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 r

r= 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 r

r= 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: