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.

Showing 1-10 of 14 results. Next

A007895 Number of terms in the Zeckendorf representation of n (write n as a sum of non-consecutive distinct Fibonacci numbers).

Original entry on oeis.org

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

Views

Author

Felix Weinstein (wain(AT)ana.unibe.ch) and Clark Kimberling

Keywords

Comments

Also number of 0's (or B's) in the Wythoff representation of n -- see the Reble link. See also A135817 for references and links for the Wythoff representation for n >= 1. - Wolfdieter Lang, Jan 21 2008; N. J. A. Sloane, Jun 28 2008
Or, a(n) is the number of applications of Wythoff's B sequence A001950 needed in the unique Wythoff representation of n >= 1. E.g., 16 = A(B(A(A(B(1))))) = ABAAB = `10110`, hence a(16) = 2. - Wolfdieter Lang, Jan 21 2008
Let M(0) = 0, M(1) = 1 and for i > 0, M(i+1) = f(concatenation of M(j), j from 0 to i - 1) where f is the morphism f(k) = k + 1. Then the sequence is the concatenation of M(j) for j from 0 to infinity. - Claude Lenormand (claude.lenormand(AT)free.fr), Dec 16 2003
From Joerg Arndt, Nov 09 2012: (Start)
Let m be the number of parts in the listing of the compositions of n into odd parts as lists of parts in lexicographic order, a(k) = (n - length(composition(k)))/2 for all k < Fibonacci(n) and all n (see example).
Let m be the number of parts in the listing of the compositions of n into parts 1 and 2 as lists of parts in lexicographic order, a(k) = n - length(composition(k)) for all k < Fibonacci(n) and all n (see example).
A000120 gives the equivalent for (all) compositions. (End)
a(n) = A104324(n) - A213911(n); row lengths in A035516 and A035516. - Reinhard Zumkeller, Mar 10 2013
a(n) is also the minimum number of Fibonacci numbers which sum to n, regardless of adjacency or duplication. - Alan Worley, Apr 17 2015
This follows from the fact that the sequence is subadditive: a(n+m) <= a(n) + a(m) for nonnegative integers n,m. See Lemma 2.1 of the Stoll link. - Robert Israel, Apr 17 2015
From Michel Dekking, Mar 08 2020: (Start)
This sequence is a morphic sequence on an infinite alphabet, i.e., (a(n)) is a letter-to-letter projection of a fixed point of a morphism tau.
The alphabet is {0,1,...,j,...}X{0,1}, and tau is given by
tau((j,0)) = (j,0) (j+1,1),
tau((j,1)) = (j,0).
The letter-to-letter map is given by the projection on the first coordinate: (j,i)->j for i=0,1.
To prove this, note first that the second coordinate of the letters generates the infinite Fibonacci word = A003849 = 0100101001001....
This implies that for all n and j one has
|tau^n(j,0)| = F(n+2),
where |w| denotes the length of a word w, and (F(n)) = A000045 are the Fibonacci numbers.
Secondly, we need the following simple, but crucial observation. Let the Zeckendorf representation of n be Z(n) = A014417(n). For example,
Z(0) = 0, Z(1) = 1, Z(2) = 10, Z(3) = 100, Z(4) = 101, Z(5) = 1000.
From the unicity of the Zeckendorf representation it follows that for the positions i = 0,1,...,F(n)-1 one has
Z(F(n+1)+i) = 10...0 Z(i),
where zeros are added to Z(i) to give the total representation length n-1.
This gives for i = 0,1,...,F(n)-1 that
a(F(n+1)+i) = a(i) + 1.
From the first observation follows that the first F(n+1) letters of tau^n(j,0) are equal to tau^{n-1}(j,0), and the last F(n) letters of tau^n(j,0) are equal to tau^{n-1}(j+1,1) = tau^{n-2}(j+1,0).
Combining this with the second observation shows that the first coordinate of the fixed point of tau, starting from (0,0), gives (a(n)).
It is of course possible to obtain a morphism tau' on the natural numbers by changing the alphabet: (j,0)-> 2j (j,1)-> 2j+1, which yields the morphism
tau'(2j) = 2j, 2j+3, tau'(2j+1) = 2j.
The fixed point of tau' starting with 0 is
u = 03225254254472544747625...
The corresponding letter-to-letter map lambda is given by lambda(2j)=j, lambda(2j+1)= j. Then lambda(u) = (a(n)).
(End)

