A048715 -id:A048715 - cp's OEIS frontend

A000930 Narayana's cows sequence: a(0) = a(1) = a(2) = 1; thereafter a(n) = a(n-1) + a(n-3).

1, 1, 1, 2, 3, 4, 6, 9, 13, 19, 28, 41, 60, 88, 129, 189, 277, 406, 595, 872, 1278, 1873, 2745, 4023, 5896, 8641, 12664, 18560, 27201, 39865, 58425, 85626, 125491, 183916, 269542, 395033, 578949, 848491, 1243524, 1822473, 2670964, 3914488, 5736961, 8407925

Offset: 0

N. J. A. Sloane

Named after a 14th-century Indian mathematician. [The sequence first appeared in the book "Ganita Kaumudi" (1356) by the Indian mathematician Narayana Pandita (c. 1340 - c. 1400). - Amiram Eldar, Apr 15 2021]
Number of compositions of n into parts 1 and 3. - Joerg Arndt, Jun 25 2011
A Lamé sequence of higher order.
Could have begun 1,0,0,1,1,1,2,3,4,6,9,... (A078012) but that would spoil many nice properties.
Number of tilings of a 3 X n rectangle with straight trominoes.
Number of ways to arrange n-1 tatami mats in a 2 X (n-1) room such that no 4 meet at a point. For example, there are 6 ways to cover a 2 X 5 room, described by 11111, 2111, 1211, 1121, 1112, 212.
Equivalently, number of compositions (ordered partitions) of n-1 into parts 1 and 2 with no two 2's adjacent. E.g., there are 6 such ways to partition 5, namely 11111, 2111, 1211, 1121, 1112, 212, so a(6) = 6. [Minor edit by Keyang Li, Oct 10 2020]
This comment covers a family of sequences which satisfy a recurrence of the form a(n) = a(n-1) + a(n-m), with a(n) = 1 for n = 0...m-1. The generating function is 1/(1-x-x^m). Also a(n) = Sum_{i=0..floor(n/m)} binomial(n-(m-1)*i, i). This family of binomial summations or recurrences gives the number of ways to cover (without overlapping) a linear lattice of n sites with molecules that are m sites wide. Special case: m=1: A000079; m=4: A003269; m=5: A003520; m=6: A005708; m=7: A005709; m=8: A005710.
a(n+2) is the number of n-bit 0-1 sequences that avoid both 00 and 010. - David Callan, Mar 25 2004 [This can easily be proved by the Cluster Method - see for example the Noonan-Zeilberger article. - N. J. A. Sloane, Aug 29 2013]
a(n-4) is the number of n-bit sequences that start and end with 0 but avoid both 00 and 010. For n >= 6, such a sequence necessarily starts 011 and ends 110; deleting these 6 bits is a bijection to the preceding item. - David Callan, Mar 25 2004
Also number of compositions of n+1 into parts congruent to 1 mod m. Here m=3, A003269 for m=4, etc. - Vladeta Jovovic, Feb 09 2005
Row sums of Riordan array (1/(1-x^3), x/(1-x^3)). - Paul Barry, Feb 25 2005
Row sums of Riordan array (1,x(1+x^2)). - Paul Barry, Jan 12 2006
Starting with offset 1 = row sums of triangle A145580. - Gary W. Adamson, Oct 13 2008
Number of digits in A061582. - Dmitry Kamenetsky, Jan 17 2009
From Jon Perry, Nov 15 2010: (Start)
The family a(n) = a(n-1) + a(n-m) with a(n)=1 for n=0..m-1 can be generated by considering the sums (A102547):
1 1 1 1 1 1 1 1 1  1  1  1  1
      1 2 3 4 5 6  7  8  9  10
            1 3 6  10 15 21 28
                   1  4  10 20
                            1
------------------------------
1 1 1 2 3 4 6 9 13 19 28 41 60
with (in this case 3) leading zeros added to each row.
(End)
Number of pairs of rabbits existing at period n generated by 1 pair. All pairs become fertile after 3 periods and generate thereafter a new pair at all following periods. - Carmine Suriano, Mar 20 2011
The compositions of n in which each natural number is colored by one of p different colors are called p-colored compositions of n. For n>=3, 2*a(n-3) equals the number of 2-colored compositions of n with all parts >= 3, such that no adjacent parts have the same color. - Milan Janjic, Nov 27 2011
