Here is a beautiful and astonishingly simple formula for π created by the French mathematician François Viète (1540-1603):
• 2 / π = √2/2 * √(2 + √2)/2 * √(2 + √(2 + √2))/2…
I can remember testing the formula on a scientific calculator that allowed simple programming. As I pressed the = key and the results began to home in on π, I felt as though I was watching a tall and elegant temple emerge through swirling mist.
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
Here’s a famous paradox (or a variant of it at least):
• There are two errers in this sentence.
The only visible error is the misspelt “errers”. But if the sentence claims to have two errors while having only one, that is another error and there are two errors after all.
Now for another variant. I’m not sure if I’ve thought this up for myself, but try this sentence:
• There are three errors in this sentence.
There are no visible errors in the sentence. Therefore it has one error: the claim that it has three errors when there is in fact no error. But if it has one error, it’s in error to claim that it has three errors. Therefore the sentence has two errors. And if it has two errors, again it’s in error, because it claims to have three errors while having only two. Therefore it has three errors after all.
The same reasoning can be applied to any integral number of errors:
• There are five errors in this sentence.
• There are 719 errors in this sentence.
• There are 1,000,000 errors in this sentence.
• There are 1,000,000,000 errors in this sentence.
No matter how large the number of errors, the sentence becomes true instantly, because each time the sentence makes a false claim, it makes another error. But those “times of error” don’t take place in time, any more than this equation does:
• 2 = 1 + 1/2 + 1/4 + 1/8 + 1/16…
So I think these sentences are instantly true:
• There are infinitely many errors in this sentence.
• There are ∞ errors in this sentence.
But there are infinitely many infinities. Ordinary infinity, the infinity of 1,2,3…, is called ℵ0 or aleph-zero. It’s a countable infinity. Above that comes ℵ1, an uncountable infinity. So does this sentence instantly become true?
• There are ℵ1 errors in this sentence.
I’m not sure. But I think I can argue for the validity of sentences claiming fractional or irrational number of errors:
• There is 1.5 errors in this sentence.
• There are π errors in this sentence.
Let’s have a look at “There is 1.5 errors in this sentence”. There are no visible errors, so there’s one error: the claim that sentence contains 1.5 errors. So now there seems to be another error: the sentence has one error but claims to have 1.5 errors. But does it therefore have two errors? No, because if it has two errors, it’s still in error and has three errors. And that generates another error and another and another, and so on for ever. The sentence becomes unstoppably and infinitely false.
So let’s go back to the point at which the sentence contains one error. Now, the difference between 1 error and 1.5 errors is small — less than a full error. So how big is the error of claiming to have 1.5 errors when having 1 error? Well, it’s obviously 0.5 of an error. So the sentence contains 1.5 errors after all.
Now for “There are π errors in this sentence”. There are no visible errors, so there’s one error: the claim that the sentence contains π errors. Therefore it contains one error. But it claims to have π errors, so it has another error. And if it has 2 errors and claims to have π errors, it has another and third error. But if it has three errors and claims to have π error, it’s still in error. But only slightly — it’s now committing a small amount of an error. How much? It can only be 0.14159265… of an error. Therefore it’s committing 3.14159265… = π errors and is a true sentence.
Now try:
• There is -1 error in this sentence.
What is a negative error? A truth. So I think that sentence is valid too. But I can’t think of how to use i, or the square root of -1, in a sentence like that.
How radians work (from Wikipedia)
• Racine carrée de 2, c’est 1,414 et des poussières… Et quelles poussières ! Des grains de sable qui empêchent d’écrire racine de 2 comme une fraction. Autrement dit, cette racine n’est pas dans Q. — Rationnel mon Q: 65 exercices de styles, Ludmilla Duchêne et Agnès Leblanc (2010)
• The square root of 2 is 1·414 and dust… And what dust! Grains of sand that stop you writing the root of 2 as a fraction. Put another way, this root isn’t in Q [the set of rational numbers].
