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Little explorations with HP calculators (no Prime)
04-10-2017, 09:30 AM (This post was last modified: 04-10-2017 09:48 AM by Dieter.)
Post: #173
RE: Little explorations with the HP calculators
(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) = 3z · f(1) + 21 · f(z) = 3z · 1 + 2 · f(z)
f(z+1) = 31 · f(z) + 2z · f(1) = 3 · f(z) + 2z · 1

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

So f(1,5) = √27 – √8 = 2,3677...

Dieter
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RE: Little explorations with the HP calculators - Dieter - 04-10-2017 09:30 AM



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