Examples

			a(46) = a(1 + 3 + 8 + 34) = 4.
From _Joerg Arndt_, Nov 09 2012: (Start)
Connection to the compositions of n into odd parts (see comment):
[ #]:  a(n)  composition into odd parts
[ 0]   [ 0]   1 1 1 1 1 1 1 1
[ 1]   [ 1]   1 1 1 1 1 3
[ 2]   [ 1]   1 1 1 1 3 1
[ 3]   [ 1]   1 1 1 3 1 1
[ 4]   [ 2]   1 1 1 5
[ 5]   [ 1]   1 1 3 1 1 1
[ 6]   [ 2]   1 1 3 3
[ 7]   [ 2]   1 1 5 1
[ 8]   [ 1]   1 3 1 1 1 1
[ 9]   [ 2]   1 3 1 3
[10]   [ 2]   1 3 3 1
[11]   [ 2]   1 5 1 1
[12]   [ 3]   1 7
[13]   [ 1]   3 1 1 1 1 1
[14]   [ 2]   3 1 1 3
[15]   [ 2]   3 1 3 1
[16]   [ 2]   3 3 1 1
[17]   [ 3]   3 5
[18]   [ 2]   5 1 1 1
[19]   [ 3]   5 3
[20]   [ 3]   7 1
Connection to the compositions of n into parts 1 or 2 (see comment):
[ #]:  a(n)  composition into parts 1 and 2
[ 0]   [0]   1 1 1 1 1 1 1
[ 1]   [1]   1 1 1 1 1 2
[ 2]   [1]   1 1 1 1 2 1
[ 3]   [1]   1 1 1 2 1 1
[ 4]   [2]   1 1 1 2 2
[ 5]   [1]   1 1 2 1 1 1
[ 6]   [2]   1 1 2 1 2
[ 7]   [2]   1 1 2 2 1
[ 8]   [1]   1 2 1 1 1 1
[ 9]   [2]   1 2 1 1 2
[10]   [2]   1 2 1 2 1
[11]   [2]   1 2 2 1 1
[12]   [3]   1 2 2 2
[13]   [1]   2 1 1 1 1 1
[14]   [2]   2 1 1 1 2
[15]   [2]   2 1 1 2 1
[16]   [2]   2 1 2 1 1
[17]   [3]   2 1 2 2
[18]   [2]   2 2 1 1 1
[19]   [3]   2 2 1 2
[20]   [3]   2 2 2 1
(End)
From _Michel Dekking_, Mar 08 2020: (Start)
The third iterate of the morphism tau generating this sequence:
      tau^3((0,0)) = (0,0)(1,1)(1,0)(1,0)(2,1)
= (a(0),0)(a(1),1)(a(2),0)(a(3),0)(a(4),1). (End)
		

References

  • Cornelius Gerrit Lekkerkerker, Voorstelling van natuurlijke getallen door een som van getallen van Fibonacci, Simon Stevin 29 (1952), 190-195.
  • F. Weinstein, The Fibonacci Partitions, preprint, 1995.
  • Édouard Zeckendorf, Représentation des nombres naturels par une somme des nombres de Fibonacci ou de nombres de Lucas, Bull. Soc. Roy. Sci. Liège 41, 179-182, 1972.

Crossrefs

Cf. A135817 (lengths of Wythoff representation), A135818 (number of 1's (or A's) in the Wythoff representation).
Record positions are in A027941.

