Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
The Wikipedia article ( notes that P(n) can be computed in log(n) multiplies. – marty cohen Sep 16 '15 at 18:59

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.

share|cite|improve this answer
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 '15 at 19:30
How come google can't find – Tito Piezas III Sep 17 '15 at 2:07
The site is – Richard Turk Sep 17 '15 at 12:25

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.