... From: markherkommer Has anybody tried n^2 - 79n + 1601 This polynomial will produce 80 primes. Mark Herkommer ... Yes, but the 80

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, Jan 2, 2006

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----- Original Message -----
From: "markherkommer" <newmetro@...>

Has anybody tried

n^2 - 79n + 1601

This polynomial will produce 80 primes.

Mark Herkommer

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Yes, but the 80 primes are symmetric respect to n = 40, so they are only 40 primes, half
of them are repetitions.

I note that (as I think you'll know) you are merely displacing the famous prime-generating
polynomial of Euler, n^2-n+41 in such a way that the 80 primes of these polynomial,
located between -39 and 40 are now located between 0 and 79:

(n-39)^2 -(n-39) + 41 =

= n^2 -78n + 1521 - n + 80 =

= n^2 - 79n + 1601

Jose Brox

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