Programs

  • Haskell
    a007895 = length . a035516_row  -- Reinhard Zumkeller, Mar 10 2013
    
  • Maple
    # With the following Maple program (not the best one), B(n) (n >= 1) yields the number of terms in the Zeckendorf representation of n.
    with(combinat): B := proc (n) local A, ct, m, j: A := proc (n) local i: for i while fibonacci(i) <= n do n-fibonacci(i) end do end proc: ct := 0; m := n: for j while 0 < A(m) do ct := ct+1: m := A(m) end do: ct+1 end proc: 0, seq(B(n), n = 1 .. 104);
    # Emeric Deutsch, Jul 05 2010
    N:= 1000: # to get a(n) for n <= N
    m:= ceil(log[(1+sqrt(5))/2](sqrt(5)*N)):
    Z:= Vector(m):
    a[0]:= 0:
    for n from 1 to N do
    if Z[1] = 0 then Z[1]:= 1; q:= 1;
    else Z[2]:= 1; Z[1]:= 0; q:= 2;
    fi;
    while Z[q+1] = 1 do
      Z[q]:= 0;
      Z[q+1]:= 0;
      Z[q+2]:= 1;
      q:= q+2;
    od:
    a[n]:= add(Z[i],i=1..m);
    od:
    seq(a[n],n=0..N); # Robert Israel, Apr 17 2015
    # alternative
    read("transforms") : # https://oeis.org/transforms.txt
    A007895 := proc(n)
        wt(A003714(n)) ;
    end proc:
    seq(A007895(n),n=0..10) ; # R. J. Mathar, Sep 22 2020
  • Mathematica
    zf[n_] := (k = 1; ff = {}; While[(fi = Fibonacci[k]) <= n, AppendTo[ff, fi]; k++]; Drop[ff, 1]); zeckRep[n_] := If[n == 0, 0, r = n; s = {}; fr = zf[n]; While[r > 0, lf = Last[fr]; If[lf <= r, r = r - lf; PrependTo[s, lf]]; fr = Drop[fr, -1]]; s]; zeckRepLen[n_] := Length[zeckRep[n]]; Table[zeckRepLen[n], {n, 0, 104}] (* Jean-François Alcover, Sep 27 2011 *)
    DigitCount[Select[Range[0, 1000], BitAnd[#, 2#] == 0 &], 2, 1] (* Jean-François Alcover, Jan 25 2018 *)
    Table[Length[DeleteCases[NestWhileList[# - Fibonacci[Floor[Log[Sqrt[5] * # + 3/2]/Log[GoldenRatio]]] &, n, # > 1 &], 0]], {n, 0, 143}] (* Alonso del Arte, May 14 2019 *)
    Flatten[Nest[{Flatten[#], #[[1]] + 1} &, {0, 1}, 9]] (* Paolo Xausa, Jun 17 2024 *)
  • PARI
    a(n,mx=0)=if(n<4,n>0,if(!mx,while(fibonacci(mx)n,mx--); 1+a(n-fibonacci(mx),mx-2)) \\ Charles R Greathouse IV, Feb 14 2013
    
  • PARI
    a(n)=if(n<4, n>0, my(k=2,s,t); while(fibonacci(k++)<=n,); while(k && n, t=fibonacci(k); if(t<=n, n-=t; s++); k--); s) \\ Charles R Greathouse IV, Sep 02 2015
    
  • Python
    from sympy import fibonacci
    def a(n):
        k=0
        x=0
        while n>0:
            k=0
            while fibonacci(k)<=n: k+=1
            x+=10**(k - 3)
            n-=fibonacci(k - 1)
        return str(x).count("1")
    print([a(n) for n in range(101)]) # Indranil Ghosh, Jun 09 2017

Formula

a(n) = A000120(A003714(n)). - Reinhard Zumkeller, May 05 2005
a(n) = A107015(n) + A107016(n). - Reinhard Zumkeller, May 09 2005
a(n) = A143299(n+1) - 1. - Filip Zaludek, Oct 31 2016
a(n) = A007953(A014417(n)). - Amiram Eldar, Oct 10 2023

Extensions

Edited by N. J. A. Sloane Jun 27 2008 at the suggestion of R. J. Mathar and Don Reble

A003622 The Wythoff compound sequence AA: a(n) = floor(n*phi^2) - 1, where phi = (1+sqrt(5))/2.

Original entry on oeis.org

1, 4, 6, 9, 12, 14, 17, 19, 22, 25, 27, 30, 33, 35, 38, 40, 43, 46, 48, 51, 53, 56, 59, 61, 64, 67, 69, 72, 74, 77, 80, 82, 85, 88, 90, 93, 95, 98, 101, 103, 106, 108, 111, 114, 116, 119, 122, 124, 127, 129, 132, 135, 137, 140, 142, 145, 148, 150, 153, 156, 158, 161, 163, 166
Offset: 1

Views

Author

Keywords

Comments

Also, integers with "odd" Zeckendorf expansions (end with ...+F_2 = ...+1) (Fibonacci-odd numbers); first column of Wythoff array A035513; from a 3-way splitting of positive integers. [Edited by Peter Munn, Sep 16 2022]
Also, numbers k such that A005206(k) = A005206(k+1). Also k such that A022342(A005206(k)) = k+1 (for all other k's this is k). - Michele Dondi (bik.mido(AT)tiscalenet.it), Dec 30 2001
Also, positions of 1's in A139764, the smallest term in Zeckendorf representation of n. - John W. Layman, Aug 25 2011
From Amiram Eldar, Sep 03 2022: (Start)
Numbers with an odd number of trailing 1's in their dual Zeckendorf representation (A104326), i.e., numbers k such that A356749(k) is odd.
The asymptotic density of this sequence is 1 - 1/phi (A132338). (End)
{a(n)} is the unique monotonic sequence of positive integers such that {a(n)} and {b(n)}: b(n) = a(n) - n form a partition of the nonnegative integers. - Yifan Xie, Jan 25 2025

References

  • A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 62.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 307-308 of 2nd edition.
  • C. Kimberling, "Stolarsky interspersions", Ars Combinatoria 39 (1995) 129-138.
  • D. R. Morrison, "A Stolarsky array of Wythoff pairs", in A Collection of Manuscripts Related to the Fibonacci Sequence. Fibonacci Assoc., Santa Clara, CA, 1980, pp. 134-136.
  • J. Roberts, Lure of the Integers, Math. Assoc. America, 1992, p. 10.
  • N. J. A. Sloane and Simon Plouffe, Encyclopedia of Integer Sequences, Academic Press, 1995: this sequence appears twice, as both M3277 and M3278.

