[seqfan] Re: A269254
Andrew N W Hone
A.N.W.Hone at kent.ac.uk
Wed Oct 25 07:41:58 CEST 2017
Continuing from my remarks below, it is possible to arrive at the following result:
Theorem. If m^2 (m+3) - 1 is not prime, then a_n = -1 for n = m^2 (m+3) - 2.
The case n=110 corresponds to m=4, which is the first value of m for which n+1 is not prime.
To see this, fix n to begin with, and suppose that the trisection u_k = s_{3k-1} can be factorized as u_k = b_k c_k,
where
b_{k+2} - n b_{k+1} + b_k = 0,
c_{k+3} - (n^2 - 1) ( c_{k+2} - c_{k+1} ) - c_k = 0.
The characteristic roots of these recurrences are z, 1/z for b_k and z^2, 1/z^2, 1 for c_k,
where
z + 1/z= n,
and we have that u_k satisfies
u_{k+2} - (n^3 - 3 n) u_{k+1} + u_k = 0,
with characteristic roots z^3, 1/z^3; and for later use we record the
values
u_{-1} = - n^3 - n^2 + 2 n + 1,
u_0 = -1,
u_1 = n^2 + n - 1,
u_2 = n^5 + n^4 - 4 n^3 - 3 n^2 + 3 n + 1.
By looking at the characteristic roots of these recurrences, it is apparent
that in general the product u_k = b_k c_k of any two solutions of the recurrences for b_k, c_k
satisfies a linear recurrence of order 4 given by
L_n L_{n^3 - 3 n} u_k = 0,
where L_N denotes the linear operator S^2 - N S + 1, with S being the shift operator
(S f_k = f_{k+1} for any f). We require that u_k satisfies the 2nd order equation
w_k := L_{n^3 - 3 n} u_k = 0
(as above), so it is enough to require that w_k =0 for any two adjacent values of k
(because w_k satisfies the 2nd order linear equation L_n w_k = 0 for all k, so if two
initial values are 0 then it vanishes for all k). This requirement puts constraints on the
total of 5 initial values required for the b_k, c_k recurrences; for instance we can
regard b_0,b_1,c_{-1},c_0,c_1 as a set of 5 initial values. Now if we choose
u_{-1},u_0,u_1,u_2 as above then we guarantee that u_k satisfies the
recurrence L_{n^3 - 3 n} u_k = 0 for k=-1, 0; hence w_{-1} = 0 =w_0,
implying that u_k satisfies the required 2nd order relation for all k.
The constraints on the 5 initial data can be written as 4 equations, that is
b_j c_j = u_j for j=-1,0,1,2,
where we regard b_1,b_2 as given linear functions of b_{-1},b_0, and c_2 is a
given function of c_{-1},c_0,c_1 (by the recurrences for b_k, c_k); so we
have 4 equations for 5 unknowns, and this should give a curve in the space
of 5 initial data. If we set
b_0 = x, b_1= y
and eliminate c_{-1},c_0,c_1, then we get a homogeneous cubic curve in
the (x,y) plane, with coefficients dependent on n. Since it is homogeneous,
we can set
y = t x
to find a union of lines through the origin, where t is a solution of the cubic
equation
t^3 + 3 t^2 - 3 (n+1) t + n^2 + n - 1 = 0
So far n was fixed, but now we can view this as a cubic curve in the (t,n)
plane. It is a nodal cubic with singularity at (t,n) = (-1,-2), so it (or its
desingularization) is a curve of genus zero, and we can obtain a rational
parametrization by setting
t = -1 + X, n = -2 + Y
so that X^3 - 3 X Y + Y^2 = 0, and then taking Y= - m X gives
t = - m (m+3) -1, n = m^2 (m+3) - 2 .
Hence when m is a positive integer we obtain trisected sequences u_k which
factorize for the corresponding value of n; and the other subsequence v_k = s_{3k}
will also factorize, by the remarks below, while the the subsequence s_{3k+1}
factorizes whenever n+1 is not prime.
________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Andrew N W Hone [A.N.W.Hone at kent.ac.uk]
Sent: 24 October 2017 07:19
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: A269254
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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