Little explorations with HP calculators (no Prime)

[Dieter]

(04-10-2017 08:00 AM)pier4r Wrote:  new day, new problem (trivial for all of you but me).
...
Now I did solve this iteratively (and I checked the f(3) with the two combinations that I could build, the result is a "small" prime number). Is there a way to solve it in a more general way, without computing every single step?

Sure. And it's very simple indeed:

f(1+3) = 3³ · f(1) + 2¹ · f(3) = 27 · 1 + 2 · f(3)
f(3+1) = 3¹ · f(3) + 2³ · f(1) = 3 · f(3) + 8 · 1

Since f(1+3) = f(3+1):

27 + 2 · f(3) = 3 · f(3) + 8
f(3) = 19

That's all.

(04-10-2017 08:00 AM)pier4r Wrote:  Also what happens if I ask for f(1.5) ? I do not have ways to compute this, like information is missing.

Why not generalize this for any f(z):

f(1+z) = 3^z · f(1) + 2¹ · f(z) = 3^z · 1 + 2 · f(z)
f(z+1) = 3¹ · f(z) + 2^z · f(1) = 3 · f(z) + 2^z · 1

3^z + 2 · f(z) = 3 · f(z) + 2^z
f(z) = 3^z – 2^z

So f(1,5) = √27 – √8 = 2,3677...

Dieter