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A214028 Entry points for the Pell sequence: smallest k such that n divides A000129(k).

%I A214028 #50 Jan 05 2025 19:51:39
%S A214028 1,2,4,4,3,4,6,8,12,6,12,4,7,6,12,16,8,12,20,12,12,12,22,8,15,14,36,
%T A214028 12,5,12,30,32,12,8,6,12,19,20,28,24,10,12,44,12,12,22,46,16,42,30,8,
%U A214028 28,27,36,12,24,20,10,20,12,31,30,12,64,21,12,68,8,44,6,70,24,36,38
%N A214028 Entry points for the Pell sequence: smallest k such that n divides A000129(k).
%C A214028 Conjecture: A175181(n)/A214027(n) = a(n). This says that the zeros appear somewhat uniformly in a period. The second zero in a period is exactly where n divides the first Lucas number, so this relationship is not really surprising.
%C A214028 From _Jianing Song_, Aug 29 2018: (Start)
%C A214028 The comment above is correct, since n divides A000129(k*a(n)) for all integers k and clearly a(n) divides A175181(n), so the zeros appear uniformly.
%C A214028 a(n) <= 4*n/3 for all n, where the equality holds iff n is a power of 3.
%C A214028 (End)
%H A214028 G. C. Greubel, <a href="/A214028/b214028.txt">Table of n, a(n) for n = 1..10000</a>
%H A214028 Bernadette Faye and Florian Luca, <a href="http://arxiv.org/abs/1508.05714">Pell Numbers whose Euler Function is a Pell Number</a>, arXiv:1508.05714 [math.NT], 2015 (called z(n)).
%H A214028 N. Robbins, <a href="https://web.archive.org/web/2024*/https://www.fq.math.ca/Scanned/22-4/robbins.pdf">On Pell numbers of the form p*x^2, where p is prime</a>, Fib. Quart. 22  (4) (1984) 340-348, definition 1.
%F A214028 If p^2 does not divide A000129(a(p)) (that is, p is not in A238736) then a(p^e) = a(p)*p^(e - 1). If gcd(m, n) = 1 then a(mn) = lcm(a(m), a(n)). - _Jianing Song_, Aug 29 2018
%e A214028 11 first divides the term A000129(12) = 13860 = 2*3*5*7*11.
%p A214028 A214028 := proc(n)
%p A214028     local a000129,k ;
%p A214028     a000129 := [1,2,5] ;
%p A214028     for k do
%p A214028         if modp(a000129[1],n) = 0 then
%p A214028             return k;
%p A214028         end if;
%p A214028         a000129[1] := a000129[2] ;
%p A214028         a000129[2] := a000129[3] ;
%p A214028         a000129[3] := 2*a000129[2]+a000129[1] ;
%p A214028     end do:
%p A214028 end proc:
%p A214028 seq(A214028(n),n=1..40); # _R. J. Mathar_, May 26 2016
%t A214028 a[n_] := With[{s = Sqrt@ 2}, ((1 + s)^n - (1 - s)^n)/(2 s)] // Simplify; Table[k = 1; While[Mod[a[k], n] != 0, k++]; k, {n, 80}] (* _Michael De Vlieger_, Aug 25 2015, after _Michael Somos_ at A000129 *)
%t A214028 Table[k = 1; While[Mod[Fibonacci[k, 2], n] != 0, k++]; k, {n, 100}] (* _G. C. Greubel_, Aug 10 2018 *)
%o A214028 (PARI) pell(n) = polcoeff(Vec(x/(1-2*x-x^2) + O(x^(n+1))), n);
%o A214028 a(n) = {k=1; while (pell(k) % n, k++); k;} \\ _Michel Marcus_, Aug 25 2015
%Y A214028 Cf. A001175, A001176, A001177, A214027, A175181.
%K A214028 nonn
%O A214028 1,2
%A A214028 _Art DuPre_, Jul 04 2012