 100005153
141421 → 10000020331
1414214 → 1000001326005
14142136 → 100000012392316
141421356 → 10000000037150046
1414213562 → 1000000000179470703
14142135624 → 100000000010876002500
141421356237 → 10000000000026940078203
1414213562373 → 1000000000000572687476751
14142135623731 → 100000000000007771272992046
141421356237310 → 10000000000000140731196136705
1414213562373095 → 1000000000000000638090771126060
14142135623730950 → 100000000000000000169466805816725
141421356237309505 → 10000000000000000087657358700327265
1414213562373095049 → 1000000000000000000987561276980703725
14142135623730950488 → 100000000000000000006832246143819194316
14142135623730950488... = √2 (without the decimal point)

When I first found these patterns, I thought I might have discovered something mathematically profound. I hadn’t. Troots are trivial. I think troots are beautiful too, but a little thought soon showed me how easily and obviously they arise. Remember that the formula for tri(n), the n-th triangular number, is tri(n) = (n^2+n)/2. As you can see above, when tri(n) is increasing in length by one digit, it rises above the next power of 10, which always begins with 1 followed by only 0s. Therefore n^2+n will begin with the digit 2 followed by some 0s, which then becomes 1 followed by some 0s as (n^2+n) is divided by 2. So n for tri(n) increasing-by-one-digit will be the first integer, n, where n^2+n yields a number with 2 as the leading digit followed by more and more 0s.

And that’s why n approximates the digits of √2·0000… and √20·0000…, for tri(n) with an odd and even number of digits, respectively. Similar trootful patterns exist in other bases and for other polygonal numbers, like the square numbers, the pentagonal numbers and so on. The troots are beautiful to see but trivial to explain. All the same, there is a sense in which you can say the mindless sequence of triangular numbers is “calculating” the digits of √2 and √20. It even rounds up the final digits when necessary:

1414214 → 1000001326005
14142136 → 100000012392316
141421356 → 10000000037150046
141421356... = √2
[...]
14142135624 → 100000000010876002500
141421356237 → 10000000000026940078203
141421356237... = √2
[...]
14142135623731 → 100000000000007771272992046
141421356237310 → 10000000000000140731196136705
1414213562373095 → 1000000000000000638090771126060
1414213562373095... = √2
[...]
1414213562373095049 → 1000000000000000000987561276980703725
14142135623730950488 → 100000000000000000006832246143819194316
14142135623730950488... = √2