For n>=2, row sums of Pascal's triangle (A007318) with triplicated diagonals. - Vladimir Shevelev, Apr 12 2012
Pisano period lengths of the sequence read mod m, m >= 1: 1, 7, 8, 14, 31, 56, 57, 28, 24, 217, 60, 56, 168, ... (A271953) If m=3, for example, the remainder sequence becomes 1, 1, 1, 2, 0, 1, 0, 0, 1, 1, 1, 2, 0, 1, 0, 0, 1, 1, 1, 2, 0, 1, 0, 0, 1, 1, 1, 2, 0, 1, 0, 0, 1, 1, ... with a period of length 8. - R. J. Mathar, Oct 18 2012
Diagonal sums of triangle A011973. - John Molokach, Jul 06 2013
"In how many ways can a kangaroo jump through all points of the integer interval [1,n+1] starting at 1 and ending at n+1, while making hops that are restricted to {-1,1,2}? (The OGF is the rational function 1/(1 - z - z^3) corresponding to A000930.)" [Flajolet and Sedgewick, p. 373] - N. J. A. Sloane, Aug 29 2013
a(n) is the number of length n binary words in which the length of every maximal run of consecutive 0's is a multiple of 3. a(5) = 4 because we have: 00011, 10001, 11000, 11111. - Geoffrey Critzer, Jan 07 2014
a(n) is the top left entry of the n-th power of the 3X3 matrix [1, 0, 1; 1, 0, 0; 0, 1, 0] or of the 3 X 3 matrix [1, 1, 0; 0, 0, 1; 1, 0, 0]. - R. J. Mathar, Feb 03 2014
a(n-3) is the top left entry of the n-th power of any of the 3 X 3 matrices [0, 1, 0; 0, 1, 1; 1, 0, 0], [0, 0, 1; 1, 1, 0; 0, 1, 0], [0, 1, 0; 0, 0, 1; 1, 0, 1] or [0, 0, 1; 1, 0, 0; 0, 1, 1]. - R. J. Mathar, Feb 03 2014
Counts closed walks of length (n+3) on a unidirectional triangle, containing a loop at one of remaining vertices. - David Neil McGrath, Sep 15 2014
a(n+2) equals the number of binary words of length n, having at least two zeros between every two successive ones. - Milan Janjic, Feb 07 2015
a(n+1)/a(n) tends to x = 1.465571... (decimal expansion given in A092526) in the limit n -> infinity. This is the real solution of x^3 - x^2 -1 = 0. See also the formula by Benoit Cloitre, Nov 30 2002. - Wolfdieter Lang, Apr 24 2015
a(n+2) equals the number of subsets of {1,2,..,n} in which any two elements differ by at least 3. - Robert FERREOL, Feb 17 2016
Let T* be the infinite tree with root 0 generated by these rules: if p is in T*, then p+1 is in T* and x*p is in T*. Let g(n) be the set of nodes in the n-th generation, so that g(0) = {0}, g(1) = {1}, g(2) = {2,x}, g(3) = {3,2x,x+1,x^2}, etc. Let T(r) be the tree obtained by substituting r for x. If a positive integer N such that r = N^(1/3) is not an integer, then the number of (not necessarily distinct) integers in g(n) is A000930(n), for n >= 1. (See A274142.) - Clark Kimberling, Jun 13 2016
a(n-3) is the number of compositions of n excluding 1 and 2, n >= 3. - Gregory L. Simay, Jul 12 2016
Antidiagonal sums of array A277627. - Paul Curtz, May 16 2019
a(n+1) is the number of multus bitstrings of length n with no runs of 3 ones. - Steven Finch, Mar 25 2020
Suppose we have a(n) samples, exactly one of which is positive. Assume the cost for testing a mix of k samples is 3 if one of the samples is positive (but you will not know which sample was positive if you test more than 1) and 1 if none of the samples is positive. Then the cheapest strategy for finding the positive sample is to have a(n-3) undergo the first test and then continue with testing either a(n-4) if none were positive or with a(n-6) otherwise. The total cost of the tests will be n. - Ruediger Jehn, Dec 24 2020