A square contains one of the great — perhaps the greatest — intellectual rites of passage. If each side of the square is 1 unit in length, how long are its diagonals? By Pythagoras’ theorem:
a^2 + b^2 = c^2
1^2 + 1^2 = 2, so c = √2
So each diagonal is √2 units long. But what is √2? It’s a new kind of number: an irrational number. That doesn’t mean that it’s illogical or against reason, but that it isn’t exactly equal to any ratio of integers like 3/2 or 17/12. When represented as decimals, the digits of all integer ratios either end or fall, sooner or later, into an endlessly repeating pattern:
3/2 = 1.5
17/12 = 1.416,666,666,666,666…
577/408 = 1.414,2156 8627 4509 8039,2156 8627 4509 8039,2156 8627 4509 8039,2156 8627 4509 8039,2156 8627 4509 8039,…
But when √2 is represented as a decimal, its digits go on for ever without any such pattern:
√2 = 1.414,213,562,373,095,048,801,688,724,209,698,078,569,671,875,376,948,073,176,679,737,990,732,478,462,107…
The intellectual rite of passage comes when you understand why √2 is irrational and behaves like that:
Proof of the irrationality of √2
1. Suppose that there is some ratio, a/b, such that
2. a and b have no factors in common and
3. a^2/b^2 = 2.
4. It follows that a^2 = 2b^2.
5. Therefore a is even and there is some number, c, such that 2c = a.
6. Substituting c in #4, we derive (2c)^2 = 4c^2 = 2b^2.
7. Therefore 2c^2 = b^2 and b is also even.
8. But #7 contradicts #2 and the supposition that a and b have no factors in common.
9. Therefore, by reductio ad absurdum, there is no ratio, a/b, such that a^2/b^2 = 2. Q.E.D.
Given that subtle proof, you might think the digits of an irrational number like √2 would be difficult to calculate. In fact, they’re easy. And one method is so easy that it’s often re-discovered by recreational mathematicians. Suppose that a is an estimate for √2 but it’s too high. Clearly, if 2/a = b, then b will be too low. To get a better estimate, you simply split the difference: a = (a + b) / 2. Then do it again and again:
a = (2/a + a) / 2
If you first set a = 1, the estimates improve like this:
(2/1 + 1) / 2 = 3/2
2 – (3/2)^2 = -0.25
(2/(3/2) + 3/2) / 2 = 17/12
2 – (17/12)^2 = -0.00694…
(2/(17/12) + 17/12) / 2 = 577/408
2 – (577/408)^2 = -0.000006007…
(2/(577/408) + 577/408) / 2 = 665857/470832
2 – (665857/470832)^2 = -0.00000000000451…
In fact, the estimate doubles in accuracy (or better) at each stage (the first digit to differ is underlined):
1.5… = 3/2 (matching digits = 1)
1.4… = √21.416… = 17/12 (m=3)
1.414… = √21.414,215… = 577/408 (m=6)
1.414,213… = √21.414,213,562,374… = 665857/470832 (m=12)
1.414,213,562,373… = √21.414,213,562,373,095,048,801,689… = 886731088897/627013566048 (m=24)
1.414,213,562,373,095,048,801,688… = √21.414,213,562,373,095,048,801,688,724,209,698,078,569,671,875,377… (m=48)
1.414,213,562,373,095,048,801,688,724,209,698,078,569,671,875,376… = √21.414,213,562,373,095,048,801,688,724,209,698,078,569,671,875,376,948,073,176,679,737,990,732,478,46
2,107,038,850,387,534,327,641,6… (m=97)
1.414,213,562,373,095,048,801,688,724,209,698,078,569,671,875,376,948,073,176,679,737,990,732,478,46
2,107,038,850,387,534,327,641,5… = √21.414,213,562,373,095,048,801,688,724,209,698,078,569,671,875,376,948,073,176,679,737,990,732,478,46
2,107,038,850,387,534,327,641,572,735,013,846,230,912,297,024,924,836,055,850,737,212,644,121,497,09
9,935,831,413,222,665,927,505,592,755,799,950,501,152,782,060,8… (m=196)
1.414,213,562,373,095,048,801,688,724,209,698,078,569,671,875,376,948,073,176,679,737,990,732,478,46
2,107,038,850,387,534,327,641,572,735,013,846,230,912,297,024,924,836,055,850,737,212,644,121,497,09
9,935,831,413,222,665,927,505,592,755,799,950,501,152,782,060,5… = √21.414,213,562,373,095,048,801,688,724,209,698,078,569,671,875,376,948,073,176,679,737,990,732,478,46
2,107,038,850,387,534,327,641,572,735,013,846,230,912,297,024,924,836,055,850,737,212,644,121,497,09
9,935,831,413,222,665,927,505,592,755,799,950,501,152,782,060,571,470,109,559,971,605,970,274,534,59
6,862,014,728,517,418,640,889,198,609,552,329,230,484,308,714,321,450,839,762,603,627,995,251,407,98
