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 A120947 a(n) = smallest m such that n-th prime divides Pell(m). 3
 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 (list; graph; refs; listen; history; text; internal format)
 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

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Last modified March 25 01:20 EDT 2023. Contains 361511 sequences. (Running on oeis4.)