			The number of compositions of 11 without any 1's and 2's is a(11-3) = a(8) = 13. The compositions are (11), (8,3), (3,8), (7,4), (4,7), (6,5), (5,6), (5,3,3), (3,5,3), (3,3,5), (4,4,3), (4,3,4), (3,4,4). - _Gregory L. Simay_, Jul 12 2016
The compositions from the above example may be mapped to the a(8) compositions of 8 into 1's and 3's using this (more generally applicable) method: replace all numbers greater than 3 with a 3 followed by 1's to make the same total, then remove the initial 3 from the composition. Maintaining the example's order, they become (1,1,1,1,1,1,1,1), (1,1,1,1,1,3), (3,1,1,1,1,1), (1,1,1,1,3,1), (1,3,1,1,1,1), (1,1,1,3,1,1), (1,1,3,1,1,1), (1,1,3,3), (3,1,1,3), (3,3,1,1), (1,3,1,3), (1,3,3,1), (3,1,3,1). - _Peter Munn_, May 31 2017

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Index entries for linear recurrences with constant coefficients, signature (1,0,1).

For Lamé sequences of orders 1 through 9 see A000045, this sequence, and A017898 - A017904.
Cf. A000073, A000213, A048715, A069241, A092526, A170954.
See also A000079, A003269, A003520, A005708, A005709, A005710.
Essentially the same as A068921 and A078012.
See also A001609, A145580, A179070, A214551 (same rule except divide by GCD).
A271901 and A271953 give the period of this sequence mod n.
Cf. A007318, A060576, A102547, A277627, A289207.
A120562 has the same recurrence for odd n.

a:=[1,1,1];; for n in [4..50] do a[n]:=a[n-1]+a[n-3]; od; a; # Muniru A Asiru, Aug 13 2018

a000930 n = a000930_list !! n
a000930_list = 1 : 1 : 1 : zipWith (+) a000930_list (drop 2 a000930_list)
-- Reinhard Zumkeller, Sep 25 2011

[1,1] cat [ n le 3 select n else Self(n-1)+Self(n-3): n in [1..50] ]; // Vincenzo Librandi, Apr 25 2015

f := proc(r) local t1,i; t1 := []; for i from 1 to r do t1 := [op(t1),0]; od: for i from 1 to r+1 do t1 := [op(t1),1]; od: for i from 2*r+2 to 50 do t1 := [op(t1),t1[i-1]+t1[i-1-r]]; od: t1; end; # set r = order
with(combstruct): SeqSetU := [S, {S=Sequence(U), U=Set(Z, card > 2)}, unlabeled]: seq(count(SeqSetU, size=j), j=3..40); # Zerinvary Lajos, Oct 10 2006
A000930 := proc(n)
    add(binomial(n-2*k,k),k=0..floor(n/3)) ;
end proc: # Zerinvary Lajos, Apr 03 2007
a:= n-> (<<1|1|0>, <0|0|1>, <1|0|0>>^n)[1,1]:
seq(a(n), n=0..50); # Alois P. Heinz, Jun 20 2008

a[0] = 1; a[1] = a[2] = 1; a[n_] := a[n] = a[n - 1] + a[n - 3]; Table[ a[n], {n, 0, 40} ]
CoefficientList[Series[1/(1 - x - x^3), {x, 0, 45}], x] (* Zerinvary Lajos, Mar 22 2007 *)
LinearRecurrence[{1, 0, 1}, {1, 1, 1}, 80] (* Vladimir Joseph Stephan Orlovsky, Feb 11 2012 *)
a[n_] := HypergeometricPFQ[{(1 - n)/3, (2 - n)/3, -n/3}, {(1 - n)/ 2, -n/2}, -27/4]; Table[a[n], {n, 0, 43}] (* Jean-François Alcover, Feb 26 2013 *)
Table[-RootSum[1 + #^2 - #^3 &, 3 #^(n + 2) - 11 #^(n + 3) + 2 #^(n + 4) &]/31, {n, 20}] (* Eric W. Weisstein, Feb 14 2025 *)

