Given the Ulam spiral with center $c = 41$ and the numbers in a clockwise direction, we have,

$$\begin{array}{cccccc} \color{red}{61}&62&63&64&\to\\ 60&\color{red}{47}&48&49&50\\ 59&46&\color{red}{\small{c=\,}41}&42&51\\ 58&45&44&\color{red}{43}&52\\ 57&56&55&54&\color{red}{53}&\downarrow\\ \end{array}$$

The main diagonal is defined by Euler's polynomial $F(n) = n^2+n+41$, and yields distinct primes for 40 consecutive $n = 0\,\text{to}\,39$.

If we let $c = 3527$ as in this old sci.math post, we get,

$$\begin{array}{cccccccc} \color{blue}{3569}&3570&3571&3572&3573&3574&\to\\ 3568&\color{red}{3547}&3548&3549&3550&3551&3552\\ 3567&3546&\color{red}{3533}&3534&3535&3536&3553\\ 3566&3545&3532&\color{red}{\small{c=\,}3527}&3528&3537&3554\\ 3565&3544&3531&3530&\color{red}{3529}&3538&3555\\ 3564&3543&3542&3541&3540&\color{red}{3539}&3556\\ 3563&3562&3561&3560&3559&3558&\color{red}{3557}&\downarrow\\ \end{array}$$

The polynomial is $G(n) = 4n^2-2n+3527$ and is prime for 23 consecutive $n = -2\,\text{to}\,20$. Its square-free discriminant is $d = -14107$ and has class number $h(d) = 11$. This is the 3rd largest (in absolute value) with that $h(d)$. The blue number, $G(-3)=3569$ is not prime.

Question: For $F(n) = n^2+n+p$, the record is held by Euler's polynomial. For the form $G(n) = 4n^2\pm 2n+p$, is there a better one?

P.S. Other polynomials such as $F(n) = 6n^2+6n+31$ are prime for $n=0\,\text{to}\,28$, but are not diagonals in the Ulam spiral.

  • $\begingroup$ The primes are arise from consecutive values of the polynomial, great. What about prime rich polynomials? I recall finding such values, don't have any idea if they set records. In any case these diagonals are evident upon inspection. That the class number h(d) = 11 is interesting! $\endgroup$ – Alan Nov 9 '13 at 0:43
  • $\begingroup$ According to the prime constellation conjecture, there should be arbitrarily large diagonal sequences. This contradicts your "For $F(n)=n^2+n+p$ the record is for $p=41$". Euler's polynomial is special because $F(n)$ is prime for $0 \le n \le \frac{1+\sqrt{(p-1)/3}}2$ if and only if $h(1-4p)=1$ if and only if $F(n)$ is prime for $0 \le n \le p-2$. Sadly, $p=41$ is the largest such prime, so it gives the most spectacular sequence (in terms of its length compared to $p$). Every other sequence fails in $O(\sqrt p)$. $\endgroup$ – mercio Nov 10 '13 at 9:24
  • $\begingroup$ I found an old piece of code that runs grids of values and reports back prime rich polynomials. So...$2753 - 810n - 36n^2 $ has 132 primes between n = -100 and n = 100. and $ 3572 - 2n + 4n^2 $ has 122 primes in the the same interval. $\endgroup$ – Alan Nov 10 '13 at 22:41
  • $\begingroup$ @mercio: The prime constellation does not seem to discuss primes in a quadratic progression. In fact, the k-Tuple Conjecture explicitly states it implies arbitrarily long arithmetic progressions of primes, but is silent on quadratic ones. $\endgroup$ – Tito Piezas III Nov 11 '13 at 2:08
  • $\begingroup$ @mercio: This MO post on prime constellation conjectures has an answer from Green and Tao's "Linear Equations in Primes". $\endgroup$ – Tito Piezas III Nov 11 '13 at 2:16

(Updated.) Knowing more Mathematica coding, I re-visited this old question. Given,

$$P(n) = 4n^2\pm2n+p\tag1$$

I searched the first $million$ primes $p$. The complete table of record $P(n)$ with at least $14$ consecutive $n$ in the range $n = -5 \to 60\,$ yielding primes are in the table below.

$$\begin{array}{|c|c|c|c|} \hline \text{#} & P(n)=an^2+bn+c & d = b^2-4ac & h(d) & Prime\; range\; n &Total\,(T)\\ 1& 4n^2-2n+41 &\color{red}{-163} & 1&-19 \to 20& 40\\ 2& 4n^2-2n+3527 &\color{blue}{-14107} & 11&-2 \to 20& 23\\ 3& 4n^2+2n+21377 &\color{red}{-85507} & 22&47 \to 64& 18\\ 4& 4n^2-2n+9281 &\color{red}{-37123} & 17& 0 \to 16& 17\\ 5& 4n^2-2n+17 &\color{blue}{-67} & 1&-7 \to 8&16\\ 6& 4n^2+2n+41201&\color{red}{-164803} & 32& 52 \to 66& 15\\ 7& 4n^2+2n+12821&-51283& 21& 8 \to 21& 14\\ 8& 4n^2-2n+3461 &\color{red}{-13843} & 10&34 \to 47&14\\ 9& 4n^2-2n+1277 &\color{blue}{-5107} & 7&19 \to 32&14\\ \hline \end{array}$$

If the discriminant $d$ in red, then it the largest $|d|$ of a class number $h(d)$. If it is in blue, then is one of the $3$ largest $|d|$ of that class.

For example, using #3, if you have an Ulam spiral with center $c=21377$, then in its main diagonal there are $18$ primes in a row. It involves $d = -85507$ which is the largest $|d|$ with $h(d) = 22$.

P.S. Considering that the millionth prime is $p(10^6) = 15485863$, it seems odd there are no large $p$ found within the search radius. However, I only searched $n = -5\to 60$, so another choice of range might yield other $P(n)$.

  • $\begingroup$ Related post : On the prime-generating polynomial $m^2+m+234505015943235329417$. $\endgroup$ – Tito Piezas III Sep 15 '15 at 2:30

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