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

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, 12, 5, 12, 30, 32, 12, 8, 6, 12, 19, 20, 28, 24, 10, 12, 44, 12, 12, 22, 46, 16, 42, 30, 8, 28, 27, 36, 12, 24, 20, 10, 20, 12, 31, 30, 12, 64, 21, 12, 68, 8, 44, 6, 70, 24, 36, 38

Offset: 1

Author

Art DuPre, Jul 04 2012

Keywords

• nonn

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)

Examples

			11 first divides the term A000129(12) = 13860 = 2*3*5*7*11.
		

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.

Crossrefs

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

Programs

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 *)

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

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