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It is well known that Euler's lucky numbers are related to the Heegner numbers, where

\begin{align} &n^2+n+p\\ \end{align}

gives primes for $n=0,\dots,p-2$ if and only if its discriminant $1-4p$ equals minus a Heegner number. This is then true for $p=2, 3, 5, 11, 17, 41.$

There seems to be another group of numbers, namely $p=2, 3, 5, 7, 13,$ that gives primes for

\begin{align} &n^2+n-p^2\\ \end{align}

for $n=1,\dots,2p-2.$

eg, $n^2+n-13^2$ for $1\leq n\leq 24$ produces

$$-167, -163, -157, -149, -139, -127, -113, -97, -79, -59, -37, \ -13,$$$$ 13, 41, 71, 103, 137, 173, 211, 251, 293, 337, 383, 431.$$

The class numbers for the discriminants are not all the same however, and I could find no reference for this finite group, though there are many sequences that begin with these primes (Mersenne exponents, Pierpont primes, etc.).

It could, of course be coincidental, simply an example of the law of small numbers. If not though, what links these numbers if not the class numbers of the discriminants?

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    $\begingroup$ There is a result saying that if $p > 0$, $n^2+n+p$ is prime for $0 \le n \le p-2$ if and only if it is prime for $0 \le n \le O(\sqrt p)$ (I don't remember the exact bound). So if a similar result holds for negative $p$, this could be a coincidence. $\endgroup$
    – mercio
    Jun 2, 2015 at 17:35
  • $\begingroup$ @mercio but in this case, $p$ is essentially a square, not a prime (so it doesn't hold for $n=0$ of course). $\endgroup$
    – martin
    Jun 2, 2015 at 23:28
  • $\begingroup$ However do note that $\mathcal O_{\mathbb Q(\sqrt{13})}$, $\mathbb Z[\sqrt{41}]$, $\mathbb Z[\sqrt{71}]$, ..., $\mathbb Z[\sqrt{431}]$ are all UFDs. The same is also true of $\mathbb Z[\sqrt{167}]$ (as opposed to $\mathcal O_{\mathbb Q(\sqrt{-167})}$), $\mathbb Z[\sqrt{163}]$, etc. $\endgroup$ Aug 15, 2018 at 4:34
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    $\begingroup$ This might be unrelated, but the five totally definite rational quaternion algebras of class number 1 have discriminants $2,3,5,7,13$, so one could say these are the "quaternionic Heegner numbers" (see for example here). $\endgroup$
    – pregunton
    Jul 5, 2019 at 17:18

1 Answer 1

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A Simple Answer

If not though, what links these numbers if not the class numbers of the discriminants?

The Occam's razor answer would be divisors of 12, then add 1. This would lead to ${2, 3, 4, 5, 7, 13}$, where notably 4 is not included in your set.

In the case of 4 a slightly different equation is used:

$$ \frac{n^2 + n - q^2}{2} $$

e.g.: $(n^2 + n - 4^2)/2$ for $1≤n≤6$ produces

$$ {-7, -5, -2, 2, 7, 13} $$

A More Complicated Answer

I feel like an unspoken aspect of your question was the observation that $p^2 - 6$ with prime $p$ in ${2, 3, 5, 7, 13}$ results in:

$$ -2, 3, 19, 43, 163 $$

which are all, essentially, Heegner numbers (ignoring the sign).

The following is an observation that may be helpful:

If you find all solutions to $12/(x-1) = y$ where $y$ is in ${12,11,10,...,3,2,1}$ you get:

$$ {2, \frac{23}{11}, \frac{11}{5}, \frac{7}{3}, \frac{5}{2}, \frac{19}{7}, 3, \frac{17}{5}, 4, 5, 7, 13} $$

which is ${2, 3, 5, 7, 13}$ with an added even number and fractional values. When you plug these same numbers into $p^2 - 6$ you get:

$$ {-2, -\frac{197}{121}, -\frac{29}{25}, -\frac{5}{9}, \frac{1}{4}, \frac{67}{49}, 3, \frac{139}{25}, 10, 19, 43, 163} $$

Notably, the prime factorization includes every Heegner number at least once. Also, four new terms appear: ${5, 29, 139, 197}$.

This is circumstantial evidence that Heegner numbers may be related.

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  • $\begingroup$ Bonus (very speculative): Table[Numerator[2*3^n + 1]], n=-4 to 4 results in {83, 29, 11, 5, 3, 7, 19, 55, 163} where 55=(43+67)/2 and 2*83 is roughly the midpoint of 139 and 197. If you interpret 55 as both 43 and 67, the only excluded Heegner numbers are 1 and 2, for the half-integers. Additionally, pi function(5) = 3, pi function(29) = 10, NumberFieldClassNumber[sqrt(-139)] = 3, NumberFieldClassNumber[sqrt(-197)] = 10. $\endgroup$ Nov 14, 2021 at 19:58
  • $\begingroup$ Largely unrelated, but possibly of interest to some is A276260. If you do the same trick where you interpret some terms as a midpoint, then the equation Table[Numerator[6^n + 61 + 10*(-1)^n]], n=1 to 4 can be interpreted as 53, 61, 107, 251, 283, 1367 in some sense. $\endgroup$ Nov 14, 2021 at 20:08

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