makelist(sum(binomial(n-2*k,k),k,0,n/3),n,0,18); /* Emanuele Munarini, May 24 2011 */

a(n)=polcoeff(exp(sum(m=1,n,((1+sqrt(1+4*x))^m + (1-sqrt(1+4*x))^m)*(x/2)^m/m)+x*O(x^n)),n) \\ Paul D. Hanna, Oct 08 2009

x='x+O('x^66); Vec(1/(1-(x+x^3))) \\ Joerg Arndt, May 24 2011

a(n)=([0,1,0;0,0,1;1,0,1]^n*[1;1;1])[1,1] \\ Charles R Greathouse IV, Feb 26 2017

from itertools import islice
def A000930_gen(): # generator of terms
    blist = [1]*3
    while True:
        yield blist[0]
        blist = blist[1:]+[blist[0]+blist[2]]
A000930_list = list(islice(A000930_gen(),30)) # Chai Wah Wu, Feb 04 2022

@CachedFunction
def a(n): # A000930
    if (n<3): return 1
    else: return a(n-1) + a(n-3)
[a(n) for n in (0..80)] # G. C. Greubel, Jul 29 2022

G.f.: 1/(1-x-x^3). - Simon Plouffe in his 1992 dissertation
a(n) = Sum_{i=0..floor(n/3)} binomial(n-2*i, i).
a(n) = a(n-2) + a(n-3) + a(n-4) for n>3.
a(n) = floor(d*c^n + 1/2) where c is the real root of x^3-x^2-1 and d is the real root of 31*x^3-31*x^2+9*x-1 (c = 1.465571... = A092526 and d = 0.611491991950812...). - Benoit Cloitre, Nov 30 2002
a(n) = Sum_{k=0..n} binomial(floor((n+2k-2)/3), k). - Paul Barry, Jul 06 2004
a(n) = Sum_{k=0..n} binomial(k, floor((n-k)/2))(1+(-1)^(n-k))/2. - Paul Barry, Jan 12 2006
a(n) = Sum_{k=0..n} binomial((n+2k)/3,(n-k)/3)*(2*cos(2*Pi*(n-k)/3)+1)/3. - Paul Barry, Dec 15 2006
a(n) = term (1,1) in matrix [1,1,0; 0,0,1; 1,0,0]^n. - Alois P. Heinz, Jun 20 2008
G.f.: exp( Sum_{n>=1} ((1+sqrt(1+4*x))^n + (1-sqrt(1+4*x))^n)*(x/2)^n/n ).
Logarithmic derivative equals A001609. - Paul D. Hanna, Oct 08 2009
a(n) = a(n-1) + a(n-2) - a(n-5) for n>4. - Paul Weisenhorn, Oct 28 2011
For n >= 2, a(2*n-1) = a(2*n-2)+a(2*n-4); a(2*n) = a(2*n-1)+a(2*n-3). - Vladimir Shevelev, Apr 12 2012
INVERT transform of (1,0,0,1,0,0,1,0,0,1,...) = (1, 1, 1, 2, 3, 4, 6, ...); but INVERT transform of (1,0,1,0,0,0,...) = (1, 1, 2, 3, 4, 6, ...). - Gary W. Adamson, Jul 05 2012
G.f.: 1/(G(0)-x) where G(k) = 1 - x^2/(1 - x^2/(x^2 - 1/G(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Dec 16 2012
G.f.: 1 + x/(G(0)-x) where G(k) = 1 - x^2*(2*k^2 + 3*k +2) + x^2*(k+1)^2*(1 - x^2*(k^2 + 3*k +2))/G(k+1); (continued fraction). - Sergei N. Gladkovskii, Dec 27 2012
a(2*n) = A002478(n), a(2*n+1) = A141015(n+1), a(3*n) = A052544(n), a(3*n+1) = A124820(n), a(3*n+2) = A052529(n+1). - Johannes W. Meijer, Jul 21 2013, corrected by Greg Dresden, Jul 06 2020
G.f.: Q(0)/2, where Q(k) = 1 + 1/(1 - x*(4*k+1 + x^2)/( x*(4*k+3 + x^2) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Sep 08 2013
a(n) = v1*w1^n+v3*w2^n+v2*w3^n, where v1,2,3 are the roots of (-1+9*x-31*x^2+31*x^3): [v1=0.6114919920, v2=0.1942540040 - 0.1225496913*I, v3=conjugate(v2)] and w1,2,3 are the roots of (-1-x^2+x^3): [w1=1.4655712319, w2=-0.2327856159 - 0.7925519925*I, w3=conjugate(w2)]. - Gerry Martens, Jun 27 2015
a(n) = (6*A001609(n+3) + A001609(n-7))/31 for n>=7. - Areebah Mahdia, Jun 07 2020
a(n+6)^2 + a(n+1)^2 + a(n)^2 = a(n+5)^2 + a(n+4)^2 + 3*a(n+3)^2 + a(n+2)^2. - Greg Dresden, Jul 07 2021
a(n) = Sum_{i=(n-7)..(n-1)} a(i) / 2. - Jules Beauchamp, May 10 2025

