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A000350
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Numbers n such that Fibonacci(n) ends with n.
(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
(list; graph; refs; listen; history; internal format)
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OFFSET
| 1,3
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COMMENTS
| Conjecture: Other than 1 and 5, there is no n such that Fibonacci(n) in binary ends with n in binary. The conjecture holds up to n=50000. - R Stephan, Aug 21 2006
The conjecture for binary numbers holds for n < 2^25. - T. D. Noe, May 14 2007
Conjecture is true. It is easy to prove (by induction on k) that if F(n) ends with n in binary, then n == 0, 1, or 5 modulo 3*2^k for any positive integer k, i.e., n must simply be equal to 0, 1, or 5. [From Max Alekseyev (maxale(AT)gmail.com), Jul 03 2009]
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REFERENCES
| M. Dunton and R. E. Grimm, Fibonacci on Egyptian fractions, Fib. Quart., 4 (1966), 339-354.
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
| T. D. Noe, Table of n, a(n) for n=1..803
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EXAMPLE
| Fibonacci(25) = 75025 ends with 25.
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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]
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CROSSREFS
| Cf. A000045, A050816, A038546, A052000, A023172.
Sequence in context: A036137 A070380 A068574 * A000221 A018612 A036127
Adjacent sequences: A000347 A000348 A000349 * A000351 A000352 A000353
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KEYWORD
| nonn,base,easy,nice
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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EXTENSIONS
| Mathematica program edited and changed by Harvey P. Dale, Sep 10 2011
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