cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A000071 a(n) = Fibonacci(n) - 1.

Original entry on oeis.org

0, 0, 1, 2, 4, 7, 12, 20, 33, 54, 88, 143, 232, 376, 609, 986, 1596, 2583, 4180, 6764, 10945, 17710, 28656, 46367, 75024, 121392, 196417, 317810, 514228, 832039, 1346268, 2178308, 3524577, 5702886, 9227464, 14930351, 24157816, 39088168, 63245985, 102334154
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Comments

a(n) is the number of allowable transition rules for passing from one change to the next (on n-1 bells) in the English art of bell-ringing. This is also the number of involutions in the symmetric group S_{n-1} which can be represented as a product of transpositions of consecutive numbers from {1, 2, ..., n-1}. Thus for n = 6 we have a(6) from (12), (12)(34), (12)(45), (23), (23)(45), (34), (45), for instance. See my 1983 Math. Proc. Camb. Phil. Soc. paper. - Arthur T. White, letter to N. J. A. Sloane, Dec 18 1986
Number of permutations p of {1, 2, ..., n-1} such that max|p(i) - i| = 1. Example: a(4) = 2 since only the permutations 132 and 213 of {1, 2, 3} satisfy the given condition. - Emeric Deutsch, Jun 04 2003 [For a(5) = 4 we have 2143, 1324, 2134 and 1243. - Jon Perry, Sep 14 2013]
Number of 001-avoiding binary words of length n-3. a(n) is the number of partitions of {1, ..., n-1} into two blocks in which only 1- or 2-strings of consecutive integers can appear in a block and there is at least one 2-string. E.g., a(6) = 7 because the enumerated partitions of {1, 2, 3, 4, 5} are 124/35, 134/25, 14/235, 13/245, 1245/3, 145/23, 125/34. - Augustine O. Munagi, Apr 11 2005
Numbers for which only one Fibonacci bit-representation is possible and for which the maximal and minimal Fibonacci bit-representations (A104326 and A014417) are equal. For example, a(12) = 10101 because 8 + 3 + 1 = 12. - Casey Mongoven, Mar 19 2006
Beginning with a(2), the "Recamán transform" (see A005132) of the Fibonacci numbers (A000045). - Nick Hobson, Mar 01 2007
Starting with nonzero terms, a(n) gives the row sums of triangle A158950. - Gary W. Adamson, Mar 31 2009
a(n+2) is the minimum number of elements in an AVL tree of height n. - Lennert Buytenhek (buytenh(AT)wantstofly.org), May 31 2010
a(n) is the number of branch nodes in the Fibonacci tree of order n-1. A Fibonacci tree of order n (n >= 2) is a complete binary tree whose left subtree is the Fibonacci tree of order n-1 and whose right subtree is the Fibonacci tree of order n-2; each of the Fibonacci trees of order 0 and 1 is defined as a single node (see the Knuth reference, p. 417). - Emeric Deutsch, Jun 14 2010
a(n+3) is the number of distinct three-strand positive braids of length n (cf. Burckel). - Maxime Bourrigan, Apr 04 2011
a(n+1) is the number of compositions of n with maximal part 2. - Joerg Arndt, May 21 2013
a(n+2) is the number of leafs of great-grandparent DAG (directed acyclic graph) of height n. A great-grandparent DAG of height n is a single node for n = 1; for n > 1 each leaf of ggpDAG(n-1) has two child nodes where pairs of adjacent new nodes are merged into single node if and only if they have disjoint grandparents and same great-grandparent. Consequence: a(n) = 2*a(n-1) - a(n-3). - Hermann Stamm-Wilbrandt, Jul 06 2014
2 and 7 are the only prime numbers in this sequence. - Emmanuel Vantieghem, Oct 01 2014
From Russell Jay Hendel, Mar 15 2015: (Start)
We can establish Gerald McGarvey's conjecture mentioned in the Formula section, however we require n > 4. We need the following 4 prerequisites.
