A269254 To find a(n), define a sequence by s(k) = n*s(k-1) - s(k-2), with s(0) = 1, s(1) = n + 1; then a(n) is the smallest index k such that s(k) is prime, or -1 if no such k exists.

1, 1, 2, 1, 2, 1, -1, 2, 2, 1, 2, 1, 2, -1, 2, 1, 3, 1, 2, 2, 2, 1, -1, 2, 6, 2, 3, 1, 3, 1, 2, 9, 9, -1, 2, 1, 6, 2, 2, 1, 2, 1, 5, 2, 2, 1, -1, 2, 5, 2, 9, 1, 2, 2, 2, 2, 6, 1, 2, 1, 14, -1, 5, 2, 2, 1, 5, 2, 3, 1, 6, 1, 8, 3, 6, 2, 3, 1, -1, 3, 18, 1, 2, 3, 2, 2, 3, 1, 2, 9, 3, 5, 2, 2, 96, 1, 3, -1, 5, 1, 2, 1, 2, 15, 14, 1, 44, 1, 3, -1

Offset: 1

Author

Arkadiusz Wesolowski, Jul 09 2016

Keywords

• sign

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

Examples

			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).
		

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.

Crossrefs

Cf. A005097, A011973, A269251, A269252, A269253.

Cf. A294099 (array used to compute this sequence).

Cf. A285992, A299107, A299109, A088165, A117522, A299100, A299101, A113501, A298675, A298677, A298878, A299045, A299071.

Programs

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;

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 *)

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

Formula

If n is prime then a(n-1) = 1.

Extensions

a(86)-a(94) from Antti Karttunen, Oct 20 2017

a(95)-a(109) appended by L. Edson Jeffery, Oct 22 2017