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

Author

Dean Hickerson, Oct 18 2001

Keywords

• nonn
• easy

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

Crossrefs

Complement of A064362. Cf. A000204.

Cf. A001177, A053032, A140409.

Programs

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 *)

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

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