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Entry points for the Pell sequence: smallest k such that n divides A000129(k).
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%I #48 Apr 25 2021 10:43:39

%S 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 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 28,27,36,12,24,20,10,20,12,31,30,12,64,21,12,68,8,44,6,70,24,36,38

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

%C 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 From _Jianing Song_, Aug 29 2018: (Start)

%C 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 a(n) <= 4*n/3 for all n, where the equality holds iff n is a power of 3.

%C (End)

%H G. C. Greubel, <a href="/A214028/b214028.txt">Table of n, a(n) for n = 1..10000</a>

%H 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 N. Robbins, <a href="http://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 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 11 first divides the term A000129(12) = 13860 = 2*3*5*7*11.

%p A214028 := proc(n)

%p local a000129,k ;

%p a000129 := [1,2,5] ;

%p for k do

%p if modp(a000129[1],n) = 0 then

%p return k;

%p end if;

%p a000129[1] := a000129[2] ;

%p a000129[2] := a000129[3] ;

%p a000129[3] := 2*a000129[2]+a000129[1] ;

%p end do:

%p end proc:

%p seq(A214028(n),n=1..40); # _R. J. Mathar_, May 26 2016

%t 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 Table[k = 1; While[Mod[Fibonacci[k, 2], n] != 0, k++]; k, {n, 100}] (* _G. C. Greubel_, Aug 10 2018 *)

%o (PARI) pell(n) = polcoeff(Vec(x/(1-2*x-x^2) + O(x^(n+1))), n);

%o a(n) = {k=1; while (pell(k) % n, k++); k;} \\ _Michel Marcus_, Aug 25 2015

%Y Cf. A001175, A001176, A001177, A214027, A175181.

%K nonn

%O 1,2

%A _Art DuPre_, Jul 04 2012