OFFSET
1,2
COMMENTS
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.
From Jianing Song, Aug 29 2018: (Start)
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.
a(n) <= 4*n/3 for all n, where the equality holds iff n is a power of 3.
(End)
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..10000
Bernadette Faye and Florian Luca, Pell Numbers whose Euler Function is a Pell Number, arXiv:1508.05714 [math.NT], 2015 (called z(n)).
N. Robbins, On Pell numbers of the form p*x^2, where p is prime, Fib. Quart. 22 (4) (1984) 340-348, definition 1.
FORMULA
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
EXAMPLE
11 first divides the term A000129(12) = 13860 = 2*3*5*7*11.
MAPLE
A214028 := proc(n)
local a000129, k ;
a000129 := [1, 2, 5] ;
for k do
if modp(a000129[1], n) = 0 then
return k;
end if;
a000129[1] := a000129[2] ;
a000129[2] := a000129[3] ;
a000129[3] := 2*a000129[2]+a000129[1] ;
end do:
end proc:
seq(A214028(n), n=1..40); # R. J. Mathar, May 26 2016
MATHEMATICA
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 *)
Table[k = 1; While[Mod[Fibonacci[k, 2], n] != 0, k++]; k, {n, 100}] (* G. C. Greubel, Aug 10 2018 *)
PROG
(PARI) pell(n) = polcoeff(Vec(x/(1-2*x-x^2) + O(x^(n+1))), n);
a(n) = {k=1; while (pell(k) % n, k++); k; } \\ Michel Marcus, Aug 25 2015
CROSSREFS
KEYWORD
nonn
AUTHOR
Art DuPre, Jul 04 2012
STATUS
approved