Crossrefs

Positions of 1's in A003849.
Complement of A022342.
The Wythoff compound sequences: Let A = A000201, B = A001950. Then AA = A003622, AB = A003623, BA = A035336, BB = A101864. The eight triples AAA, AAB, ..., BBB are A134859, A134860, A035337, A134862, A134861, A134863, A035338, A134864, resp.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021

Programs

  • Haskell
    a003622 n = a003622_list !! (n-1)
    a003622_list = filter ((elem 1) . a035516_row) [1..]
    -- Reinhard Zumkeller, Mar 10 2013
    
  • Maple
    A003622 := proc(n)
        n+floor(n*(1+sqrt(5))/2)-1 ;
    end proc: # R. J. Mathar, Jan 25 2015
    # Maple code for the Wythoff compound sequences, from N. J. A. Sloane, Mar 30 2016
    # The Wythoff compound sequences: Let A = A000201, B = A001950. Then AA = A003622, AB = A003623, BA = A035336, BB = A101864. The eight triples AAA, AAB, ..., BBB are A134859, A134860, A035337, A134862, A134861, A134863, A035338, A134864, resp.
    # Assume files out1, out2 contain lists of the terms in the base sequences A and B from their b-files
    read out1; read out2; b[0]:=b1: b[1]:=b2:
    w2:=(i,j,n)->b[i][b[j][n]];
    w3:=(i,j,k,n)->b[i][b[j][b[k][n]]];
    for i from 0 to 1 do
    lprint("name=",i);
    lprint([seq(b[i][n],n=1..100)]):
    od:
    for i from 0 to 1 do for j from 0 to 1 do
    lprint("name=",i,j);
    lprint([seq(w2(i,j,n),n=1..100)]);
    od: od:
    for i from 0 to 1 do for j from 0 to 1 do for k from 0 to 1 do
    lprint("name=",i,j,k);
    lprint([seq(w3(i,j,k,n),n=1..100)]);
    od: od: od:
  • Mathematica
    With[{c=GoldenRatio^2},Table[Floor[n c]-1,{n,70}]] (* Harvey P. Dale, Jun 11 2011 *)
    Range[70]//Floor[#*GoldenRatio^2]-1& (* Waldemar Puszkarz, Oct 10 2017 *)
  • PARI
    a(n)=floor(n*(sqrt(5)+3)/2)-1
    
  • PARI
    a(n) = (sqrtint(n^2*5)+n*3)\2 - 1; \\ Michel Marcus, Sep 17 2022
    
  • Python
    from sympy import floor
    from mpmath import phi
    def a(n): return floor(n*phi**2) - 1 # Indranil Ghosh, Jun 09 2017
    
  • Python
    from math import isqrt
    def A003622(n): return (n+isqrt(5*n**2)>>1)+n-1 # Chai Wah Wu, Aug 11 2022

Formula

a(n) = floor(n*phi) + n - 1. [Corrected by Jianing Song, Aug 18 2022]
a(n) = floor(floor(n*phi)*phi) = A000201(A000201(n)). [See the Mathematics Stack Exchange link for a proof of the equivalence of the definition. - Jianing Song, Aug 18 2022]
a(n) = 1 + A022342(1 + A022342(n)).
G.f.: 1 - (1-x)*Sum_{n>=1} x^a(n) = 1/1 + x/1 + x^2/1 + x^3/1 + x^5/1 + x^8/1 + ... + x^F(n)/1 + ... (continued fraction where F(n)=n-th Fibonacci number). - Paul D. Hanna, Aug 16 2002
a(n) = A001950(n) - 1. - Philippe Deléham, Apr 30 2004
a(n) = A022342(n) + n. - Philippe Deléham, May 03 2004
a(n) = a(n-1) + 2 + A005614(n-2); also a(n) = a(n-1) + 1 + A001468(n-1). - A.H.M. Smeets, Apr 26 2024

A022342 Integers with "even" Zeckendorf expansions (do not end with ...+F_2 = ...+1) (the Fibonacci-even numbers); also, apart from first term, a(n) = Fibonacci successor to n-1.