Name expanded by N. J. A. Sloane, Sep 07 2012

A003714 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). Also numbers whose binary representation contains no two adjacent 1's.

Original entry on oeis.org

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, 261, 264

Offset: 0

N. J. A. Sloane

The name "Fibbinary" is due to Marc LeBrun.
"... 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]
From Benoit Cloitre, Mar 08 2003: (Start)
A number m is in the sequence if and only if C(3m, m) (or equally, C(3m, 2m)) is odd.
a(n) == A003849(n) (mod 2). (End)
Numbers m such that m XOR 2*m = 3*m. - Reinhard Zumkeller, May 03 2005. [This implies that A003188(2*a(n)) = 3*a(n) holds for all n.]
Numbers whose base-2 representation contains no two adjacent ones. For example, m = 17 = 10001_2 belongs to the sequence, but m = 19 = 10011_2 does not. - Ctibor O. Zizka, May 13 2008
m is in the sequence if and only if the central Stirling number of the second kind S(2*m, m) = A007820(m) is odd. - O-Yeat Chan (math(AT)oyeat.com), Sep 03 2009
A000120(3*a(n)) = 2*A000120(a(n)); A002450 is a subsequence.
Every nonnegative integer can be expressed as the sum of two terms of this sequence. - Franklin T. Adams-Watters, Jun 11 2011
Subsequence of A213526. - Arkadiusz Wesolowski, Jun 20 2012
This is also the union of A215024 and A215025 - see the Comment in A014417. - N. J. A. Sloane, Aug 10 2012
The binary representation of each term m contains no two adjacent 1's, so we have (m XOR 2m XOR 3m) = 0, and thus a two-player Nim game with three heaps of (m, 2m, 3m) stones is a losing configuration for the first player. - V. Raman, Sep 17 2012
Positions of zeros in A014081. - John Keith, Mar 07 2022
These numbers are similar to Fibternary numbers A003726, Tribbinary numbers A060140 and Tribternary numbers. This sequence is a subsequence of Fibternary numbers A003726. The number of Fibbinary numbers less than any power of two is a Fibonacci number. We can generate this sequence recursively: start with 0 and 1; then, if x is in the sequence add 2x and 4x+1 to the sequence. The Fibbinary numbers have the property that the n-th Fibbinary number is even if the n-th term of the Fibonacci word is a. Respectively, the n-th Fibbinary number is odd (of the form 4x+1) if the n-th term of the Fibonacci word is b. Every number has a Fibbinary multiple. - Tanya Khovanova and PRIMES STEP Senior, Aug 30 2022
This is the ordered set S of numbers defined recursively by: 0 is in S; if x is in S, then 2*x and 4*x + 1 are in S. See Kimberling (2006) Example 3, in references below. - Harry Richman, Jan 31 2024

