I think we can go a little deeper.
Just my observation. I'm not an expert in coding, I tried to create an alternative brute-force algorithm, but I could not achieve the required speed, but, nevertheless, maybe it will give something, or maybe I wrote nonsense.
We have Pk formula = 2^((n-1)+(x^(n-1))) where (x^(n-1)) is rem.
We know "n" and we can get "x" from previous outputs.
For example:
26x=0.9858782811338544
27x=0.9883972936421218
28x=0.989990412459008
.
.
61x=0.980482112...
62x=0.995538469...
63x=0.999401211...
55 30045390491869460 55.737993190511333 3.0122337265934062 x=0.994389489 4.7925492576909791
56 44218742292676575 56.2955075095680657 1.4717313228011646 x=0.978079087 4.8507162216308491 +0.05816696393987
57 138245758910846492 57.9400128353741358 3.1264064001599271 x=0.998895936 4.8350969232074259 -0.0156192984234232
58 199976667976342049 58.4726092982930311 1.4465302194572587 x=0.986937187 4.883935984044099 +0.0488390608366731
59 525070384258266191 59.8652884381768158 2.6256582308911352 x=0.997508413 4.8876200153560635 +0.0036840313119645
60 1135041350219496382 60.9774505646692825 2.1616937162108232 x=0.999613506 4.907663434230911 +0.0200434188748475
61 1425787542618654982 61.3064647289927286 1.2561547139608029 x=0.980482112 4.9693921656109065 +0.0617287313799955
62 3908372542507822062 62.7612736982093203 2.7412026165760701 x=0.995538469 4.9630994671280154 -0.0062926984828911
63 8993229949524469768 63.96354506567706 2.3010165616796422 x=0.999401211 4.9784771429516494 +0.015377675823634
First of all, I want to draw your attention, I think that it was the "x" that was randomly generated, and here we need to look for an ace.
Even if not, then iterating over the "x" value itself, raising it into a formula, and then performing AES operations saves a lot of resources, because changing "x" by 1 part changes Decimal to millions.
I hope if this helps to find something you will not forget about me.
17HHd3M5ennd12CEUZXQ75W7gC6UgTkoUz