Original entry on oeis.org

0, 2, 3, 5, 7, 8, 10, 11, 13, 15, 16, 18, 20, 21, 23, 24, 26, 28, 29, 31, 32, 34, 36, 37, 39, 41, 42, 44, 45, 47, 49, 50, 52, 54, 55, 57, 58, 60, 62, 63, 65, 66, 68, 70, 71, 73, 75, 76, 78, 79, 81, 83, 84, 86, 87, 89, 91, 92, 94, 96, 97, 99, 100, 102, 104, 105, 107
Offset: 1

Views

Author

Keywords

Comments

The Zeckendorf expansion of n is obtained by repeatedly subtracting the largest Fibonacci number you can until nothing remains; for example, 100 = 89 + 8 + 3.
The Fibonacci successor to n is found by replacing each F_i in the Zeckendorf expansion by F_{i+1}; for example, the successor to 100 is 144 + 13 + 5 = 162.
If k appears, k + (rank of k) does not (10 is the 7th term in the sequence but 10 + 7 = 17 is not a term of the sequence). - Benoit Cloitre, Jun 18 2002
From Michele Dondi (bik.mido(AT)tiscalenet.it), Dec 30 2001: (Start)
a(n) = Sum_{k in A_n} F_{k+1}, where a(n)= Sum_{k in A_n} F_k is the (unique) expression of n as a sum of "noncontiguous" Fibonacci numbers (with index >= 2).
a(10^n) gives the first few digits of g = (sqrt(5)+1)/2.
The sequences given by b(n+1) = a(b(n)) obey the general recursion law of Fibonacci numbers. In particular the (sub)sequence (of a(-)) yielded by a starting value of 2=a(1), is the sequence of Fibonacci numbers >= 2. Starting points of all such subsequences are given by A035336.
a(n) = floor(phi*n+1/phi); phi = (sqrt(5)+1)/2. a(F_n)=F_{n+1} if F_n is the n-th Fibonacci number.
(End)
From Amiram Eldar, Sep 03 2022: (Start)
Numbers with an even number of trailing 1's in their dual Zeckendorf representation (A104326), i.e., numbers k such that A356749(k) is even.
The asymptotic density of this sequence is 1/phi (A094214). (End)

Examples

			The successors to 1, 2, 3, 4=3+1 are 2, 3, 5, 7=5+2.
		

References

  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 307-308 of 2nd edition.
  • E. Zeckendorf, Représentation des nombres naturels par une somme des nombres de Fibonacci ou de nombres de Lucas, Bull. Soc. Roy. Sci. Liège 41, 179-182, 1972.

Crossrefs

Positions of 0's in A003849.
Complement of A003622.
The following sequences are all essentially the same, in the sense that they are simple transformations of each other, with A000201 as the parent: A000201, A001030, A001468, A001950, A003622, A003842, A003849, A004641, A005614, A014675, A022342, A088462, A096270, A114986, A124841. - N. J. A. Sloane, Mar 11 2021

Programs

  • Haskell
    a022342 n = a022342_list !! (n-1)
    a022342_list = filter ((notElem 1) . a035516_row) [0..]
    -- Reinhard Zumkeller, Mar 10 2013
    
  • Magma
    [Floor(n*(Sqrt(5)+1)/2)-1: n in [1..100]]; // Vincenzo Librandi, Feb 16 2015
    
  • Maple
    A022342 := proc(n)
          local g;
          g := (1+sqrt(5))/2 ;
        floor(n*g)-1 ;
    end proc: # R. J. Mathar, Aug 04 2013
  • Mathematica
    With[{t=GoldenRatio^2},Table[Floor[n*t]-n-1,{n,70}]] (* Harvey P. Dale, Aug 08 2012 *)
  • PARI
    a(n)=floor(n*(sqrt(5)+1)/2)-1
    
  • PARI
    a(n)=(sqrtint(5*n^2)+n-2)\2 \\ Charles R Greathouse IV, Feb 27 2014
    
  • Python
    from math import isqrt
    def A022342(n): return (n+isqrt(5*n**2)>>1)-1 # Chai Wah Wu, Aug 17 2022

Formula

a(n) = floor(n*phi^2) - n - 1 = floor(n*phi) - 1 = A000201(n) - 1, where phi is the golden ratio.
a(n) = A003622(n) - n. - Philippe Deléham, May 03 2004
a(n+1) = A022290(2*A003714(n)). - R. J. Mathar, Jan 31 2015
For n > 1: A035612(a(n)) > 1. - Reinhard Zumkeller, Feb 03 2015
a(n) = A000201(n) - 1. First differences are given in A014675 (or A001468, ignoring its first term). - M. F. Hasler, Oct 13 2017
a(n) = a(n-1) + 1 + A005614(n-2) for n > 1; also a(n) = a(n-1) + A014675(n-2) = a(n-1) + A001468(n-1). - A.H.M. Smeets, Apr 26 2024

Extensions

Name edited by Peter Munn, Dec 07 2021

A035517 Triangular array read by rows, formed from Zeckendorf expansion of integers: repeatedly subtract the largest Fibonacci number you can until nothing remains. Row n give Z. expansion of n.