			From _Joerg Arndt_, Jun 11 2011: (Start)
In the following, dots are used for zeros in the binary representation:
  a(n)  binary(a(n))  n
    0:    .......     0
    1:    ......1     1
    2:    .....1.     2
    4:    ....1..     3
    5:    ....1.1     4
    8:    ...1...     5
    9:    ...1..1     6
   10:    ...1.1.     7
   16:    ..1....     8
   17:    ..1...1     9
   18:    ..1..1.    10
   20:    ..1.1..    11
   21:    ..1.1.1    12
   32:    .1.....    13
   33:    .1....1    14
   34:    .1...1.    15
   36:    .1..1..    16
   37:    .1..1.1    17
   40:    .1.1...    18
   41:    .1.1..1    19
   42:    .1.1.1.    20
   64:    1......    21
   65:    1.....1    22
(End)

Donald E. Knuth, The Art of Computer Programming: Fundamental Algorithms, Vol. 1, 2nd ed., Addison-Wesley, 1973, pp. 85, 493.

G. C. Greubel and T. D. Noe, Table of n, a(n) for n = 0..5000 (terms 0 to 1363 by T. D. Noe)
J.-P. Allouche, J. Shallit, and G. Skordev, Self-generating sets, integers with missing blocks and substitutions, Discrete Math., Vol. 292, No. 1-3 (2005), pp. 1-15.
Joerg Arndt, Matters Computational (The Fxtbook), pp. 74-77, pp. 754-756.
Robert Baillie and Thomas Schmelzer, Summing Kempner's Curious (Slowly-Convergent) Series, Mathematica Notebook kempnerSums.nb, Wolfram Library Archive, 2008.
Marc Chamberland and Karl Dilcher, A Binomial Sum Related to Wolstenholme's Theorem, J. Number Theory, Vol. 171, Issue 11 (Nov. 2009), pp. 2659-2672. See Lemma 4.2 p. 2668.
O-Yeat Chan and Dante Manna, Divisibility properties of Stirling numbers of the second kind.
F. Michel Dekking, Morphisms, Symbolic Sequences, and Their Standard Forms, Journal of Integer Sequences, Vol. 19 (2016), Article 16.1.1.
David Eppstein, Generating fibbinary numbers, three ways, 2021.
Clark Kimberling, Affinely recursive sets and orderings of languages, Discrete Math., Vol. 274, No. 1-3 (2004), pp. 147-160.
Roman S. Klyujkov, Quick Fibbinary Numbers Addition, C++ function with test program.
Ron Knott, Rabbit Sequence in Zeckendorf Expansion (A003714).
Linus Lindroos, Andrew Sills, and Hua Wang, Odd fibbinary numbers and the golden ratio, Fib. Q., Vol. 52, No. 1 (2014), pp. 61-65; alternative link.
A. J. Macfarlane, On the fibbinary numbers and the Wythoff array, arXiv:2405.18128 [math.CO], 2024. See pages 1, 5.
Douglas M. McKenna, Fibbinary Zippers in a Monoid of Toroidal Hamiltonian Cycles that Generate Hilbert-Style Square-Filling Curves, ECA 2:2 (2022) Article S2R13.
Joris Nieuwveld, Fractions, Functions and Folding. A Novel Link between Continued Fractions, Mahler Functions and Paper Folding, Master's Thesis, arXiv:2108.11382 [math.NT], 2021.
Yaron Shany and Amit Berman, The Generating Idempotent Is a Minimum-Weight Codeword for Some Binary BCH Codes, arXiv:2408.08218 [cs.IT], 2024. See p. 10.
N. J. A. Sloane, Table of n, a(n) (base 10), a(n) (base 2), for n = 0..1000.
Chai Wah Wu, Record values in appending and prepending bitstrings to runs of binary digits, arXiv:1810.02293 [math.NT], 2018.
Index entries for 2-automatic sequences.
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.

A007088(a(n)) = A014417(n) (same sequence in binary). Complement: A004780. Char. function: A085357. Even terms: A022340, odd terms: A022341. First difference: A129761.
Other sequences based on similar restrictions on binary expansion: A003726 & A278038, 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.
Cf. also A215022-A215025, A242407.
Cf. A014081 (count 11 bits).

import Data.Set (Set, singleton, insert, deleteFindMin)
a003714 n = a003714_list !! n
a003714_list = 0 : f (singleton 1) where
   f :: Set Integer -> [Integer]
   f s = m : (f $ insert (4*m + 1) $ insert (2*m) s')
         where (m, s') = deleteFindMin s
-- Reinhard Zumkeller, Jun 03 2012, Feb 07 2012

