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A065156
Numbers n such that some Lucas number (A000204) is divisible by n.
4
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.
Sequence in context: A173254 A015851 A225529 * A097987 A049149 A332555
KEYWORD
nonn,easy
AUTHOR
Dean Hickerson, Oct 18 2001
STATUS
approved