Original entry on oeis.org

0, 1, 2, 3, 1, 3, 5, 1, 5, 2, 5, 8, 1, 8, 2, 8, 3, 8, 1, 3, 8, 13, 1, 13, 2, 13, 3, 13, 1, 3, 13, 5, 13, 1, 5, 13, 2, 5, 13, 21, 1, 21, 2, 21, 3, 21, 1, 3, 21, 5, 21, 1, 5, 21, 2, 5, 21, 8, 21, 1, 8, 21, 2, 8, 21, 3, 8, 21, 1, 3, 8, 21, 34, 1, 34, 2, 34, 3, 34, 1, 3, 34, 5, 34, 1, 5, 34, 2, 5, 34
Offset: 0

Views

Author

Keywords

Comments

Row n has A007895(n) terms.
With the 2nd Maple program, B(n) yields the number of terms in the Zeckendorf expansion of n, while Z(n) yields the expansion itself. For example, B(100)=3 and Z(100)=3, 8, 89. [Emeric Deutsch, Jul 05 2010]

Examples

			0=0; 1=1; 2=2; 3=3; 4=1+3; 5=5; 6=1+5; 7=2+5; 8=8; 9=1+8; 10=2+8; ... so triangle begins
  0;
  1;
  2;
  3;
  1, 3;
  5;
  1, 5;
  2, 5;
  8;
  1, 8;
  2, 8;
  3, 8;
  1, 3, 8;
		

References

  • Zeckendorf, E., Représentation des nombres naturels par une somme des nombres de Fibonacci ou de nombres de Lucas, Bull. Soc. Roy. Sci. Liège 41, 179-182, 1972.

Crossrefs

Programs

  • Haskell
    a035517 n k = a035517_tabf !! n !! k
    a035517_row n = a035517_tabf !! n
    a035517_tabf = map reverse a035516_tabf
    -- Reinhard Zumkeller, Mar 10 2013
    
  • Maple
    with(combinat): B := proc (n) local A, ct, m, j: A := proc (n) local i: for i while fibonacci(i) <= n do n-fibonacci(i) end do end proc: ct := 0: m := n: for j while 0 < A(m) do ct := ct+1: m := A(m) end do: ct+1 end proc: F := proc (n) local i: for i while fibonacci(i) <= n do fibonacci(i) end do end proc: Z := proc (n) local j, z: for j to B(n) do z[j] := F(n-add(z[i], i = 1 .. j-1)) end do: seq(z[B(n)+1-k], k = 1 .. B(n)) end proc: for n to 25 do Z(n) end do;
    # Emeric Deutsch, Jul 05 2010
    # yields sequence in triangular form; end of this Maple program
  • Mathematica
    f[n_] := (k=1; ff={}; While[(fi = Fibonacci[k]) <= n, AppendTo[ff, fi]; k++]; Drop[ff,1]); ro[n_] := If[n == 0, 0, r = n; s = {}; fr = f[n];
    While[r > 0, lf = Last[fr]; If[lf <= r, r = r - lf; PrependTo[s, lf]]; fr = Drop[fr,-1]]; s]; Flatten[ro /@ Range[0, 42]] (* Jean-François Alcover, Jul 23 2011 *)
  • Python
    zeck, fib = [], [0, 1]
    from itertools import count, islice
    def agen(): # generator of terms
        for r in count(0):
            while fib[-1] < r:
                fib.append(fib[-2] + fib[-1])
            i = 1
            while fib[-i] > r: i += 1
            bigfib = fib[-i]
            zeck.append( ([] if r == bigfib else zeck[r-bigfib]) + [bigfib] )
            yield from zeck[r]  # row r of the triangle
    print(list(islice(agen(), 90))) # Michael S. Branicky, Apr 04 2022

Extensions

More terms from James Sellers, Dec 13 1999

A087172 Greatest Fibonacci number that does not exceed n.

