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%I #28 Aug 28 2015 21:10:39
%S 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,
%T 51,34,108,55,28,126,132,17,140,75,150,79,164,166,87,36,91,190,96,9,
%U 18,212,74,76,23,116,14,40,84,64,262,15,270,139,140,284,49,308,310,78,159,332
%N a(n) = smallest m such that n-th prime divides Pell(m).
%C 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.
%C 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
%H Alois P. Heinz and Robert Israel, <a href="/A120947/b120947.txt">Table of n, a(n) for n = 1..10000</a> (n = 1 .. 1000 from Alois P. Heinz)
%H J. L. Schiffman, <a href="http://archives.math.utk.edu/ICTCM/i/24/C027.html">Exploring the Fibonacci sequence of order two with CAS technology</a>, Paper C027, Electronic Proceedings of the Twenty-fourth Annual International Conference on Technology in Collegiate Mathematics, Orlando, Florida, March 22-25, 2012.
%e a(4)=6 because the 6th Pell number, 70, is the first that is divisible by the 4th prime (=7).
%p p:= proc(n) p(n):=`if`(n<2, n, 2*p(n-1)+p(n-2)) end:
%p a:= proc(n) local k, t; t:= ithprime(n);
%p for k while irem(p(k), t)>0 do od; k
%p end:
%p seq(a(n), n=1..100); # _Alois P. Heinz_, Mar 28 2014
%p f:= proc(n)
%p local p, r, G;
%p uses numtheory;
%p p:= ithprime(n);
%p if quadres(2,p)=1 then
%p r:= msqrt(2,p);
%p order(-3-2*r, p)
%p else
%p G:= GF(p, 2, r^2-2);
%p G:-order( G:-ConvertIn(-3-2*r));
%p fi
%p end proc:
%p 2, seq(f(n), n=2..100); # _Robert Israel_, Aug 28 2015
%t 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_ *)
%o (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
%Y Cf. A000129 (Pell numbers), A001602 (equivalent sequence with Fibonacci numbers), A239111, A024704, A024705.
%K nonn
%O 1,1
%A _Ralf Stephan_, Aug 19 2006
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