(1) a(n) = F(n) - 1, with {F(n)}A000045.%20(2)%20(Binet%20form)%20F(n)%20=%20(d%5En%20-%20e%5En)/sqrt(5)%20with%20d%20=%20phi%20and%20e%20=%201%20-%20phi,%20de%20=%20-1%20and%20d%20+%20e%20=%201.%20It%20follows%20that%20a(n)%20=%20(d(n)%20-%20e(n))/sqrt(5)%20-%201.%20(3)%20To%20prove%20floor(x)%20=%20y%20is%20equivalent%20to%20proving%20that%20x%20-%20y%20lies%20in%20the%20half-open%20interval%20%5B0,%201).%20(4)%20The%20series%20%7Bs(n)%20=%20c1%20x%5En%20+%20c2%7D">{n >= 1} the Fibonacci numbers A000045. (2) (Binet form) F(n) = (d^n - e^n)/sqrt(5) with d = phi and e = 1 - phi, de = -1 and d + e = 1. It follows that a(n) = (d(n) - e(n))/sqrt(5) - 1. (3) To prove floor(x) = y is equivalent to proving that x - y lies in the half-open interval [0, 1). (4) The series {s(n) = c1 x^n + c2}{n >= 1}, with -1 < x < 0, and c1 and c2 positive constants, converges by oscillation with s(1) < s(3) < s(5) < ... < s(6) < s(4) < s(2). If follows that for any odd n, the open interval (s(n), s(n+1)) contains the subsequence {s(t)}_{t >= n + 2}. Using these prerequisites we can analyze the conjecture.
Using prerequisites (2) and (3) we see we must prove, for all n > 4, that d((d^(n-1) - e^(n-1))/sqrt(5) - 1) - (d^n - e^n)/sqrt(5) + 1 + c lies in the interval [0, 1). But de = -1, implying de^(n-1) = -e^(n-2). It follows that we must equivalently prove (for all n > 4) that E(n, c) = (e^(n-2) + e^n)/sqrt(5) + 1 - d + c = e^(n-2) (e^2 + 1)/sqrt(5) + e + c lies in [0, 1). Clearly, for any particular n, E(n, c) has extrema (maxima, minima) when c = 2*(1-d) and c = (1+d)*(1-d). Therefore, the proof is completed by using prerequisite (4). It suffices to verify E(5, 2*(1-d)) = 0, E(6, 2*(1-d)) = 0.236068, E(5, (1-d)*(1+d)) = 0.618034, E(6, (1-d)*(1+d)) = 0.854102, all lie in [0, 1).
(End)
a(n) can be shown to be the number of distinct nonempty matchings on a path with n vertices. (A matching is a collection of disjoint edges.) - Andrew Penland, Feb 14 2017
Also, for n > 3, the lexicographically earliest sequence of positive integers such that {phi*a(n)} is located strictly between {phi*a(n-1)} and {phi*a(n-2)}. - Ivan Neretin, Mar 23 2017
From Eric M. Schmidt, Jul 17 2017: (Start)
Number of sequences (e(1), ..., e(n-2)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) != e(j) <= e(k). [Martinez and Savage, 2.5]
Number of sequences (e(1), ..., e(n-2)), 0 <= e(i) < i, such that there is no triple i < j < k with e(i) >= e(j) <= e(k) and e(i) != e(k). [Martinez and Savage, 2.5]
(End)
Numbers whose Zeckendorf (A014417) and dual Zeckendorf (A104326) representations are the same: alternating digits of 1 and 0. - Amiram Eldar, Nov 01 2019
a(n+2) is the length of the longest array whose local maximum element can be found in at most n reveals. See link to the puzzle by Alexander S. Kulikov. - Dmitry Kamenetsky, Aug 08 2020
a(n+2) is the number of nonempty subsets of {1,2,...,n} that contain no consecutive elements. For example, the a(6)=7 subsets of {1,2,3,4} are {1}, {2}, {3}, {4}, {1,3}, {1,4} and {2,4}. - Muge Olucoglu, Mar 21 2021
a(n+3) is the number of allowed patterns of length n in the even shift (that is, a(n+3) is the number of binary words of length n in which there are an even number of 0s between any two occurrences of 1). For example, a(7)=12 and the 12 allowed patterns of length 4 in the even shift are 0000, 0001, 0010, 0011, 0100, 0110, 0111, 1000, 1001, 1100, 1110, 1111. - Zoran Sunic, Apr 06 2022
Conjecture: for k a positive odd integer, the sequence {a(k^n): n >= 1} is a strong divisibility sequence; that is, for n, m >= 1, gcd(a(k^n), a(k^m)) = a(k^gcd(n,m)). - Peter Bala, Dec 05 2022
In general, the sum of a second-order linear recurrence having signature (c,d) will be a third-order recurrence having a signature (c+1,d-c,-d). - Gary Detlefs, Jan 05 2023
a(n) is the number of binary strings of length n-2 whose longest run of 1's is of length 1, for n >= 3. - Félix Balado, Apr 03 2025