Original entry on oeis.org

1, 2, 3, 3, 5, 5, 5, 8, 8, 8, 8, 8, 13, 13, 13, 13, 13, 13, 13, 13, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55
Offset: 1

Views

Author

Sam Alexander, Oct 19 2003

Keywords

Comments

Also the largest term in Zeckendorf representation of n; starting at Fibonacci positions the sequence is repeated again and again in A107017: A107017(A000045(n)+k) = a(k) with 0 < k < A000045(n-1). - Reinhard Zumkeller, May 09 2005
Fibonacci(n) occurs Fibonacci(n-1) times, for n >= 2. - Benoit Cloitre, Dec 15 2022

Crossrefs

Programs

  • Haskell
    a087172 = head . a035516_row -- Reinhard Zumkeller, Mar 10 2013
  • Maple
    with(combinat):
    A087172 := proc (n) local j: for j while fibonacci(j) <= n do fibonacci(j) end do: fibonacci(j-1) end proc:
    seq(A087172(n), n = 1 .. 40); # Emeric Deutsch, Nov 11 2014
    # Alternative
    N:= 100: # to get a(n) for n from 1 to N
    Fibs:= [seq(combinat:-fibonacci(i), i = 1 .. ceil(log[(1 + sqrt(5))/2](sqrt(5)*N)))]:
    A:= Vector(N):
    for i from 1 to nops(Fibs)-1 do
      A[Fibs[i] .. min(N,Fibs[i+1]-1)]:= Fibs[i]
    od:
    convert(A,list); # Robert Israel, Nov 11 2014
  • Mathematica
    With[{rf=Reverse[Fibonacci[Range[10]]]},Flatten[Table[ Select[ rf,n>=#&, 1],{n,80}]]] (* Harvey P. Dale, Dec 08 2012 *)
    Flatten[Map[ConstantArray[Fibonacci[#],Fibonacci[#-1]]&,Range[15]]] (* Peter J. C. Moses, May 02 2022 *)
  • PARI
    a(n)=my(k=log(n)\log((1+sqrt(5))/2)); while(fibonacci(k)<=n, k++); fibonacci(k--) \\ Charles R Greathouse IV, Jul 24 2012
    

Formula

a(n) = Fibonacci(A130233(n)) = Fibonacci(A130234(n+1)-1). - Hieronymus Fischer, May 28 2007
a(n) = A035516(n, 0) = A035517(n, A007895(n)-1). - Reinhard Zumkeller, Mar 10 2013
a(n) = n - A066628(n). - Michel Marcus, Feb 02 2016
Sum_{n>=1} 1/a(n)^2 = Sum_{n>=1} Fibonacci(n)/Fibonacci(n+1)^2 = 1.7947486789... . - Amiram Eldar, Aug 16 2022

A107015 Number of even terms in Zeckendorf representation of n.

Original entry on oeis.org

0, 1, 0, 0, 0, 0, 1, 1, 1, 2, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 2, 3, 2, 2, 1, 1, 2, 1, 1, 1, 1, 2, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 2, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 2, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 2, 1, 1, 0, 0, 1, 0
Offset: 1

Views

Author

Reinhard Zumkeller, May 09 2005

Keywords

Comments

a(n) = A007895(n) - A107016(n).
a(A107228(n)) = 0. - Reinhard Zumkeller, May 15 2005

Examples

			n = 77 = 55+21+1 -> a(77) = #{} = 0;
n = 88 = 55+21+8+3+1 -> a(88) = #{8} = 1;
n = 99 = 89+8+2 -> a(99) = #{2, 8} = 2.
		

Crossrefs

Programs

A107016 Number of odd terms in Zeckendorf representation of n.

Original entry on oeis.org

1, 0, 1, 2, 1, 2, 1, 0, 1, 0, 1, 2, 1, 2, 1, 2, 3, 2, 3, 2, 1, 2, 1, 2, 3, 2, 3, 2, 1, 2, 1, 2, 3, 0, 1, 0, 1, 2, 1, 2, 1, 0, 1, 0, 1, 2, 1, 2, 1, 2, 3, 2, 3, 2, 1, 2, 1, 2, 3, 2, 3, 2, 1, 2, 1, 2, 3, 2, 3, 2, 3, 4, 3, 4, 3, 2, 3, 2, 3, 4, 3, 4, 3, 2, 3, 2, 3, 4, 1, 2, 1, 2, 3, 2, 3, 2, 1, 2, 1, 2, 3, 2, 3, 2, 3
Offset: 1

Views

Author

Reinhard Zumkeller, May 09 2005

Keywords

Comments

a(n) = A007895(n) - A107015(n).
a(A107227(n)) = 0. - Reinhard Zumkeller, May 15 2005

Examples

			n = 77 = 55+21+1 -> a(77) = #{1, 21, 55} = 3;
n = 88 = 55+21+8+3+1 -> a(88) = #{1, 3, 21, 55} = 4;
n = 99 = 89+8+2 -> a(99) = #{89} = 1.
		

