[seqfan] Re: A269254

[Bob Selcoe]

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/
>