[seqfan] Re: A269254 and A034807

Andrew N W Hone A.N.W.Hone at kent.ac.uk
Thu Oct 26 05:40:56 CEST 2017

Just a few thoughts on the cases where a_n = -1:

The terms of the sequence (s_k) begin

s_{-1} = -1, s_0 = 1, s_1 = n+1, s_2 = n^2+n-1, s_3=n^3+n^2-2n-1, ...

If we want to find values of n for which there are no prime terms in the sequence (s_k), then first of all we can remove all values of n for which s_1 = n+1 is prime, then
remove all values of n for which s_2=n^2+n-1 is prime (that is, A045546),

The family n = j^2 -2 (with j>3), for which Brad proved a factorization of the sequence (s_k), is a specific sequence of n for which s_1=n+1 factors, i.e. s_1 = (j+1)(j-1).

Once we exclude the case that n+1 is prime, then by the extension of Neil's comment on the case n=110, to show that a_n = -1 it is sufficient to prove that
the trisected sequence (u_k) given by

u_k = s_{3k-1}

can be factorized (for all k, positive and negative), since we have s_{3k} = - s_{-3k-1} and s_{3k+1} = 0 mod n+1.

The family n = j^3 - 3j (j>4), corresponding to the theorem in my previous post (which also included the proof - maybe I didn't state this clearly), is a specific
sequence for which s_2 = n^2 + n - 1 factors, i.e. s_2 = ( j^2 + j -1 )( j^4 - j^3 - 4 j^2 + 4 j + 1 ); but we have to exclude the values of j for which s_1 = n+1 is prime
in order to have a_n = -1.

If we continue with the assumption that n+1 is not prime, then (as already mentioned) it is enough to focus on the sequence (u_k), which
satisfies

u_{k+2} - (n^3 - 3 n) u_{k+1} + u_k = 0

with initial values

u_0 = -1, u_1 = s_2 = n^2 + n - 1.

So we want to further assume that u_1 = s_2 is not prime. It is not hard to see that the sequence (u_k) has
period 5 mod u_1, with the repeated pattern mod u_1 being

-1, 0, 1, - (n+1), n+1, ...

Thus we should look at the pentasection of the trisection (!), and take the subsequences ( u_{5k+r} ) for r=0,1,2,3,4.
However, we can ignore the case r=1 since u_{5k+1}=0 mod u_1. Also, I expect that looking at the cases
at r=0 and r=3 may be enough, due to the symmetry (up to signs) of the above values mod u_1.

All the best,
Andy
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Hans Havermann [gladhobo at bell.net]
Sent: 26 October 2017 03:36
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: A269254 and A034807

> especially if a(2525) holds

I'm starting to go deeper on some of my initial unknowns, the ones that failed to find a prime in 20 minutes. I've resolved a couple already: a(721) = 7358 and a(735) = 3623. I can see some of these ending up in the tens of thousands.

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