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

%I #25 Dec 13 2023 07:32:59

%S 3,5,7,113,11,13,43,17,19,61,23,73,79,29,31,97,103,37,1241463763,41,

%T 43,664973,47,2593,151,53,163,14972833,59,61,4217,193,67,23801,71,73,

%U 223,229,79,241,83,7561,61068909859,89,271,277,283,97,10193,101,103,313

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

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

%C From _Robert Israel_, Sep 26 2017: (Start)

%C 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.

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

%C 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)

%H Robert Israel, <a href="/A292848/b292848.txt">Table of n, a(n) for n = 1..306</a>

%e 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.

%p f:= proc(n) local a,b,t;

%p a:= 1; b:= 1;

%p do

%p t:= a; a:= 2*a + (2*n-1)*b;

%p if isprime(a) then return a fi;

%p b:= t;

%p od

%p end proc:

%p map(f, [$1..100]); # _Robert Israel_, Sep 26 2017

%t f[n_, k_] := ((1 + Sqrt[n])^k + (1 - Sqrt[n])^k)/2;

%t Table[k = 1; While[! PrimeQ[Expand@f[2n, k]], k++]; Expand@f[2n, k], {n, 52}]

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

%K nonn

%O 1,1

%A _XU Pingya_, Sep 24 2017

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