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Looking at Eulers prime generating polynomial

$\large n^2-n+41$

it generates these primes $n\in$ $[0,40]$

  41, 41, 43, 47, 53, 61,..., 1447, 1523, 1601

I started to look for my own polynomial of the form

$an^2-bn+c$

and noticed that Eulers primes can be labeled with sequential odd numbers

  1   3   5   7   9  ...  75    77    79

41, 43, 47, 53, 61,..., 1447, 1523, 1601

Setting the last pair $b=79$ and $c=1601$ yields this formula

$n^2-79n+1601$

and found it produced $80$ non-distinct primes for $n\in [0,79]$ which is the longest prime chain I could find.

Questions

Why does the last Euler number $1601$ paired with an odd number $79$ yield

$\large n^2-79n+1601$

Do other polynomials follow a similar pattern to create prime chains greater than $80$?

vengy
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    $ 1^2 - 4 \cdot 41 = -163. $ Your version $79^2 - 4 \cdot 1601 = -163. $ You have simply taken a shift of the original, and your primes are the same as those given by $n^2 - n + 41$ with something like $-39 < n < 39,$ considerable duplication... – Will Jagy May 03 '22 at 03:06
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    Your polynomial is simply $,f(39-x),,$ for $,f(x) = x^2+x+41,,$ i.e. as here in the dupe, it's a shift (and reflection) of the common Euler polynomial $,f(x),$ in the linked dupe. See the Theorem there for generalizations. – Bill Dubuque May 03 '22 at 07:36

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