Total Score

The number 23 is always (and trivially) equal to some running total of the digits of its roots in base 2. In other bases, that’s not always true (n.b. numbers inside square brackets represent single digits in that base):

√23 = 23^(1/2) = 100.1100101110111011100111010101110111000001000... in base 2
23 = digsum(100.110010111011101110011101010111011)
23^(1/2) = 11.21011101110011111122022101121121... in base 3
23 = digsum(11.2101110111001111112202)
23^(1/2) = 4.8832850[10]89028... in base 11
23 = digsum(4.883)
23^(1/2) = 4.[14]5[15]53[14]0[12]0[14]5[13]... in base 18
23 = digsum(4.[14]5)
23^(1/2) = 4.[19]29[13][19]4[11][23][19][11][20]... in base 24
23 = digsum(4.[19])
23^(1/2) = 4.[19][22]9[21][17]5[12][10]456... in base 25
23 = digsum(4.[19])

23^(1/3) = 10.11011000000001111010101010011000101000110000001100000010010000101011... in base 2
23 = digsum(10.1101100000000111101010101001100010100011000000110000001001)
23^(1/3) = 2.21121001121111121022212100220... in base 3
23 = digsum(2.2112100112111112102)
23^(1/3) = 2.312000132222212022030003... in base 4
23 = digsum(2.31200013222221)
23^(1/3) = 2.6600365246121403... in base 8
23 = digsum(2.660036)
23^(1/3) = 2.753154453877080... in base 9
23 = digsum(2.75315)
23^(1/3) = 2.93120691571[10]001[10]... in base 11
23 = digsum(2.931206)
23^(1/3) = 2.[12]9[13]0[11]74[11]61[14]2... in base 15
23 = digsum(2.[12]9)
23^(1/3) = 2.[13]807[10][10]98[10]303... in base 16
23 = digsum(2.[13]8)
23^(1/3) = 2.[21]2[10][10][13][11][21][23][15][24][21]... in base 25
23 = digsum(2.[21])
23^(1/3) = 2.[21][24][11][20][24][22][23][25]0[11][11]... in base 26
23 = digsum(2.[21])

23^(1/4) = 10.0011000010011111110100101010011000001001011110001110101... in base 2
23 = digsum(10.001100001001111111010010101001100000100101111)
23^(1/4) = 2.1411772251404570... in base 8
23 = digsum(2.141177)
23^(1/4) = 2.1634161832077814... in base 9
23 = digsum(2.163416)
23^(1/4) = 2.33[15]2[14][13]967[10]6[12]5... in base 17
23 = digsum(2.33[15])
23^(1/4) = 2.6[15][19][11][31][17][10][18][21]30[27]... in base 34
23 = digsum(2.6[15])
23^(1/4) = 2.[12]9[63][18][41][32][37][56][58][60]1[17]... in base 64
23 = digsum(2.[12]9)
23^(1/4) = 2.[21]9[26]6[54][21][20]3[64][86][110]... in base 111
23 = digsum(2.[21])
23^(1/4) = 2.[21][30][66][22][73][19]3[15][51][24]8... in base 112
23 = digsum(2.[21])
23^(1/4) = 2.[21][52][36][111][32][104][66][40][95][33]5... in base 113
23 = digsum(2.[21])
23^(1/4) = 2.[21][74][50][62][27]19[100][70][48][89]... in base 114
23 = digsum(2.[21])
23^(1/4) = 2.[21][96][108]2[101][62][43][18][71][113][37]... in base 115
23 = digsum(2.[21])

23^(1/5) = 1.110111110100011010011101000111111011111011000... in base 2
23 = digsum(1.11011111010001101001110100011111101)
23^(1/5) = 1.313310122131013323323010... in base 4
23 = digsum(1.31331012213101)
23^(1/5) = 1.[10]5714140[10][11][11]61... in base 12
23 = digsum(1.[10]57)
23^(1/5) = 1.[11]45210[12]3974[12]0[11]... in base 13
23 = digsum(1.[11]452)
23^(1/5) = 1.[22][17][15]788[12][20][10][16]5... in base 26
23 = digsum(1.[22])

And in base 10:

23^(1/7) = 1.565065607960239...
23 = digsum(1.56506)

23^(1/11) = 1.32982177397055...
23 = digsum(1.3298)

23^(1/25) = 1.133624213096260543...
23 = digsum(1.13362421)

23^(1/43) = 1.075642836327515...
23 = digsum(1.07564)

23^(1/51) = 1.0634095245502272...
23 = digsum(1.063409)

23^(1/59) = 1.054581462032154...
23 = digsum(1.05458)

23^(1/74) = 1.043282031364111825...
23 = digsum(1.04328203)

23^(1/78) = 1.041017545329593513...
23 = digsum(1.04101754)

23^(1/81) = 1.039468791371841...
23 = digsum(1.03946)

23^(1/85) = 1.037576979258809...
23 = digsum(1.03757)

23^(1/86) = 1.0371320245405187874...
23 = digsum(1.037132024)

23^(1/101) = 1.031531403111493041428...
23 = digsum(1.03153140311)

Self-Raising Power

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:
autoroot