OFFSET
1,3
COMMENTS
The s(k) sequences can be viewed in A294099, where they appear as rows. - Peter Munn, Aug 31 2020
For n >= 3, a(n) is that positive integer k yielding the smallest prime of the form (x^y - 1/x^y)/(x - 1/x), where x = (sqrt(n+2) +- sqrt(n-2))/2 and y = 2*k + 1, or -1 if no such k exists.
Every positive term belongs to A005097.
When n=7, the sequence {s(k)} is A033890, which is Fibonacci(4i+2), and since x|y <=> F_x|F_y, and 2i+1|4i+2, A033890 is never prime, and so a(7)=-1. For the other -1 terms below 100, see the theorem below and the Klee link - N. J. A. Sloane, Oct 20 2017 and Oct 22 2017
Theorem (Brad Klee): For all n > 2, a(n^2 - 2) = -1. See Klee link for a proof. - L. Edson Jeffery, Oct 22 2017
Theorem (Based on work of Hans Havermann, L. Edson Jeffery, Brad Klee, Don Reble, Bob Selcoe, and N. J. A. Sloane) a(110) = -1. [For proof see link. - N. J. A. Sloane, Oct 23 2017]
From Bob Selcoe, Oct 24 2017, edited by N. J. A. Sloane, Oct 27 2017: (Start)
Suppose n = m^2 - 2, where m >= 3, and let j = m-2, with j >= 1.
For this value of n, the sequence s(k) satisfies s(k) = (c(k) + d(k))*(c(k) - d(k)), where c(0) = 1, d(0) = 0; and for k >= 1: c(k) = (j+2)*c(k-1) - d(k-1), and d(k) = c(k-1). So (as Brad Klee already proved) a(n) = -1 .
We have s(0) = 1 and s(1) = n+1 = j^2 + 4j + 3. In general, the coefficients of s(k) when expanded in powers of j are given by the (4k+2)-th row of A011973 (the triangle of coefficients of Fibonacci polynomials) in reverse order. For example, s(2) = j^4 + 8j^3 + 21j^2 + 20j + 5, s(3) = j^6 + 12j^5 + 55j^4 + 120j^3 + 126j^2 + 56j + 7, etc.
Perhaps the above comments could be generalized to apply to a(110) or to other n for which a(n) = -1?
(End)
For detailed theory, see [Hone]. - L. Edson Jeffery, Feb 09 2018
LINKS
Hans Havermann, Table of n, a(n) for n = 1..946
Hans Havermann, Table of n, a(n) for n = 1..10000
C. K. Caldwell, Top Twenty page, Lehmer number
Andrew N. W. Hone, et al., On a family of sequences related to Chebyshev polynomials, arXiv:1802.01793 [math.NT], 2018.
Brad Klee, Proof for A269254, Sequence Fans Mailing List, October 2017.
N. J. A. Sloane et al., Proof that a(110) = -1
Wikipedia, Lehmer number.
FORMULA
If n is prime then a(n-1) = 1.
EXAMPLE
Let b(k) be the recursive sequence defined by the initial conditions b(0) = 1, b(1) = 16, and the recursive equation b(k) = 15*b(k-1) - b(k-2). a(15) = 2 because b(2) = 239 is the smallest prime in b(k).
Let c(k) be the recursive sequence defined by the initial conditions c(0) = 1, c(1) = 18, and the recursive equation c(k) = 17*c(k-1) - c(k-2). a(17) = 3 because c(3) = 5167 is the smallest prime in c(k).
MATHEMATICA
kmax = 100;
a[1] = a[2] = 1;
a[n_ /; IntegerQ[Sqrt[n+2]]] = -1;
a[n_] := Module[{s}, s[0] = 1; s[1] = n+1; s[k_] := s[k] = n s[k-1] - s[k-2]; For[k=1, k <= kmax, k++, If[PrimeQ[s[k]], Return[k]]]; Print["For n = ", n, ", k = ", k, " exceeds the limit kmax = ", kmax]; -1];
Array[a, 110] (* Jean-François Alcover, Aug 05 2018 *)
PROG
(Magma) lst:=[]; for n in [1..85] do if n gt 2 and IsSquare(n+2) then Append(~lst, -1); else a:=n+1; c:=1; t:=1; if IsPrime(a) then Append(~lst, t); else repeat b:=n*a-c; c:=a; a:=b; t+:=1; until IsPrime(a); Append(~lst, t); end if; end if; end for; lst;
(PARI)
allocatemem(2^30);
default(primelimit, (2^31)+(2^30));
s(n, k) = if(0==k, 1, if(1==k, (1+n), ((n*s(n, k-1)) - s(n, k-2))));
A269254(n) = { my(k=1); if((n>2)&&issquare(2+n), -1, while(!isprime(s(n, k)), k++); (k)); }; \\ Antti Karttunen, Oct 20 2017
CROSSREFS
KEYWORD
sign
AUTHOR
Arkadiusz Wesolowski, Jul 09 2016
EXTENSIONS
a(86)-a(94) from Antti Karttunen, Oct 20 2017
a(95)-a(109) appended by L. Edson Jeffery, Oct 22 2017
STATUS
approved