"Couple of observations to speed up iterating through values of $a$ and $b$:\n",
"1. $0^2 + 0a + b = b$ must be prime. Additionally, since $|b| \\leq 1000$, the largest possible value for $b$ is 997.\n",
"2. $1^2 + 1a + b = 1 + a + b$ must be prime. Therefore $a = p - b - 1$ for some prime $p$. Furthermore, since $|a| < 1000$, it follows that $-1000 < p - b - 1 < 1000 \\implies b - 999 < p < b + 1001$, so $p$ must be less than $997 + 1001 = 1998$ in the most extreme case.\n",
"If you iterate through $n=0,1,2,\\ldots$ for the resulting quadratic and the other formula given, $n^2 - 79n + 1601$, you'll see that these polynomials don't actually generate any new prime numbers compared to Euler's formula - they just repeat primes that Euler's formula already gives.\n",
"What's going on? Let $f(n) = n^2 + n + 41$. An interesting property of [prime-generating functions](https://mathworld.wolfram.com/Prime-GeneratingPolynomial.html) is if $p(n)$ generates primes for $0 \\leq n \\leq u$, then $p(u - n)$ also will. To understand why, try evaluating $p(u-n)$ at $n=0,1,2,\\ldots,u$; you get $p(u),p(u-1),p(u-2),\\ldots,p(0)$, which are the same values given by $p(n)$, just in reverse.\n",
"\n",
"However, that property alone doesn't explain why shifting $f$ right generates *more* (non-distinct) primes than $f(n)$. To understand that aspect, let's look at a plot of $f(n)$."
"This quadratic is symmetric across $x=-0.5$. Because of this, $f(-1) = f(0)$, $f(-2) = f(1)$, and so on; in general, $f(n) = f(-1-n)$. However, since the axis of symmetry is just to the left of the $y$-axis, when we iterate through $n=0,1,2,\\ldots$, we only output values on one side of the axis, so we don't output any duplicated values.\n",
"\n",
"But when we transform this function by shifting it $k$ units to the right, $f(n - k)$, we cause the part of the function that's left of the axis of symmetry to *also* be output when we iterate through $n=0,1,2,\\ldots$. Since these values are all equal to values we were already outputting, they'll also be prime, but they won't be *new* primes. Since $f$ outputs primes from $n=0$ to $39$ (40 distinct values), we can shift it up to 40 units to the right and have every value from $n=0$ to $79$ be prime.\n",
"\n",
"Because of the symmetry of $f$, the transformation $f(u-n)$ is actually the *same* as shifting $f$ to the right $u+1$ units (this can be shown algebraically); then, after outputting the primes in reverse, the vertex is reached and the function starts increasing again, repeating the primes that were already output.\n",
"* Primes generated by Euler's formula: [A005846](https://oeis.org/A005846)\n",
"\n",
"#### Copyright (C) 2025 filifa\n",
"\n",
"This work is licensed under the [Creative Commons Attribution-ShareAlike 4.0 International license](https://creativecommons.org/licenses/by-sa/4.0/) and the [BSD Zero Clause license](https://spdx.org/licenses/0BSD.html)."
