A120947 a(n) = smallest m such that n-th prime divides Pell(m).

2, 4, 3, 6, 12, 7, 8, 20, 22, 5, 30, 19, 10, 44, 46, 27, 20, 31, 68, 70, 36, 26, 84, 44, 48, 51, 34, 108, 55, 28, 126, 132, 17, 140, 75, 150, 79, 164, 166, 87, 36, 91, 190, 96, 9, 18, 212, 74, 76, 23, 116, 14, 40, 84, 64, 262, 15, 270, 139, 140, 284, 49, 308, 310, 78, 159, 332

Offset: 1

Author

Ralf Stephan, Aug 19 2006

Keywords

• nonn

Comments

For all divisors d of n>0, Pell(d) divides Pell(n), so if a prime divides the n-th Pell number, so does it for all multiples of n.

For n > 1, a(n) is the multiplicative order of -3-2*sqrt(2), in GF(prime(n)) if 2 is a quadratic residue (mod prime(n)) or GF(prime(n)^2) otherwise. Thus a(n) divides prime(n)-1 if prime(n) == 1 or 7 (mod 8), i.e. n is in A024704, and a(n) divides prime(n)+1 if prime(n) == 3 or 5 (mod 8), i.e. n is 2 or is in A024705. - Robert Israel, Aug 28 2015

Examples

			a(4)=6 because the 6th Pell number, 70, is the first that is divisible by the 4th prime (=7).
		

Links

• Alois P. Heinz and Robert Israel, Table of n, a(n) for n = 1..10000 (n = 1 .. 1000 from Alois P. Heinz)
• J. L. Schiffman, Exploring the Fibonacci sequence of order two with CAS technology, Paper C027, Electronic Proceedings of the Twenty-fourth Annual International Conference on Technology in Collegiate Mathematics, Orlando, Florida, March 22-25, 2012.

Crossrefs

Cf. A000129 (Pell numbers), A001602 (equivalent sequence with Fibonacci numbers), A239111, A024704, A024705.

Programs

Maple

p:= proc(n) p(n):=`if`(n<2, n, 2*p(n-1)+p(n-2)) end:
a:= proc(n) local k, t; t:= ithprime(n);
      for k while irem(p(k), t)>0 do od; k
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Mar 28 2014
f:= proc(n)
local p, r, G;
uses numtheory;
p:= ithprime(n);
if quadres(2,p)=1 then
   r:= msqrt(2,p);
   order(-3-2*r, p)
else
   G:= GF(p, 2, r^2-2);
   G:-order( G:-ConvertIn(-3-2*r));
fi
end proc:
2, seq(f(n), n=2..100); # Robert Israel, Aug 28 2015

Mathematica

p[n_] := p[n] = If[n<2, n, 2*p[n-1] + p[n-2]]; a[n_] := Module[{k, t}, t = Prime[n]; For[k=1, Mod[p[k], t]>0, k++]; k]; Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Jun 16 2015, after Alois P. Heinz *)

PARI

a(n,p=prime(n))=my(cur=Mod(1,p),last,m=1); while(cur, m++; [last,cur]=[cur,2*cur+last]); m \\ Charles R Greathouse IV, Jun 16 2015