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A005652
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Lexicographically least increasing sequence, starting with 1, such that the sum of 2 distinct terms is never a Fibonacci number.
(Formerly M2517)
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25
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1, 3, 6, 8, 9, 11, 14, 16, 17, 19, 21, 22, 24, 27, 29, 30, 32, 35, 37, 40, 42, 43, 45, 48, 50, 51, 53, 55, 56, 58, 61, 63, 64, 66, 69, 71, 74, 76, 77, 79, 82, 84, 85, 87, 90, 92, 95, 97, 98, 100, 103, 105, 106, 108, 110, 111, 113, 116, 118, 119, 121, 124, 126, 129, 131
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OFFSET
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1,2
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COMMENTS
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Also, k such that k = 2*ceiling(k*phi) - ceiling(k*sqrt(5)) where phi = (1+sqrt(5))/2. - Benoit Cloitre, Dec 05 2002
The Chow-Long paper gives a connection with continued fractions, as well as generalizations and other references for this and related sequences.
Also positive integers k such that {k*r} > 1/2, where r = golden ratio = (1 + sqrt(5))/2 and { } = fractional part. - Clark Kimberling and Jianing Song, Sep 12 2019
The lexicographically least property can be proved with the Walnut theorem prover. - Jeffrey Shallit, Nov 20 2023
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REFERENCES
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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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FORMULA
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The set of all k such that the integer multiple of (1+sqrt(5))/2 nearest k is greater than k (Chow-Long).
Numbers k such that 2*{k*phi} - {2k*phi} = 1, where { } denotes fractional part. - Clark Kimberling, Jan 01 2007
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MATHEMATICA
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f[n_] := Block[{k = Floor[n/GoldenRatio]}, If[n - k*GoldenRatio > (k + 1)*GoldenRatio - n, 1, 0]]; Select[ Range[131], f[ # ] == 1 &]
r = (1 + Sqrt[5])/2; z = 300;
t = Table[Floor[2 n*r] - 2 Floor[n*r], {n, 1, z}] (* {A078588(n) : n > 0} *)
Flatten[Position[t, 0]] (* A005653 *)
Flatten[Position[t, 1]] (* this sequence *)
r = GoldenRatio;
t = Table[If[FractionalPart[n*r] < 1/2, 0, 1 ], {n, 1, 120}] (* {A078588(n) : n > 0} *)
Flatten[Position[t, 0]] (* A005653 *)
Flatten[Position[t, 1]] (* this sequence *)
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CROSSREFS
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KEYWORD
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nonn,easy,nice
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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