# [seqfan] Re: A269254

Andrew N W Hone A.N.W.Hone at kent.ac.uk
Tue Oct 24 08:19:50 CEST 2017

```Just a general comment: Neil's observation that we should look at the trisection is valid for any n.

It is convenient to start the recurrence

s_{k+2} - n s_{k+1} + s_{k} = 0

with initial values s_{-1}=-1, s_{0}=1, which ensures s_{1}=n+1 as required.

Shifting the recurrence implies

s_{k+3} - (n+1) ( s_{k+2} - s_{k+1} ) - s_{k}=0,

so the sequence has period 3 when taken mod n+1.

In particular, s_{3j+1} = 0 mod n+1, so all terms of this subsequence are composite unless
n=p-1 for p prime, so we only need to consider the subsequences
with indices equal 0 or 2 mod 3.

Let u_{j} = s_{3 j -1} and v_{j}= s_{3 j}.

The initial conditions for these two subsequences are

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

v_0 = 1,  v_1 = n^3 + n^2 - 2 n - 1,

and they both satisfy the same 3-term relation, i.e.

u_{j+2} - ( n^3 - 3 n ) u_{j+1} + u_{j} = 0.

Then observe that v_0 = - u_0 and v_{-1} = - u_1; hence it follows that

v_{j} = - u_{-j}

for all j. So to see if the subsequences have composite terms, it is sufficient to consider the sequence (u_j),
but for both positive and negative indices.

Andy

________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Bob Selcoe [rselcoe at entouchonline.net]
Sent: 23 October 2017 03:05
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: A269254

>>>>This might help.  m^2 -, m >= 2 == j^2 + 4j + 2, j >= 1.  So for these
terms:

rather m^2 - 2, m >= 2...

Bob

--------------------------------------------------
From: "Bob Selcoe" <rselcoe at entouchonline.net>
Sent: Sunday, October 22, 2017 8:50 PM
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: A269254

> Hi Don, Brad and Seqfans,
>
> Don - interesting!  As Brad showed, a(n) = -1 when n = m^2 - 2, m>= 2.  I
> was wondering if there might be other terms as well.
>
> This might help.  m^2 -, m >= 2 == j^2 + 4j + 2, j >= 1.  So for these
> terms:
>
> b(0) = 1;
> b(1) = n+1 = j^2 + 4j + 3;
> b(2) = n(n+1) - 1 = (j^2 + 4j +2)*(j^2 + 4j +3) - 1 = j^4 + 8j^3 + 21j^2 +
> 20j +5
>
> The coefficients of b(0), b(1) and b(2) are the 2nd, 6th and 10th rows of
> the Fibonacci coefficient triangle A011973; but note the degrees of the
> polynomials are in reverse order.  Generally, b(k) is obtained by the
> (4k+2)-th row of the triangle (in reverse order) for all k.  So:
>
> b(3) = j^6 + 12j^5 + 55j^4 + 120j^3 + 126j^2 + 56j + 7;
> b(4) = j^8 + 16j^7 + 105j^6 ... + 9;
> etc.
>
> These polynomials can be described as the product (c+d)*(c-d), where c(k)
> = (j+2)*c(k-1) - d(k-1) and d(k) = c(k-1). for example,  j=2; n=14:
>
>                                    c         d
>
> b(0)  =          1            1         0
> b(1)  =        15           4          1
> b(2)  =      209          15        4
> b(3) =     2911         56       15
> b(4) =   40545       209      56
> b(5) = 546719       780    209
> etc.
>
> This satisfies that condition that all b(k) are non-prime for all j
> (unless I missed something in Brad's proof, this seems to be an alternate
> way of showing that a(n) = -1 for n = m^2 - 2).
>
> So if a(n) = -1 when n != m^2 - 2, perhaps it can be shown that other
> polynomials correspond to other reverse-order rows in A011973 which
> generate only non-primes?
>
> Regards,
> Bob Selcoe
>
>
> --------------------------------------------------
> From: "Don Reble" <djr at nk.ca>
> Sent: Sunday, October 22, 2017 4:15 PM
> To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
> Subject: [seqfan] Re: A269254
>
>>> could a(110) be -1?
>>
>>    It could!
>>    Trisect sequence-110 to get:
>>
>> s(0) -1 = -1 * 1
>> s(3) 12209 = 29 * 421
>> s(6) 16246150031 = 3191 * 5091241
>> s(9) 21618264461738561 = 350981 * 61593831181
>> s(12) 28766775971285404815839 = 38604719 * 745162164534481
>> s(15) 38279085781688731361830743569 = 4246168109 * 9014971804944317941
>> s(18) 50936831077090977385276030140145391
>>      = 467039887271 * 109063128151054193913721
>> s(21) 67780113009314371791483566295225436698401
>>      = 51370141431701 * 1319445715356481833023876701
>> s(24) 90192962978053418280696346184682355821321113279
>>      = 5650248517599839 * 15962654155319589064868666412961
>>
>>    So far, this subsequence is the product of a Lucas sequence
>>    [t(n+2) = 110 t(n+1) - t(n)] and an integer sequence.
>>
>>
>> s(1) 1 = 1 * 1
>> s(4) 1342879 = 139 * 9661
>> s(7) 1786928798929 = 15289 * 116876761
>> s(10) 2377812544869509551 = 1681651 * 1413975042901
>> s(13) 3164083819079723345430241 = 184966321 * 17106269952137521
>> s(16) 4210351415532437651518789281919 = 20344613659 *
>> 206951652466984684141
>> s(19) 5602588318103384725927427610425725489
>>      = 2237722536169 * 2503701074439310756598281
>> s(22) 7455196197246420601834317660713681347165711
>>      = 246129134364931 * 30289775391615129066341317381
>> s(25) 9920405923784291913924768096855946930622570930881
>>      = 27071967057606241 * 366445700184058757005286501075041
>>
>>    ... product of a Lucas sequence [t(n+2) = 110 t(n+1) - t(n)]
>>    and an integer sequence.
>>
>>
>> s(2) 111 = 111 * 1
>> s(5) 147704481 = 111 * 1330671
>> s(8) 196545921732159 = 111 * 1770683979569
>> s(11) 261537761671184312049 = 111 * 2356196051091750559
>> s(14) 348020453322798282592510671 = 111 * 3135319399304489032364961
>> s(17) 463100376622786452935704990267521
>>      = 111 * 4172075465070148224645990903311
>> s(20) 616233778160295228874631761116689658399
>>      = 111 * 5551655659101758818690376226276483409
>> s(23) 820003801584096951829983459112209722751529809
>>      = 111 * 7387421635892765331801652784794682186950719
>> s(26) 1091154458653294057113443794307969480012661481283631
>>      = 111 * 9830220348227874388409403552324049369483436768321
>>
>>    ... multiples of 111. (But there's another factorization.)
>>
>> --
>> Don Reble  djr at nk.ca
>>
>>
>> --
>> Seqfan Mailing list - http://list.seqfan.eu/
>>
>
> --
> Seqfan Mailing list - http://list.seqfan.eu/
>

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