A182625 Numbers n for which Fibonacci(n) mod n is a Fibonacci number.

1, 2, 3, 4, 5, 6, 8, 10, 11, 12, 14, 19, 20, 21, 22, 24, 25, 29, 31, 32, 33, 36, 38, 41, 42, 48, 54, 55, 56, 58, 59, 60, 61, 62, 71, 72, 76, 77, 79, 80, 82, 89, 92, 93, 95, 96, 101, 104, 105, 108, 109, 110, 118, 119, 120, 121, 122, 123, 124, 125, 131, 133, 139, 142

Offset: 1

Author

Carmine Suriano, Mar 30 2011

Keywords

• nonn
• easy

Examples

			Fibonacci(12) = 144, 144 mod 12 = 0, and 0 is a Fibonacci number. Therefore 12 is in the sequence.
Fibonacci(14) = 377, 377 mod 14 = 13, and 13 is a Fibonacci number. Therefore 14 is in the sequence.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 2241 terms from Klaus Brockhaus)

Crossrefs

Cf. A000045, A002708.

Programs

Maple

isA000045 := proc(n) local F,i; for i from 0 do F := combinat[fibonacci](i) ; if F> n then return false; elif F = n then return true; end if; end do;end proc:
isA182625 := proc(n) isA000045(combinat[fibonacci](n) mod n) ; end proc:
for n from 1 to 300 do if isA182625(n) then printf("%d,",n) ; end if; end do: # R. J. Mathar, Apr 02 2011
# second Maple program:
b:= proc(n) local r, M, p; r, M, p:=
      <<1|0>, <0|1>>, <<0|1>, <1|1>>, n;
      do if irem(p, 2, 'p')=1 then r:= r.M mod n fi;
         if p=0 then break fi; M:= M.M mod n
      od; r[1, 2]
    end:
a:= proc(n) option remember; local k;
      for k from 1+`if`(n=1, 0, a(n-1)) while (t->
         not (issqr(t+4) or issqr(t-4)))(5*b(k)^2)
      do od; k
    end:
seq(a(n), n=1..100);  # Alois P. Heinz, Nov 26 2016

Mathematica

nn=12; f=Table[Fibonacci[n], {n,0,nn}]; Select[Range[f[[-1]]], MemberQ[f, Mod[Fibonacci[#],#]]&] (* T. D. Noe, Apr 02 2011 *)

PARI

is(n)=my(k=(fibonacci(n)%n)^2);k+=(k+1)<<2; issquare(k) || issquare(k-8) \\ Charles R Greathouse IV, Jul 30 2012

Formula

{n: A002708(n) in A000045}. - R. J. Mathar, Apr 02 2011