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Given the roots of $x^3=x^2+1$, we have sequence A001609,

$M(n) = x_1^n+x_2^n+x_3^n = \,_3F_2\left(\frac{-n}{3}, \frac{1-n}{3}, \frac{2-n}{3};\; \frac{1-n}{2}, \frac{2-n}{2};\; -\frac{3^3}{2^2}\right) = 1, 1, 4, 5, 6, 10, 15, 21,\dots$

for $n = {1,2,3,\dots}$

Question: Given $y^3=y+1$, is there any similar generalized hypergeometric formula for the Perrin numbers?

$P(n) = y_1^n+y_2^n+y_3^n = 0,2,3,2,5,5,7,10,\dots$

The closest I found is the binomial sum,

$ \begin{aligned}P(n) &= n\sum_{k=1}^{n/2} \frac{\binom k{n-2k}}{k} = 0,2,3,2,5,5,7,10,\dots\end{aligned}$

where both start with $n = 1,2,3,\dots$ Anyone knows how to translate that into the generalized hypergeometric function?

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The Wikipedia article ( notes that P(n) can be computed in log(n) multiplies. – marty cohen Sep 16 at 18:59

1 Answer 1

A solution for calculating P(n) when n is prime using the hypergeometric function can be found at (Chapter 15)

hypergeometric function and Perrin(n)

This equation is derived from an incomplete beta function giving the nth term of the Perrin sequence when n is prime.

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While this link may answer the question, it is better to include the essential parts of the answer here and provide the link for reference. Link-only answers can become invalid if the linked page changes. – TravisJ Sep 16 at 19:30
How come google can't find – Tito Piezas III Sep 17 at 2:07
The site is – Richard Turk Sep 17 at 12:25

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