A003714 := proc(n)
    option remember;
    if n < 3 then
        n ;
    else
        2^(A072649(n)-1) + procname(n-combinat[fibonacci](1+A072649(n))) ;
    end if;
end proc:
seq(A003714(n),n=0..10) ;
# To produce a table giving n, a(n) (base 10), a(n) (base 2) - from N. J. A. Sloane, Sep 30 2018
# binary: binary representation of n, in human order
binary:=proc(n) local t1,L;
if n<0 then ERROR("n must be nonnegative"); fi;
if n=0 then return([0]); fi;
t1:=convert(n,base,2); L:=nops(t1);
[seq(t1[L+1-i],i=1..L)];
end;
for n from 0 to 100 do t1:=A003714(n); lprint(n, t1, binary(t1)); od:

fibBin[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[fibBin[n], {n, 0, 61}] (* Robert G. Wilson v, Sep 18 2004 *)
Select[Range[0, 270], ! MemberQ[Partition[IntegerDigits[#, 2], 2, 1], {1, 1}] &] (* Harvey P. Dale, Jul 17 2011 *)
Select[Range[256], BitAnd[#, 2 #] == 0 &] (* Alonso del Arte, Jun 18 2012 *)
With[{r = Range[10^5]}, Pick[r, BitAnd[r, 2 r], 0]] (* Eric W. Weisstein, Aug 18 2017 *)
Select[Range[0, 299], SequenceCount[IntegerDigits[#, 2], {1, 1}] == 0 &] (* Requires Mathematica version 10 or later. -- Harvey P. Dale, Dec 06 2018 *)

msb(n)=my(k=1); while(k<=n, k<<=1); k>>1
for(n=1,1e4,k=bitand(n,n<<1);if(k,n=bitor(n,msb(k)-1),print1(n", "))) \\ Charles R Greathouse IV, Jun 15 2011

select( is_A003714(n)=!bitand(n,n>>1), [0..266])
{(next_A003714(n,t)=while(t=bitand(n+=1,n<<1), n=bitor(n,1<A003714(t)) \\ M. F. Hasler, Nov 30 2021

for n in range(300):
    if 2*n & n == 0:
        print(n, end=",") # Alex Ratushnyak, Jun 21 2012

def A003714(n):
    tlist, s = [1,2], 0
    while tlist[-1]+tlist[-2] <= n:
        tlist.append(tlist[-1]+tlist[-2])
    for d in tlist[::-1]:
        s *= 2
        if d <= n:
            s += 1
            n -= d
    return s # Chai Wah Wu, Jun 14 2018

def fibbinary():
    x = 0
    while True:
        yield x
        y = ~(x >> 1)
        x = (x - y) & y # Falk Hüffner, Oct 23 2021
(C++)
/* start with x=0, then repeatedly call x=next_fibrep(x): */
ulong next_fibrep(ulong x)
{
    // 2 examples:         //  ex. 1             //  ex.2
    //                     // x == [*]0 010101   // x == [*]0 01010
    ulong y = x | (x>>1);  // y == [*]? 011111   // y == [*]? 01111
    ulong z = y + 1;       // z == [*]? 100000   // z == [*]? 10000
    z = z & -z;            // z == [0]0 100000   // z == [0]0 10000
    x ^= z;                // x == [*]0 110101   // x == [*]0 11010
    x &= ~(z-1);           // x == [*]0 100000   // x == [*]0 10000
    return x;
}
/* Joerg Arndt, Jun 22 2012 */

(0 to 255).filter(n => (n & 2 * n) == 0) // Alonso del Arte, Apr 12 2020
(C#)
public static bool IsFibbinaryNum(this int n) => ((n & (n >> 1)) == 0) ? true : false; // Frank Hollstein, Jul 07 2021

