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Search: id:A003714
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| A003714 |
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Fibbinary numbers: if n = F_i1+F_i2+...+F_ik is the Zeckendorf representation of n (i.e. write n in Fibonacci number system) then a(n) = 2^{i1-2}+2^{i2-2}+...+2^{ik-2}. |
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+0 56
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| 0, 1, 2, 4, 5, 8, 9, 10, 16, 17, 18, 20, 21, 32, 33, 34, 36, 37, 40, 41, 42, 64, 65, 66, 68, 69, 72, 73, 74, 80, 81, 82, 84, 85, 128, 129, 130, 132, 133, 136, 137, 138, 144, 145, 146, 148, 149, 160, 161, 162, 164, 165, 168, 169, 170, 256, 257, 258, 260
(list; graph; listen)
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OFFSET
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0,3
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COMMENT
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The name "Fibbinary" is due to Marc LeBrun (mlb(AT)well.com)
"... integers whose binary representation contains no consecutive ones and noticed that the number of such numbers with n bits was fibonacci(n)" posting to sci.math by Bob Jenkins (bob_jenkins(AT)burtleburtle.net) Jul 17 2002.
n is in the sequence if and only if C(3n,2n) is odd; also a(n) (mod 2) = A003849(n) - Benoit Cloitre, Mar 8 2003
Numbers m such that m XOR 2*m = 3*m. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), May 03 2005. This implies that A003188(2*a(n)) = 3*a(n) holds for all n.
A116361(a(n)) <= 1. - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Feb 04 2006
Numbers n>=1 such that base 2 representation contains no block of 1's longer than 1. For example, n=17 in binary 10001 belongs to the sequence, but n=19 in binary 10011 is not in the sequence as it contains 11 which is a block of 1's longer then 1. - Ctibor O. Zizka (ctibor.zizka(AT)seznam.cz), May 13 2008
n is in the sequence if and only if the central Stirling number of the second kind S(2n,n)=A007820(n) is odd. [From O-Yeat Chan (math(AT)oyeat.com), Sep 03 2009]
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REFERENCES
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O-Yeat Chan, http://www.oyeat.com/papers/stirling9.pdf [From O-Yeat Chan (math(AT)oyeat.com), Sep 03 2009]
D. E. Knuth, Art of Comp. Programming, Vol. 1, 2nd ed., pp. 85, 493.
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LINKS
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T. D. Noe, Table of n, a(n) for n=0..1363
Joerg Arndt, Fxtbook
J.-P. Allouche, J. Shallit and G. Skordev, Self-generating sets, integers with missing blocks and substitutions, Discrete Math. 292 (2005) 1-15.
R. Knott, Rabbit Sequence in Zeckendorf Expansion (A003714)
Index entries for sequences defined by congruent products between domains N and GF(2)[X]
Index entries for sequences defined by congruent products under XOR
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FORMULA
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No two adjacent 1's in binary expansion.
Let f(x) := sum(x^Fibbinary(n), n, 0, inf). Then f satisfies the functional equation f(x) = x f(x^4) + f(x^2).
a(0)=0, a(1)=1, a(2)=2, a(n) = 2^(A072649(n)-1) + a(n - A000045(1+A072649(n))) - Antti Karttunen
It appears that sequence gives n such that A082759(3n) is odd; or, probably equivalently, n such that A037011(3n)=1 - Benoit Cloitre (benoit7848c(AT)orange.fr), Jun 20 2003
If n is in the sequence then so are 2n and 4n+1. - Henry Bottomley (se16(AT)btinternet.com), Jan 11 2005
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MAPLE
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with(combinat, fibonacci); A003714 := proc(n) option remember; if(n < 3) then RETURN(n); else RETURN((2^(A072649(n)-1))+A003714(n-fibonacci(1+A072649(n)))); fi; end;
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MATHEMATICA
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f[n_Integer] := Block[{k = Ceiling[ Log[ GoldenRatio, n*Sqrt[5]]], t = n, fr = {}}, While[k > 1, If[t >= Fibonacci[k], AppendTo[fr, 1]; t = t - Fibonacci[k], AppendTo[fr, 0]]; k-- ]; FromDigits[ fr, 2]]; Table[ f[n], {n, 0, 61}] (from Robert G. Wilson v Sep 18 2004)
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CROSSREFS
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A007088(a(n)) = A014417(n) (same sequence in binary). Complement: A004780. Char. function: A085357. Even terms: A022340, Odd terms: A022341.
Other sequences based on similar restrictions on binary expansion: A003754, A048715, A048718, A107907, A107909.
Cf. A000045, A005203, A005590, A007895, A037011, A048728, A048679, A056017, A060112, A072649, A083368, A089939, A106027, A106028, A116361.
3*a(n) is in A001969.
Sequence in context: A045702 A005658 A166021 this_sequence A010402 A010443 A035269
Adjacent sequences: A003711 A003712 A003713 this_sequence A003715 A003716 A003717
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KEYWORD
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nonn,nice
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AUTHOR
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N. J. A. Sloane (njas(AT)research.att.com).
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EXTENSIONS
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Edited Feb 21 2006 by Antti Karttunen (His-Firstname.His-Surname(AT)gmail.com).
Cross reference to A007820 added (into O-Y.C. comment) by Jason Kimberley (Jason.Kimberley(AT)newcastle.edu.au), Sep 14 2009
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