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 1.
  • GCHQ, The GCHQ Puzzle Book, Penguin, 2016. See page 28.
  • M. Kauers and P. Paule, The Concrete Tetrahedron, Springer 2011, p. 64.
  • D. E. Knuth, The Art of Computer Programming, Vol. 3, 2nd edition, Addison-Wesley, Reading, MA, 1998, p. 417.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 155.
  • 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).
  • J. L. Yucas, Counting special sets of binary Lyndon words, Ars Combin., 31 (1991), 21-29.

Links

Crossrefs

Cf. A000032, A000045, A000119, A001611, A001654, A001911, A005968, A005969.
Cf. A006327, A014417, A054761, A081006, A081007, A081008, A081009, A083368.
Cf. A098531, A098532, A098533, A101220, A104326, A105488, A105489, A119282.
Cf. A128697, A157725, A157726, A157727, A157728, A157729, A158950, A167616.
Cf. A001622, A228074.
Antidiagonal sums of array A004070.
Right-hand column 2 of triangle A011794.
Related to sum of Fibonacci(kn) over n. Cf. A099919, A058038, A138134, A053606.
Subsequence of A226538. Also a subsequence of A061489.

Programs

  • Haskell
    a000071 n = a000071_list !! n
a000071_list = map (subtract 1) $ tail a000045_list
-- Reinhard Zumkeller, May 23 2013
  • Magma
    [Fibonacci(n)-1: n in [1..60]]; // Vincenzo Librandi, Apr 04 2011
  • Maple
    A000071 := proc(n) combinat[fibonacci](n)-1 ; end proc; # R. J. Mathar, Apr 07 2011
a:= n-> (Matrix([[1, 1, 0], [1, 0, 0], [1, 0, 1]])^(n-1))[3, 2]; seq(a(n), n=1..50); # Alois P. Heinz, Jul 24 2008
  • Mathematica
    Fibonacci[Range[40]] - 1 (* or *) LinearRecurrence[{2, 0, -1}, {0, 0, 1}, 40] (* Harvey P. Dale, Aug 23 2013 *)
Join[{0}, Accumulate[Fibonacci[Range[0, 39]]]] (* Alonso del Arte, Oct 22 2017, based on Giorgi Dalakishvili's formula *)
  • PARI
    {a(n) = if( n<1, 0, fibonacci(n)-1)};
  • SageMath
    [fibonacci(n)-1 for n in range(1,60)] # G. C. Greubel, Oct 21 2024

