A292848 a(n) is the smallest prime of form (1/2)*((1 + sqrt(2*n))^k + (1 - sqrt(2*n))^k).

3, 5, 7, 113, 11, 13, 43, 17, 19, 61, 23, 73, 79, 29, 31, 97, 103, 37, 1241463763, 41, 43, 664973, 47, 2593, 151, 53, 163, 14972833, 59, 61, 4217, 193, 67, 23801, 71, 73, 223, 229, 79, 241, 83, 7561, 61068909859, 89, 271, 277, 283, 97, 10193, 101, 103, 313

Offset: 1

Author

XU Pingya, Sep 24 2017

Keywords

• nonn

Comments

When 2n + 1 = p is prime, a(n) = p.

From Robert Israel, Sep 26 2017: (Start)

a(n) is also the first prime in the sequence defined by the recursion x(k+2)=2*x(k+1)+(2*n-1)*x(k) with x(0)=x(1)=1.

a(307), if it exists, has more than 10000 digits.

It appears that x(n*k) is divisible by x(k) if n is odd. Thus a(n) (if it exists) must be x(k) where k is either a power of 2 or a prime. (End)

Examples

			For k = {1, 2, 3, 4}, (1/2)((1 + sqrt(8))^k + (1 - sqrt(8))^k) = {1, 9, 25, 113}. 113 is prime, so a(4) = 113.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..306

Crossrefs

Cf. A001333, A026150, A046717, A084057, A002533, A083098, A083100, A003665, A002535, A133294, A090042, A125816, A133343, A133345, A120612, A133356, A125818.

Programs

Maple

f:= proc(n) local a,b,t;
  a:= 1; b:= 1;
  do
    t:= a; a:= 2*a + (2*n-1)*b;
    if isprime(a) then return a fi;
    b:= t;
  od
end proc:
map(f, [$1..100]); # Robert Israel, Sep 26 2017

Mathematica

f[n_, k_] := ((1 + Sqrt[n])^k + (1 - Sqrt[n])^k)/2;
Table[k = 1; While[! PrimeQ[Expand@f[2n, k]], k++]; Expand@f[2n, k], {n, 52}]