A232570 Numbers k that divide tribonacci(k) (A000073(k)).

1, 8, 16, 19, 32, 47, 53, 64, 103, 112, 128, 144, 155, 163, 192, 199, 208, 221, 224, 256, 257, 269, 272, 299, 311, 368, 397, 401, 419, 421, 448, 499, 512, 587, 599, 617, 640, 683, 757, 768, 773, 784, 863, 883, 896, 907, 911, 929, 936, 991, 1021, 1024

Offset

1,2

Comments

Inspired by A023172 (numbers k such that k divides Fibonacci(k)).

Includes all primes p such that x^3-x^2-x-1 has 3 distinct roots in the field GF(p) (A106279). - Robert Israel, Feb 07 2018

Includes 2^k for k >= 3. - Robert Israel, Jul 26 2024

Links

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Maple

with(LinearAlgebra[Modular]):
T:= (n, m)-> MatrixPower(m, Mod(m, <<0|1|0>,
    <0|0|1>, <1|1|1>>, float[8]), n)[1, 3]:
a:= proc(n) option remember; local k; if n=1
      then 1 else for k from 1+a(n-1)
      while T(k$2)>0 do od; k fi
    end:
seq(a(n), n=1..70);  # Alois P. Heinz, Feb 05 2018

Mathematica

trib = LinearRecurrence[{1, 1, 1}, {0, 0, 1}, 2000]; Reap[Do[If[Divisible[ trib[[n+1]], n], Print[n]; Sow[n]], {n, 1, Length[trib]-1}]][[2, 1]] (* Jean-François Alcover, Mar 22 2019 *)

Prog

(Ruby)
require 'matrix'
def power(a, n, mod)
  return Matrix.I(a.row_size) if n == 0
  m = power(a, n >> 1, mod)
  m = (m * m).map{|i| i % mod}
  return m if n & 1 == 0
  (m * a).map{|i| i % mod}
end
def f(m, n)
  ary0 = Array.new(m, 0)
  ary0[0] = 1
  v = Vector.elements(ary0)
  ary1 = [Array.new(m, 1)]
  (0..m - 2).each{|i|
    ary2 = Array.new(m, 0)
    ary2[i] = 1
    ary1 << ary2
  }
  a = Matrix[*ary1]
  mod = n
  (power(a, n, mod) * v)[m - 1]
end
def a(n)
  (1..n).select{|i| f(3, i) == 0}
end

Crossrefs

Cf. A000073, A023172, A106279.

Sequence in context: A232724 A260409 A257509 * A029522 A033309 A114435

Adjacent sequences: A232567 A232568 A232569 * A232571 A232572 A232573

Keyword

nonn

Author

Seiichi Manyama, Jun 17 2016

Status

approved