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
KEYWORD
nonn,easy
AUTHOR
Dean Hickerson, Oct 18 2001
STATUS
approved