A035515 -id:A035515 - cp's OEIS frontend

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

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

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

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)

			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)

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.

T. D. Noe, Table of n, a(n) for n = 0..10000
Joerg Arndt, Matters Computational (The Fxtbook), pp. 754-756.
Paul Baird-Smith, Alyssa Epstein, Kristen Flint, and Steven J. Miller, The Zeckendorf Game, arXiv:1809.04881 [math.NT], 2018.
D. E. Daykin, Representation of natural numbers as sums of generalized Fibonacci numbers, J. London Math. Soc. 35 (1960) 143-160.
Michel Dekking, Points of increase of the sum of digits function of the base phi expansion, arXiv:2003.14125 [math.CO], 2020.
F. Michel Dekking, The sum of digits functions of the Zeckendorf and the base phi expansions, Theoretical Computer Science (2021) Vol. 859, 70-79.
Damien Jamet, Pierre Popoli, and Thomas Stoll, Maximum order complexity of the sum of digits function in Zeckendorf base and polynomial subsequences, arXiv:2106.09959 [math.NT], 2021, see p. 5.
C. G. Lekkerkerker, Voorstelling van natuurlijke getallen door een som van getallen van Fibonacci, Stichting Mathematisch Centrum, Zuivere Wiskunde, 1951.
A. J. Macfarlane, On the fibbinary numbers and the Wythoff array, arXiv:2405.18128 [math.CO], 2024. See p. 10.
I. Nemes, Fibonacci representations of multiples of Fibonacci numbers.
Don Reble, Zeckendorf vs. Wythoff representations: Comments on A007895.
Thomas Stoll, Combinatorial constructions for the Zeckendorf sum of digits of polynomial values, The Ramanujan Journal November 2013, Volume 32, Issue 2, pp 227-243.
F. V. Weinstein, Notes on Fibonacci partitions, arXiv:math/0307150 [math.NT], 2003-2018.

Cf. A000045, A007953, A035514, A035515, A035516, A035517, A105446, A189920, A213676, A000120, A001950, A003714, A007015, A007016, A104324, A182535, A213911, A014417, A003849.
Cf. A135817 (lengths of Wythoff representation), A135818 (number of 1's (or A's) in the Wythoff representation).
Record positions are in A027941.

a007895 = length . a035516_row  -- Reinhard Zumkeller, Mar 10 2013

# 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

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 *)

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

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

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

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

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

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

N. J. A. Sloane

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]

			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;

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.

T. D. Noe, Rows n = 0..1000 of triangle, flattened
D. E. Knuth, Fibonacci multiplication, Appl. Math. Lett. 1 (1988), 57-60.
N. J. A. Sloane, Classic Sequences

Cf. A014417, A007895, A035514, A035515, A035516.

a035517 n k = a035517_tabf !! n !! k
a035517_row n = a035517_tabf !! n
a035517_tabf = map reverse a035516_tabf
-- Reinhard Zumkeller, Mar 10 2013

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

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 *)

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

More terms from James Sellers, Dec 13 1999

A035516 Triangular array formed from Zeckendorf expansion of integers: repeatedly subtract the largest Fibonacci number you can until nothing remains.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane

Row n has A007895(n) terms.

			16 = 13 + 3, so row 16 is 13, 3. [Corrected by _Sean A. Irvine_, Oct 14 2020]
The first few rows are:
  0;
  1;
  2;
  3;
  3, 1;
  5;
  5, 1;
  5, 2;
  8;
  8, 1;
  8, 2;
  ...
Row 1000000 is 832040,121393,46368,144,55. Indeed, the Maple program yields in no time Z(1000000) = {55,144,46368,121393,832040}. - _Emeric Deutsch_, Oct 22 2014

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.

T. D. Noe, Rows n=0..1000 of triangle, flattened
N. J. A. Sloane, Classic Sequences

Cf. A035517, A035514, A035515, A000045, A106530, A273156.

a035516 n k = a035516_tabf !! n !! k
a035516_tabf = map a035516_row [0..]
a035516_row 0 = [0]
a035516_row n = z n $ reverse $ takeWhile (<= n) a000045_list where
   z 0 _              = []
   z x (f:fs'@(_:fs)) = if f <= x then f : z (x - f) fs else z x fs'
-- Reinhard Zumkeller, Mar 10 2013

with(combinat): Z := proc (n) local F, LF, A, m: F := proc (n) options operator, arrow: fibonacci(n) end proc: LF := proc (m) local i: for i from 0 while F(i) <= m do  end do: F(i-1) end proc: A := {}: m := n: while 0 < m do A := `union`(A, {LF(m)}): m := m-LF(m) end do: A end proc: # The Maple program, with the command Z(n), yields the set of the Fibonacci numbers in the Zeckendorf representation of n (terms in {} are in reverse order). - Emeric Deutsch, Oct 21 2014

t = Fibonacci /@ Range@ 12; Table[If[MemberQ[t, n], {n}, Most@ MapAt[# + 1 &, Abs@ Differences@ FixedPointList[# - First@ Reverse@ TakeWhile[t, Function[k, # >= k]] &, n], -1]], {n, 41}] // Flatten (* faster, or *)
t = Fibonacci /@ Range@ 12; {{0}}~Join~Table[First@ Select[ Select[ IntegerPartitions@ n, Times @@ Boole@ Map[MemberQ[t, #] &, #] == 1 &], Times @@ Boole@ Map[# > 1 &, Abs@ Differences@ Map[Position[t, #][[1, 1]] &, #, {1}]] == 1 &], {n, 41}] // Flatten (* Michael De Vlieger, May 17 2016 *)

More terms from James Sellers, Dec 13 1999

cp's OEIS Frontend

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

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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.

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A035516 Triangular array formed from Zeckendorf expansion of integers: repeatedly subtract the largest Fibonacci number you can until nothing remains.

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

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