Formula

a(n) = A000045(n) - 1.
a(0) = -1, a(1) = 0; thereafter a(n) = a(n-1) + a(n-2) + 1.
a(n) = A101220(1, 1, n-2), for n > 1.
G.f.: x^3/((1-x-x^2)*(1-x)). - Simon Plouffe in his 1992 dissertation, dropping initial 0's
a(n) = 2*a(n-1) - a(n-3). - R. H. Hardin, Apr 02 2011
Partial sums of Fibonacci numbers. - Wolfdieter Lang
a(n) = -1 + (A*B^n + C*D^n)/10, with A, C = 5 +- 3*sqrt(5), B, D = (1 +- sqrt(5))/2. - Ralf Stephan, Mar 02 2003
a(1) = 0, a(2) = 0, a(3) = 1, then a(n) = ceiling(phi*a(n-1)) where phi is the golden ratio (1 + sqrt(5))/2. - Benoit Cloitre, May 06 2003
Conjecture: for all c such that 2*(2 - Phi) <= c < (2 + Phi)*(2 - Phi) we have a(n) = floor(Phi*a(n-1) + c) for n > 4. - Gerald McGarvey, Jul 22 2004. This is true provided n > 3 is changed to n > 4, see proof in Comments section. - Russell Jay Hendel, Mar 15 2015
a(n) = Sum_{k = 0..floor((n-2)/2)} binomial(n-k-2, k+1). - Paul Barry, Sep 23 2004
a(n+3) = Sum_{k = 0..floor(n/3)} binomial(n-2*k, k)*(-1)^k*2^(n-3*k). - Paul Barry, Oct 20 2004
a(n+1) = Sum(binomial(n-r, r)), r = 1, 2, ... which is the case t = 2 and k = 2 in the general case of t-strings and k blocks: a(n+1, k, t) = Sum(binomial(n-r*(t-1), r)*S2(n-r*(t-1)-1, k-1)), r = 1, 2, ... - Augustine O. Munagi, Apr 11 2005
a(n) = Sum_{k = 0..n-2} k*Fibonacci(n - k - 3). - Ross La Haye, May 31 2006
a(n) = term (3, 2) in the 3 X 3 matrix [1, 1, 0; 1, 0, 0; 1, 0, 1]^(n-1). - Alois P. Heinz, Jul 24 2008
For n >= 4, a(n) = ceiling(phi*a(n-1)), where phi is the golden ratio. - Vladimir Shevelev, Jul 04 2010
Closed-form without two leading zeros g.f.: 1/(1 - 2*x - x^3); ((5 + 2*sqrt(5))*((1 + sqrt(5))/2)^n + (5 - 2*sqrt(5))*((1 - sqrt(5))/2)^n - 5)/5; closed-form with two leading 0's g.f.: x^2/(1 - 2*x - x^3); ((5 + sqrt(5))*((1 + sqrt(5))/2)^n + (5 - sqrt(5))*((1 - sqrt(5))/2)^n - 10)/10. - Tim Monahan, Jul 10 2011
A000119(a(n)) = 1. - Reinhard Zumkeller, Dec 28 2012
a(n) = A228074(n - 1, 2) for n > 2. - Reinhard Zumkeller, Aug 15 2013
G.f.: Q(0)*x^2/2, where Q(k) = 1 + 1/(1 - x*(4*k + 2 - x^2)/( x*(4*k + 4 - x^2) + 1/Q(k+1) )); (continued fraction). - Sergei N. Gladkovskii, Aug 30 2013
A083368(a(n+3)) = n. - Reinhard Zumkeller, Aug 10 2014
E.g.f.: 1 - exp(x) + 2*exp(x/2)*sinh(sqrt(5)*x/2)/sqrt(5). - Ilya Gutkovskiy, Jun 15 2016
a(n) = A000032(3+n) - 1 mod A000045(3+n). - Mario C. Enriquez, Apr 01 2017
a(n) = Sum_{i=0..n-2} Fibonacci(i). - Giorgi Dalakishvili (mcnamara_gio(AT)yahoo.com), Apr 02 2005 [corrected by Doug Bell, Jun 01 2017]
a(n+2) = Sum_{j = 0..floor(n/2)} Sum_{k = 0..j} binomial(n - 2*j, k+1)*binomial(j, k). - Tony Foster III, Sep 08 2017
From Peter Bala, Nov 12 2021: (Start)
a(4*n) = Fibonacci(2*n+1)*Lucas(2*n-1) = A081006(n);
a(4*n+1) = Fibonacci(2*n)*Lucas(2*n+1) = A081007(n);
a(4*n+2) = Fibonacci(2*n)*Lucas(2*n+2) = A081008(n);
a(4*n+3) = Fibonacci(2*n+2)*Lucas(2*n+1) = A081009(n). (End)
G.f.: x^3/((1 - x - x^2)*(1 - x)) = Sum_{n >= 0} (-1)^n * x^(n+3) *( Product_{k = 1..n} (k - x)/Product_{k = 1..n+2} (1 - k*x) ) (a telescoping series). - Peter Bala, May 08 2024
Product_{n>=4} (1 + (-1)^n/a(n)) = 3*phi/4, where phi is the golden ratio (A001622). - Amiram Eldar, Nov 28 2024

