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Numbers n such that some Lucas number (A000204) is divisible by n.
4

%I #36 Oct 05 2020 01:04:32

%S 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,

%T 59,62,67,71,76,79,81,82,83,86,94,98,101,103,107,116,118,121,123,124,

%U 127,129,131,134,139,142,151,158,161,162,163,166,167,179,181,191,199

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

%C From _A.H.M. Smeets_, Sep 20 2020 (Start)

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

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

%H A.H.M. Smeets, <a href="/A065156/b065156.txt">Table of n, a(n) for n = 1..20000</a> (terms 1..1000 from T. D. Noe)

%H B. Avila and T. Khovanova, <a href="http://arxiv.org/abs/1403.4614">Free Fibonacci Sequences</a>, arXiv preprint arXiv:1403.4614 [math.NT], 2014 and <a href="https://cs.uwaterloo.ca/journals/JIS/VOL17/Avila/avila4.html">J. Int. Seq. 17 (2014) # 14.8.5</a>

%F 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

%t 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 ]

%t Take[Flatten[Divisors/@LucasL[Range[200]]]//Union,70] (* _Harvey P. Dale_, Jun 07 2020 *)

%o (Python)

%o a, n = 0, 0

%o while n < 1000:

%o a, f0, f1, i = a+1, 1, 2, 1

%o if f1%a == 0:

%o n = n+1

%o print(n,a)

%o else:

%o while f0%a != 0 and i <= a:

%o f0, f1, i = f0+f1, f0, i+1

%o if i <= a:

%o n = n+1

%o print(n,a) # _A.H.M. Smeets_, Sep 20 2020

%Y Complement of A064362. Cf. A000204.

%Y Cf. A001177, A053032, A140409.

%K nonn,easy

%O 1,2

%A _Dean Hickerson_, Oct 18 2001