Crossrefs

Programs

A035514 Zeckendorf expansion of n: repeatedly subtract the largest Fibonacci number you can until nothing remains. Big-endian concatenation of decimals.

Original entry on oeis.org

0, 1, 2, 3, 31, 5, 51, 52, 8, 81, 82, 83, 831, 13, 131, 132, 133, 1331, 135, 1351, 1352, 21, 211, 212, 213, 2131, 215, 2151, 2152, 218, 2181, 2182, 2183, 21831, 34, 341, 342, 343, 3431, 345, 3451, 3452, 348, 3481, 3482, 3483, 34831, 3413, 34131, 34132
Offset: 0

Views

Author

Keywords

Examples

			16 = 13 + 3, so a(16)=13_3 => 133.
		

References

  • Zeckendorf, E., Représentation des nombres naturels par une somme des nombres de Fibonacci ou de nombres de Lucas, Bull. Soc. Roy. Sci. Liège 41, 179-182, 1972.

Crossrefs

Programs

  • Haskell
    a035514 n = a035514_list !! (n-1)
    a035514_list = map (read . concatMap show) a035516_tabf :: [Integer]
    -- Reinhard Zumkeller, Mar 10 2013

Extensions

More terms from James Sellers, Dec 13 1999

A035515 Zeckendorf expansion of n: repeatedly subtract the largest Fibonacci number you can until nothing remains. Little-endian concatenation of decimals.

Original entry on oeis.org

0, 1, 2, 3, 13, 5, 15, 25, 8, 18, 28, 38, 138, 13, 113, 213, 313, 1313, 513, 1513, 2513, 21, 121, 221, 321, 1321, 521, 1521, 2521, 821, 1821, 2821, 3821, 13821, 34, 134, 234, 334, 1334, 534, 1534, 2534, 834, 1834, 2834, 3834, 13834, 1334, 11334, 21334
Offset: 0

Views

Author

Keywords

Examples

			16 = 13 + 3, so a(16)=3_13 => 313.
		

References

  • Zeckendorf, E., Représentation des nombres naturels par une somme des nombres de Fibonacci ou de nombres de Lucas, Bull. Soc. Roy. Sci. Liège 41, 179-182, 1972.

Crossrefs

Programs

  • Haskell
    a035515 n = a035515_list !! (n-1)
    a035515_list = map (read . concatMap show) a035517_tabf :: [Integer]
    -- Reinhard Zumkeller, Mar 10 2013

Extensions

More terms from James Sellers, Dec 13 1999

A106530 Least common multiple of all parts in Zeckendorf representation of n.

Original entry on oeis.org

1, 2, 3, 3, 5, 5, 10, 8, 8, 8, 24, 24, 13, 13, 26, 39, 39, 65, 65, 130, 21, 21, 42, 21, 21, 105, 105, 210, 168, 168, 168, 168, 168, 34, 34, 34, 102, 102, 170, 170, 170, 136, 136, 136, 408, 408, 442, 442, 442, 1326, 1326, 2210, 2210, 2210, 55, 55, 110, 165, 165, 55, 55, 110, 440, 440
Offset: 1

Views

Author

Reinhard Zumkeller, May 08 2005

Keywords

Comments

Records: A106531(n) = a(A106532(n)).
a(n) = lcm_{k=0..A007895(n)-1} A035516(n,k). - Reinhard Zumkeller, Mar 10 2013

Examples

			n=33: 21+8+3+1 = 33 -> a(33) = lcm(21,8,3,1) = (21*8*3*1)/3 = 168.
		

Crossrefs

Programs

  • Haskell
    a106530 = foldr lcm 1 . a035516_row -- Reinhard Zumkeller, Mar 10 2013
  • Maple
    F:= combinat[fibonacci]:
    a:= proc(n) option remember; local j;
          if n=0 then 1
        else for j from 2 while F(j+1)<=n do od;
             ilcm(a(n-F(j)), F(j))
          fi
        end:
    seq(a(n), n=1..64);  # Alois P. Heinz, Mar 17 2018
  • Mathematica
    t = Fibonacci /@ Range@ 21; Table[LCM @@ If[MemberQ[t, n], {n}, Most@ MapAt[# + 1 &, Abs@ Differences@ FixedPointList[# - First@ Reverse@ TakeWhile[t, Function[k, # >= k]] &, n], -1]], {n, 61}] (* Michael De Vlieger, May 17 2016 *)
Showing 1-10 of 14 results. Next