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

Author

Seiichi Manyama, Jun 17 2016

Keywords

• nonn

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

Crossrefs

Cf. A000073, A023172, A106279.

Programs

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 *)

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