A000350 Numbers m such that Fibonacci(m) ends with m.

0, 1, 5, 25, 29, 41, 49, 61, 65, 85, 89, 101, 125, 145, 149, 245, 265, 365, 385, 485, 505, 601, 605, 625, 649, 701, 725, 745, 749, 845, 865, 965, 985, 1105, 1205, 1249, 1345, 1445, 1585, 1685, 1825, 1925, 2065, 2165, 2305, 2405, 2501, 2545, 2645, 2785, 2885

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• base
• easy
• nice

Comments

Conjecture: Other than 1 and 5, there is no m such that Fibonacci(m) in binary ends with m in binary. The conjecture holds up to m=50000. - Ralf Stephan, Aug 21 2006

The conjecture for binary numbers holds for m < 2^25. - T. D. Noe, May 14 2007

Conjecture is true. It is easy to prove (by induction on k) that if Fibonacci(m) ends with m in binary, then m == 0, 1, or 5 (mod 3*2^k) for any positive integer k, i.e., m must simply be equal to 0, 1, or 5. - Max Alekseyev, Jul 03 2009

Examples

			Fibonacci(25) = 75025 ends with 25.
		

References

• N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..1034 (terms n = 1..803 from T. D. Noe)
• G. R. Deily, Terminal Digit Coincidences Between Fibonacci Numbers and Their Indices, The Fibonacci Quarterly, 4.2 (1966) 151.
• M. Dunton and R. E. Grimm, Fibonacci on Egyptian fractions, Fib. Quart., 4 (1966), 339-354.
• D. A. Lind, Extended Computations of Terminal Digit Coincidences, Fibonacci Quarterly, 5.2 April 1967 pp. 183-184.

Crossrefs

Cf. A000045, A050816, A038546, A052000, A023172.

Programs

Haskell

import Data.List (isSuffixOf, elemIndices)
import Data.Function (on)
a000350 n = a000350_list !! (n-1)
a000350_list = elemIndices True $
               zipWith (isSuffixOf `on` show) [0..] a000045_list
-- Reinhard Zumkeller, Apr 10 2012

Mathematica

a=0;b=1;c=1;lst={}; Do[a=b;b=c;c=a+b;m=Floor[N[Log[10,n]]]+1; If[Mod[c,10^m]==n,AppendTo[lst,n]],{n,3,5000}]; Join[{0,1},lst] (* edited and changed by Harvey P. Dale, Sep 10 2011 *)
fnQ[n_]:=Mod[Fibonacci[n],10^IntegerLength[n]]==n; Select[Range[ 0,2900],fnQ] (* Harvey P. Dale, Nov 03 2012 *)

PARI

for(n=0,1e4,if(((Mod([1,1;1,0],10^#Str(n)))^n)[1,2]==n,print1(n", "))) \\ Charles R Greathouse IV, Apr 10 2012

Python

from sympy import fibonacci
[i for i in range(1000) if str(fibonacci(i))[-len(str(i)):]==str(i)] # Nicholas Stefan Georgescu, Feb 27 2023