No two adjacent 1's in binary expansion.
Let f(x) := Sum_{n >= 0} x^Fibbinary(n). (This is the generating function of the characteristic function of this sequence.) 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 this sequence gives m such that A082759(3*m) is odd; or, probably equivalently, m such that A037011(3*m) = 1. - Benoit Cloitre, Jun 20 2003
If m is in the sequence then so are 2*m and 4*m + 1. - Henry Bottomley, Jan 11 2005
A116361(a(n)) <= 1. - Reinhard Zumkeller, Feb 04 2006
A085357(a(n)) = 1; A179821(a(n)) = a(n). - Reinhard Zumkeller, Jul 31 2010
a(n)/n^k is bounded (but does not tend to a limit), where k = 1.44... = A104287. - Charles R Greathouse IV, Sep 19 2012
a(n) = a(A193564(n+1))*2^(A003849(n) + 1) + A003849(n) for n > 0. - Daniel Starodubtsev, Aug 05 2021
There are Fibonacci(n+1) terms with up to n bits in this sequence. - Charles R Greathouse IV, Oct 22 2021
Sum_{n>=1} 1/a(n) = 3.704711752910469457886531055976801955909489488376627037756627135425780134020... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 12 2022

Edited by Antti Karttunen, Feb 21 2006
Cross reference to A007820 added (into O-Y.C. comment) by Jason Kimberley, Sep 14 2009
Typo corrected by Jeffrey Shallit, Sep 26 2014

If bit i is 1, then bits i+-2 must be 0. All terms satisfy A048725(n) = 5*n.
It appears that n is in the sequence if and only if C(5n,n) is odd (cf. A003714). - Benoit Cloitre, Mar 09 2003
Yes, as remarked in A048715, "This is easily proved using the well-known result that the multiplicity with which a prime p divides C(n+m,n) is the number of carries when adding n+m in base p." - Jason Kimberley, Dec 21 2011
A116361(a(n)) <= 2. - Reinhard Zumkeller, Feb 04 2006

From Robert Israel, Dec 27 2016: (Start)
7*n <= a(n) < 8*n.
a(n) = 7*n if and only if n is in A048715.
It appears that a(n) = 8*n-1 if and only if n = (4*8^j+2*8^k+3)/7 for some j and k. (End)

For n = binary n[k],n[k-1],...,n[0], bits a(n) = binary b[k+1],b[k],...,b[0] are b[i] = 1 when n[i-1] + n[i-2] + n[i-3] >= 2, so the majority bit 0 or 1 among the 3 bits of n below position i (with 0 bits below the radix point of n as necessary). This is since 7*n = 4*n + 2*n + n is n[i-1] + n[i-2] + n[i-3] at position i-1, and 4*n XOR 2*n XOR n is the same but no carry, so b[i] is the carry only. - Kevin Ryde, Mar 26 2021

Equivalently, numbers n such that 9*n = 9 X n, i.e., 8*n XOR n = 9*n. Here * stands for ordinary multiplication and X means carryless (GF(2)[X]) multiplication (A048720).
Equivalently, numbers n such that the binomial coefficient C(9n,n) (A169958) is odd. - Zak Seidov, Aug 06 2010
The equivalence of these three definitions follows from Lucas's theorem on binomial coefficients. - N. J. A. Sloane, Sep 01 2010
Clearly all numbers k*2^i for 1 <= k <= 7 have this property. - N. J. A. Sloane, Sep 01 2010
A116361(a(n)) <= 3. - Reinhard Zumkeller, Feb 04 2006

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A350215 A048715, written in binary.

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A000930 Narayana's cows sequence: a(0) = a(1) = a(2) = 1; thereafter a(n) = a(n-1) + a(n-3).

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A003714 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). Also numbers whose binary representation contains no two adjacent 1's.

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A048716 Numbers n such that binary expansion matches ((0)*00(1?)1)*(0*).

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A048718 Binary expansion matches ((0)*0001)*(0*); or, Zeckendorf-like expansion of n using recurrence f(n) = f(n-1) + f(n-4).

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A178894 a(n) = n OR 7n, where OR is bitwise OR.

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A048716 Numbers n such that binary expansion matches ((0)00(1?)1)(0*).

A048718 Binary expansion matches ((0)0001)(0*); or, Zeckendorf-like expansion of n using recurrence f(n) = f(n-1) + f(n-4).

A115424 Integers n > 0 such that n XOR 62n = 63n.