9,687,253,396,546,331,808,829,640,620,615,258,352,395,054,745,750,287,759,961,729,835,575,220,337,53
1,857,011,354,374,603,43… (m=392)
1.414,213,562,373,095,048,801,688,724,209,698,078,569,671,875,376,948,073,176,679,737,990,732,478,46
2,107,038,850,387,534,327,641,572,735,013,846,230,912,297,024,924,836,055,850,737,212,644,121,497,09
9,935,831,413,222,665,927,505,592,755,799,950,501,152,782,060,571,470,109,559,971,605,970,274,534,59
6,862,014,728,517,418,640,889,198,609,552,329,230,484,308,714,321,450,839,762,603,627,995,251,407,98
9,687,253,396,546,331,808,829,640,620,615,258,352,395,054,745,750,287,759,961,729,835,575,220,337,53
1,857,011,354,374,603,40… = √21.414,213,562,373,095,048,801,688,724,209,698,078,569,671,875,376,948,073,176,679,737,990,732,478,46
2,107,038,850,387,534,327,641,572,735,013,846,230,912,297,024,924,836,055,850,737,212,644,121,497,09
9,935,831,413,222,665,927,505,592,755,799,950,501,152,782,060,571,470,109,559,971,605,970,274,534,59
6,862,014,728,517,418,640,889,198,609,552,329,230,484,308,714,321,450,839,762,603,627,995,251,407,98
9,687,253,396,546,331,808,829,640,620,615,258,352,395,054,745,750,287,759,961,729,835,575,220,337,53
1,857,011,354,374,603,408,498,847,160,386,899,970,699,004,815,030,544,027,790,316,454,247,823,068,49
2,936,918,621,580,578,463,111,596,668,713,013,015,618,568,987,237,235,288,509,264,861,249,497,715,42
1,833,420,428,568,606,014,682,472,077,143,585,487,415,565,706,967,765,372,022,648,544,701,585,880,16
2,075,847,492,265,722,600,208,558,446,652,145,839,889,394,437,092,659,180,031,138,824,646,815,708,26
3,010,059,485,870,400,318,648,034,219,489,727,829,064,104,507,263,688,131,373,985,525,611,732,204,02
4,509,122,770,022,694,112,757,362,728,049,574… (m=783)
1.414,213,562,373,095,048,801,688,724,209,698,078,569,671,875,376,948,073,176,679,737,990,732,478,46
2,107,038,850,387,534,327,641,572,735,013,846,230,912,297,024,924,836,055,850,737,212,644,121,497,09
9,935,831,413,222,665,927,505,592,755,799,950,501,152,782,060,571,470,109,559,971,605,970,274,534,59
6,862,014,728,517,418,640,889,198,609,552,329,230,484,308,714,321,450,839,762,603,627,995,251,407,98
9,687,253,396,546,331,808,829,640,620,615,258,352,395,054,745,750,287,759,961,729,835,575,220,337,53
1,857,011,354,374,603,408,498,847,160,386,899,970,699,004,815,030,544,027,790,316,454,247,823,068,49
2,936,918,621,580,578,463,111,596,668,713,013,015,618,568,987,237,235,288,509,264,861,249,497,715,42
1,833,420,428,568,606,014,682,472,077,143,585,487,415,565,706,967,765,372,022,648,544,701,585,880,16
2,075,847,492,265,722,600,208,558,446,652,145,839,889,394,437,092,659,180,031,138,824,646,815,708,26
3,010,059,485,870,400,318,648,034,219,489,727,829,064,104,507,263,688,131,373,985,525,611,732,204,02
4,509,122,770,022,694,112,757,362,728,049,573… = √2
Here’s √2 in base 2:
√2 = 1.01101010000010011110... (base=2)
And in base 3:
√2 = 1.10201122122200121221... (base=3)
And in bases 4, 5, 6, 7, 8, 9 and 10:
√2 = 1.12220021321212133303... (b=4)
√2 = 1.20134202041300003420... (b=5)
√2 = 1.22524531420552332143... (b=6)
√2 = 1.26203454521123261061... (b=7)
√2 = 1.32404746317716746220... (b=8)
√2 = 1.36485805578615303608... (b=9)
√2 = 1.41421356237309504880... (b=10)
And here’s π in the same bases:
π = 11.00100100001111110110... (b=2)
π = 10.01021101222201021100... (b=3)
π = 03.02100333122220202011... (b=4)
π = 03.03232214303343241124... (b=5)
π = 03.05033005141512410523... (b=6)
π = 03.06636514320361341102... (b=7)
π = 03.11037552421026430215... (b=8)
π = 03.12418812407442788645... (b=9)
π = 03.14159265358979323846... (b=10)
Mathematicians know that in all standard bases, the digits of √2 and π go on for ever, without falling into any regular pattern. These numbers aren’t merely irrational but transcedental. But are they also normal? That is, in each base b, do the digits 0 to [b-1] occur with the same frequency 1/b? (In general, a sequence of length l will occur in a normal number with frequency 1/(b^l).) In base 2, are there as many 1s as 0s in the digits of √2 and π? In base 3, are there as many 2s as 1s and 0s? And so on.
