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A000350
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Numbers m such that Fibonacci(m) ends with m.
(Formerly M3935 N1619)
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7
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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
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graph;
refs;
listen;
history;
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OFFSET
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1,3
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COMMENTS
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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
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REFERENCES
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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).
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LINKS
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EXAMPLE
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Fibonacci(25) = 75025 ends with 25.
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MATHEMATICA
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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 *)
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PROG
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(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
(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
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CROSSREFS
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KEYWORD
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nonn,base,easy,nice
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AUTHOR
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STATUS
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approved
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