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

1,1

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

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.

Example

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

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 *)

Prog

(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

Crossrefs

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

Sequence in context: A334384 A225055 A341194 * A222600 A046793 A182940

Adjacent sequences: A120944 A120945 A120946 * A120948 A120949 A120950

Keyword

nonn

Author

Ralf Stephan, Aug 19 2006

Status

approved