2232.11 – From a Progression to a Quadratic

Consider the following list of integers:

$P_0 = 41, P_1 = 43, P_2 = 47, P_3 = 53, \ldots,$

in which $P_n$ is obtained by adding $2n$ to $P_{n-1}$ . It so happens that there is a quadratic function $F(x)$ with the property that $P_n = F(n)$ for all non-negative $n$ . Find $F$ .

Once you have $F$ , consider this question: Is $F(n)$ a prime number for every non-negative value of $n$ ?

Solution

The trick is to rewrite each integer in terms of the first value (41) and look for a pattern that resembles a quadratic function:

\begin{aligned} P_0 &=& 41 &=& 41 + 0 &=& 41 + 0 \\ P_1 &=& 43 &=& 41 + 2 &=& 41 + 1^2 + 1 \\ P_2 &=& 47 &=& 41 + 6 &=& 41 + 2^2 + 2 \\ P_3 &=& 53 &=& 41 + 12 &=& 41 + 3^2 + 3 \\ P_4 &=& 61 &=& 41 + 20 &=& 41 + 4^2 + 4 \\ &&& \ldots &&& \end{aligned}

Thus,

$P_n = 41 + n^2 + n = n^2 + n + 41.$

Although the instances we're seeing are all prime, when we reach $n = 41$ ,

$P_{41} = 41^2 + 41 + 41 = 41(41 + 2) = 41 \cdot 43,$

the result is not prime.

This sequence is famous because for a sequence with a simple formula it generates primes for a long time. It was Euler (in 1772) who first noticed it.