It’s a simple question, but so far it’s proved impossible to answer. Another question starts very simple but quickly gets very difficult. Here are the answers so far at the Online Encyclopedia of Integer Sequences (OEIS):
2, 572, 8410815, 59609420837337474 – A049364
The sequence is defined as the “Smallest number that is digitally balanced in all bases 2, 3, … n”. In base 2, the number 2 is 10, which has one 1 and one 0. In bases 2 and 3, 572 = 1000111100 and 210012, respectively. 1000111100 has five 1s and five 0s; 210012 has two 2s, two 1s and two 0s. Here are the numbers of A049364 in the necessary bases:
10 (n=2)
1000111100, 210012 (n=572)
100000000101011010111111, 120211022110200, 200011122333 (n=8410815)
11010011110001100111001111010010010001101011100110000010, 101201112000102222102011202221201100, 3103301213033102101223212002, 1000001111222333324244344 (n=59609420837337474)
But what number, a(6), satisfies the definition for bases 2, 3, 4, 5 and 6? According to the notes at the OEIS, a(6) > 5^434. That means finding a(6) is way beyond the power of present-day computers. But I assume a quantum computer could crack it. And maybe someone will come up with a short-cut or even an algorithm that supplies a(b) for any base b. Either way, I think we’ll get there, π and by.
Squares are often thought to be the most boring of all shapes. Yet every square holds a stunning secret – something that in legend prompted a mathematical cult to murder a traitor. If each side of a square is one unit long, how long is the square’s diagonal, that is, the line from one corner to the opposite corner?
By Pythagoras’ theorem, the answer is this:
• x^2 = 1^2 + 1^2
• x^2 = 2
• x = √2
But what is √2? Pythagoras and his followers thought that all numbers could be represented as either whole numbers or ratios of whole numbers. To their dismay, so it’s said, they discovered that they were wrong. √2 is an irrational number – it can’t be represented as a ratio. In modern notation, it’s an infinitely decimal that never repeats:
• √2 = 1·414213562373095048801688724209698…
A modern story, unattested in ancient records, says that the irrationality of √2 was a closely guarded secret in the Pythagorean cult. When Hippasus of Metapontum betrayed the secret, he was drowned at sea by enraged fellow cultists. Apocryphal or not, the story shows that squares aren’t so boring after all.
