A065156 Numbers n such that some Lucas number (A000204) is divisible by n.

1, 2, 3, 4, 6, 7, 9, 11, 14, 18, 19, 22, 23, 27, 29, 31, 38, 41, 43, 44, 46, 47, 49, 54, 58, 59, 62, 67, 71, 76, 79, 81, 82, 83, 86, 94, 98, 101, 103, 107, 116, 118, 121, 123, 124, 127, 129, 131, 134, 139, 142, 151, 158, 161, 162, 163, 166, 167, 179, 181, 191, 199

Offset

1,2

Comments

From A.H.M. Smeets, Sep 20 2020 (Start)

For the Fibonacci numbers, each natural number divides some Fibonacci number (see A001177).

If, for some number m, m divides some Lucas number L_i (=A000204(i)), then, the smallest i satisfies i <= m. (End)

Links

A.H.M. Smeets, Table of n, a(n) for n = 1..20000 (terms 1..1000 from T. D. Noe)

B. Avila and T. Khovanova, Free Fibonacci Sequences, arXiv preprint arXiv:1403.4614 [math.NT], 2014 and J. Int. Seq. 17 (2014) # 14.8.5

Formula

Equals {1,2,4} union {p^e | p in A140409 and e > 0} union {2*p^e | p in A140409 and e > 0} union {4*p | p in A053032} union {4*p*q | p, q in A053032}. - A.H.M. Smeets, Sep 20 2020

Mathematica

test[ n_ ] := For[ a=1; b=3, True, t=b; b=Mod[ a+b, n ]; a=t, If[ b==0, Return[ True ] ]; If[ a==2&&b==1, Return[ False ] ] ]; Select[ Range[ 200 ], test ]
Take[Flatten[Divisors/@LucasL[Range[200]]]//Union, 70] (* Harvey P. Dale, Jun 07 2020 *)

Prog

(Python)
a, n = 0, 0
while n < 1000:
    a, f0, f1, i = a+1, 1, 2, 1
    if f1%a == 0:
        n = n+1
        print(n, a)
    else:
        while f0%a != 0 and i <= a:
            f0, f1, i = f0+f1, f0, i+1
        if i <= a:
            n = n+1
            print(n, a) # A.H.M. Smeets, Sep 20 2020

Crossrefs

Complement of A064362. Cf. A000204.

Cf. A001177, A053032, A140409.

Sequence in context: A173254 A015851 A225529 * A097987 A049149 A332555

Adjacent sequences: A065153 A065154 A065155 * A065157 A065158 A065159

Keyword

nonn,easy

Author

Dean Hickerson, Oct 18 2001

Status

approved