Extensions

Edited by N. J. A. Sloane, Apr 04 2011

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

Views

Author

N. J. A. Sloane

Keywords

Comments

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

Examples

			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)

References

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

Links

Crossrefs

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).

Programs

  • Haskell
    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
  • Maple
    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:
  • Mathematica
    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 *)
  • PARI
    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
  • PARI
    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
  • Python
    for n in range(300):
    if 2*n & n == 0:
        print(n, end=",") # Alex Ratushnyak, Jun 21 2012
  • Python
    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
  • Python
    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 */
  • Scala
    (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

Formula

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

Extensions

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

A200648 Length of Stolarsky representation of n.

Original entry on oeis.org

1, 1, 2, 2, 3, 3, 3, 4, 3, 4, 4, 4, 5, 4, 4, 5, 4, 5, 5, 5, 6, 4, 5, 5, 5, 6, 5, 5, 6, 5, 6, 6, 6, 7, 5, 5, 6, 5, 6, 6, 6, 7, 5, 6, 6, 6, 7, 6, 6, 7, 6, 7, 7, 7, 8, 5, 6, 6, 6, 7, 6, 6, 7, 6, 7, 7, 7, 8, 6, 6, 7, 6, 7, 7, 7, 8, 6, 7, 7, 7, 8, 7, 7, 8, 7, 8, 8
Offset: 1

Views

Author

Casey Mongoven, Nov 19 2011

Keywords

Comments

For the Stolarsky representation of n, see the C. Mongoven link.
Conjecture: a(n) is the sum of row n-1 of A385886. To obtain it, first take maximal anti-run lengths of binary indices of each nonnegative integer (giving A384877), then remove all duplicate rows (giving A385886), and finally take the sum of each remaining row. For length instead of sum we appear to have A200649. - Gus Wiseman, Jul 21 2025

Examples

			The Stolarsky representation of 19 is 11101. This is of length 5. So a(19) = 5.

Links

Crossrefs

Cf. A135817, A200651.
Counting just ones gives A200649.
Counting just zeros gives A200650.
Stolarsky representation is listed by A385888, ranks A200714.
A000120 counts 1's in binary expansion.
A384890 counts maximal anti-runs of binary indices, ranks A385816.
A385886 lists maximal anti-run lengths of binary indices.
Cf. A023758, A048793, A052499, A069010, A072649, A083368, A135818, A245562, A245563, A348366, A384877, A384893.

Programs

  • Mathematica
    stol[n_] := stol[n] = If[n == 1, {}, If[n != Round[Round[n/GoldenRatio]*GoldenRatio], Join[stol[Floor[n/GoldenRatio^2] + 1], {0}], Join[stol[Round[n/GoldenRatio]], {1}]]];
a[n_] := If[n == 1, 1, Length[stol[n]]]; Array[a, 100] (* Amiram Eldar, Jul 07 2023 *)
  • PARI
    stol(n) = {my(phi=quadgen(5)); if(n==1, [], if(n != round(round(n/phi)*phi), concat(stol(floor(n/phi^2) + 1), [0]), concat(stol(round(n/phi)), [1])));}
a(n) = if(n == 1, 1, #stol(n)); \\ Amiram Eldar, Jul 07 2023

Formula

a(n) = A200649(n) + A200650(n). - Michel Marcus, Mar 14 2023

Extensions

More terms from Amiram Eldar, Jul 07 2023

A385817 Irregular triangle read by rows listing the lengths of maximal runs (sequences of consecutive elements increasing by 1) of binary indices, duplicate rows removed.