Nor are they boring when they’re caught in the fract. Divide one square into nine smaller copies of itself:
Discard three of the copies like this:
Stage 1
Retain squares 1, 2, 4, 6, 8, 9 (reading left-to-right, bottom-to-top)
Then do the same to each of the sub-squares:
Stage 1
And repeat:
Stage 3
Stage 4
Stage 5
Stage 6
The result is a fractal of endlessly subdividing contingent hexagons:
Animated vesion
Retain squares 1, 2, 4, 6, 8, 9 (reading left-to-right, bottom-to-top)
Here are a few more of the fractals you can create by squaring and paring:
Retain squares 1, 3, 5, 7, 9 (reading left-to-right, bottom-to-top)
Retain squares 2, 4, 5, 6, 8
Retain squares 1, 2, 4, 5, 6, 8, 9
Retain squares 1, 4, 6, 7, 10, 11, 13, 16
Retain squares 1, 3, 6, 7, 8, 9, 10, 11, 14, 16
Retain squares 2, 3, 5, 6, 8, 9, 11, 12, 14, 15
Retain squares 1, 3, 5, 7, 9, 11, 15, 17, 19, 21, 23, 25
Retain squares 1, 3, 7, 8, 11, 12, 14, 15, 18, 19, 23, 25
Retain squares 1, 5, 7, 8, 9, 12, 14, 17, 18, 19, 21, 25
Retain squares 2, 3, 4, 6, 7, 9, 10, 11, 15, 16, 17, 19, 20, 22, 23, 24
Retain squares 1, 2, 5, 6, 7, 9, 13, 17, 19, 20, 21, 24, 25
Previously pre-posted (please peruse):
The square root of 2 is the number that, raised to the power of 2, equals 2. That is, if r^2 = r * r = 2, then r = √2. The cube root of 2 is the number that, raised to the power of 3, equals 2. That is, if r^3 = r * r * r = 2, then r = [3]√2.
But what do you call the number that, raised to the power of itself, equals 2? I suggest “the auto-root of 2”. Here, if r^r = 2, then r = [r]√2. I don’t know a quick way to calculate the auto-root, but you can adapt a well-known algorithm for approximating the square root of a number. The square-root algorithm looks like this:
n = 2
r = 1
for c = 1 to 20
r = (r + n/r) / 2
next c
print rr = 1.414213562…
Note the fourth line of the algorithm: r = (r + n/r) / 2. When r is an over-estimate of √2, then 2/r will be an under-estimate (and vice versa). (r + 2/r) / 2 splits the difference and refines the estimate. Using the lines above as the model, the auto-root algorithm looks like this:
n = 2
r = 1
for c = 1 to 20
r = (r + [r]√n) / 2[*]
next c
print rr = 1.559610469…
*This is equivalent to r = (r + n^(1/r)) / 2
Here are the first 100 digits of [r]√2 = r in base 10:
1, 5, 5, 9, 6, 1, 0, 4, 6, 9, 4, 6, 2, 3, 6, 9, 3, 4, 9, 9, 7, 0, 3, 8, 8, 7, 6, 8, 7, 6, 5, 0, 0, 2, 9, 9, 3, 2, 8, 4, 8, 8, 3, 5, 1, 1, 8, 4, 3, 0, 9, 1, 4, 2, 4, 7, 1, 9, 5, 9, 4, 5, 6, 9, 4, 1, 3, 9, 7, 3, 0, 3, 4, 5, 4, 9, 5, 9, 0, 5, 8, 7, 1, 0, 5, 4, 1, 3, 4, 4, 4, 6, 9, 1, 2, 8, 3, 9, 7, 3…
And here is [r]√n = r for n = 2..20:
autopower(2) = 1.5596104694623693499703887…
autopower(3) = 1.8254550229248300400414692…
autopower(4) = 2
autopower(5) = 2.1293724827601566963803119…
autopower(6) = 2.2318286244090093673920215…
autopower(7) = 2.3164549587856123013255030…
autopower(8) = 2.3884234844993385564187215…
autopower(9) = 2.4509539280155796306228059…
autopower(10) = 2.5061841455887692562929409…
autopower(11) = 2.5556046121008206152514542…
autopower(12) = 2.6002950000539155877172082…
autopower(13) = 2.6410619164843958084118390…
autopower(14) = 2.6785234858912995813011990…
autopower(15) = 2.7131636040042392095764012…
autopower(16) = 2.7453680235674634847098492…
autopower(17) = 2.7754491049442334313328329…
autopower(18) = 2.8036632456580215496843618…
autopower(19) = 2.8302234384970308956026277…
autopower(20) = 2.8553085030012414128332189…
I assume that the auto-root is always an irrational number, except when n is a perfect power of suitable form, i.e. n = p^p for some integer p. For example, autoroot(4) = 2, because 2^2 = 4, autoroot(27) = 3, because 3^3 = 27, and so on.
And here is the graph of autoroot(n) for n = 2..10000:
Yale tablet, YBC 7289, c. 1700 BC.
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