[seqfan] Re: {Spam?} A007996, A096264

[franktaw at netscape.net]

Any prime == 1 (mod 6) divides numbers of the form n^2 - n + 1. And 
dividing one means dividing infinitely many, since p|n^2-n+1 implies 
p|(n+p)^2-(n+p)+1.

2, of course, only divides f(1).

Franklin T. Adams-Watters

-----Original Message-----
From: M. F. Hasler <oeis at hasler.fr>
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
Sent: Thu, Apr 24, 2014 5:47 pm
Subject: [seqfan] Re: {Spam?} A007996, A096264


Indeed at least 19 = A096264(4) divides several among the first terms 
of n^2-n+1
(and the definition f(n+1) = ... does not make that much sense if
non-recursive as it stands).

On Tue, Apr 22, 2014 at 7:10 PM, Don Reble <djr at nk.ca> wrote:
>> %I A007996
>> %S 2,3,7,13,43,73,139,181,547,...
>> %N Primes that divide at least one term of the sequence f given by
>>    f(1) = 2, f(n+1) = n^2-n+1.
>
>
>> %I A096264
>> %S 5,11,17,19,23,29,31,37,41,...
>> %N Primes that do not divide any terms of the sequence f given by
>>    f(1) = 2, f(n+1) = n^2-n+1.
>
>
>    Perhaps they should be f(n+1) = f(n)^2 - f(n) + 1. See also 
A007018.
>    The stated sequence begins 2,3,7,13,19,31,...; it looks rather like
>    A007645. (But f(1)=2 instead of 1 puts 2 into the sequence.)
>
> --
> Don Reble  djr at nk.ca
>
>
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--
Maximilian

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