Original entry on oeis.org

1, 2, 1, 1, 3, 2, 1, 1, 2, 4, 1, 1, 1, 3, 1, 2, 2, 1, 3, 5, 2, 1, 1, 1, 2, 1, 4, 1, 1, 1, 2, 3, 2, 2, 3, 1, 4, 6, 1, 1, 1, 1, 3, 1, 1, 2, 2, 1, 1, 3, 1, 5, 1, 2, 1, 2, 1, 2, 2, 4, 2, 1, 1, 3, 3, 3, 2, 4, 1, 5, 7, 2, 1, 1, 1, 1, 2, 1, 1, 4, 1, 1, 1, 1, 2, 1
Offset: 0

Views

Author

Gus Wiseman, Jul 14 2025

Keywords

Comments

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.
This is the triangle A245563, except all duplicates after the first instance of each composition are removed. It lists all compositions in order of their first appearance as a row of A245563.

Examples

			The binary indices of 53 are {1,3,5,6}, with maximal runs ((1),(3),(5,6)), with lengths (1,1,2). After removing duplicates, this is our row 16.
Triangle begins:
   0: .
   1: 1
   2: 2
   3: 1 1
   4: 3
   5: 2 1
   6: 1 2
   7: 4
   8: 1 1 1
   9: 3 1
  10: 2 2
  11: 1 3
  12: 5
  13: 2 1 1
  14: 1 2 1
  15: 4 1
  16: 1 1 2
  17: 3 2
  18: 2 3
  19: 1 4
  20: 6
  21: 1 1 1 1

Crossrefs

In the following references, "before" is short for "before removing duplicate rows".
Positions of singleton rows appear to be A000071 = A000045-1, before A023758.
Positions of firsts appearances appear to be A001629.
Positions of rows of the form (1,1,...) appear to be A055588 = A001906+1.
First term of each row appears to be A083368.
Row sums appear to be A200648, before A000120.
Row lengths after the first row appear to be A200650+1, before A069010 = A037800+1.
Before the removals we had A245563 (except first term), see A245562, A246029, A328592.
For anti-run ranks we have A385816, before A348366, firsts A052499.
Standard composition numbers of rows are A385818, before A385889.
For anti-runs we have A385886, before A384877, firsts A384878.
Cf. A044813, A048793, A164707, A200649, A242882, A243815, A384175, A384890.

Programs

  • Mathematica
    DeleteDuplicates[Table[Length/@Split[Join@@Position[Reverse[IntegerDigits[n,2]],1],#2==#1+1&],{n,0,100}]]

A385892 In the sequence of run lengths of binary indices of each positive integer (A245563), remove all duplicate rows after the first and take the last term of each remaining row.

Original entry on oeis.org

1, 2, 1, 3, 1, 2, 4, 1, 1, 2, 3, 5, 1, 1, 1, 2, 2, 3, 4, 6, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 5, 7, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 6, 8, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6, 7
Offset: 1

Views

Author

Gus Wiseman, Jul 18 2025

Keywords

Comments

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

Examples

			The binary indices of 53 are {1,3,5,6}, with maximal runs ((1),(3),(5,6)), with lengths (1,1,2), which is the 16th row of A385817, so a(16) = 2.

Crossrefs

In the following references, "before" is short for "before removing duplicate rows".
Positions of firsts appearances appear to be A000071.
Without the removals we have A090996.
For sum instead of last we have A200648, before A000120.
For length instead of last we have A200650+1, before A069010 = A037800+1.
Last term of row n of A385817 (ranks A385818, before A385889), first A083368.
A245563 gives run lengths of binary indices, see A245562, A246029, A328592.
A384877 gives anti-run lengths of binary indices, A385816.
Cf. A001629, A044813, A048793, A164707, A200649, A384878, A384890, A385886.

Programs

  • Mathematica
    Last/@DeleteDuplicates[Table[Length/@Split[Join@@Position[Reverse[IntegerDigits[n,2]],1],#2==#1+1&],{n,100}]]
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