Short & Sweet Math Challenge #21: Powers that be

[J-F Garnier]

(11-07-2016 10:23 PM)Valentin Albillo Wrote:  How many did you identify ? How do you know they aren't all ?

I didn't attempt to count them, but I know that I didn't identify all constants with Emu71, because I missed at least one: the 1.3802775691 value that Gerson reported.

When trying to check if the powers of this value get closer and closer to an integer, I got:
I 1.3802775691^I
51 13749532.9553
52 18978171.9238
53 26195145.0089
54 36156571.0751
55 49906104.0305
56 68884275.9545
57 95079420.9637
58 131235992.039
59 181142096.07
60 250026372.026
61 345105792.99
62 476341785.031
63 657483881.103
64 907510253.132
65 1252616046.13
66 1728957831.16
67 2386441712.27
68 3293951965.42
69 4546568011.56
70 6275525842.74
...
The closest values are around power 60, then the lack of accuracy makes the fractional part no more significant.
My criteria was that the value must be close to an integer with 0.01 tolerance for 3 successive power values, to have a good confidence.
I could have relax my criteria, but what if other values had an even worst behaviour?

With Free42, I had the right accuracy, with 35 digits.
I first solved the equation x^4-x^3-1=0 with the solver to have the root X with full accuracy, then the powers (frac part) are:
FP(X^75) = 0.98147851
FP(X^76) = 0.98907003
FP(X^77) = 0.01158426
FP(X^78) = 0.01475045
FP(X^79) = 0.99622896
FP(X^80) = 0.98529899
FP(X^81) = 0.99688326
FP(X^82) = 0.01163371
FP(X^83) = 0.00786267
FP(X^84) = 0.99316166
FP(X^85) = 0.99004492
FP(X^86) = 0.00167862
FP(X^87) = 0.00954129
FP(X^88) = 0.00270295
FP(X^89) = 0.99274787
FP(X^90) = 0.99442649
FP(X^91) = 0.00396778
FP(X^92) = 0.00667073
FP(X^93) = 0.99941859
FP(X^94) = 0.99384508
FP(X^95) = 0.99781286
Here my criteria is met around power 85.
But since there is no polynomial root solver on the HP-42S (and I didn't want to write one), I had to use the equation solver, and I wasn't sure that the reported root was the correct, largest one.

Regarding the participation to the challenge, maybe others met the same difficulties.
Unless there is another, better way to solve the challenge :-)

J-F