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 vasyannyasha 2019-06-15 11:45

Searching many formulas to one limit

Hi guys!
This is formula for nth r-gonal pyramidal number
[TEX]P_n^r= \frac{3n^2 + n^3(r-2) - n(r-5)}{6}[/TEX]
I have , for example 2<r<1000.
And i want to search all this numbers to one constant border(for example 10[SUP]15[/SUP])
How i can find such n's in C++?
Pre-thanks

 vasyannyasha 2019-06-16 08:43

And i forget about searching from some minimal border.
How to find n[SUB]min[/SUB] without computing previous [TEX]P_n^r[/TEX]?
Pre-thx

 lavalamp 2019-06-24 15:12

What exactly are you searching for?

 vasyannyasha 2019-06-25 06:34

Im searching for numbers that n-gonal pyramidal number and n-angular number. Cannonball problem for different bases

 Dr Sardonicus 2019-06-25 16:37

[QUOTE=vasyannyasha;520023]Im searching for numbers that n-gonal pyramidal number and n-angular number. Cannonball problem for different bases[/QUOTE]

For a given r, an r-gonal number is of the form

$$p_r^{n} \;=\; n((r-2)*n-(r-4))/2$$

Multiplying by 1/2*(r - 2) and adding 1/16*r^2 - 1/2*r + 1 gives a square y^2.

(Note that when r = 4 you multiply by 1 and add 0).

Thus, for a given r you can write

$$(r - 2)P_r^{n}/2 \; + \; \frac{r^{2}}{16}\;-\;\frac{r}{2}\;+\;1\;=\;y^{2}$$

where the P is the n[sup]th[/sup] r-pyramidal number, which is cubic in n. For any given r, this is an elliptic curve. This gives a mighty bludgeon to use on the problem.

For r = 4, there are elementary proofs that n = 70 is the only n > 1 giving a square value for the cannonball problem.

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