A035612 -id:A035612 - cp's OEIS frontend

A003603 Fractal sequence obtained from Fibonacci numbers (or Wythoff array).

1, 1, 1, 2, 1, 3, 2, 1, 4, 3, 2, 5, 1, 6, 4, 3, 7, 2, 8, 5, 1, 9, 6, 4, 10, 3, 11, 7, 2, 12, 8, 5, 13, 1, 14, 9, 6, 15, 4, 16, 10, 3, 17, 11, 7, 18, 2, 19, 12, 8, 20, 5, 21, 13, 1, 22, 14, 9, 23, 6, 24, 15, 4, 25, 16, 10, 26, 3, 27, 17, 11, 28, 7, 29, 18, 2, 30, 19, 12, 31, 8, 32, 20, 5, 33

Offset: 1

N. J. A. Sloane, Mira Bernstein

Length of n-th row = A000045(n); last term of n-th row = A094967(n-1); sum of n-th row = A033192(n-1). - Reinhard Zumkeller, Jan 26 2012

			In the recurrence for making new rows, we get row 5 from row 4 thus: write row 4: 1,3,2, and then place 4 right after 1, and place 5 right after 2, getting 1,4,3,2,5. - _Clark Kimberling_, Oct 29 2009

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Reinhard Zumkeller, Rows n = 1..20 of triangle, flattened
J. H. Conway and N. J. A. Sloane, Notes on the Para-Fibonacci and related sequences.
David Garth and Joseph Palmer, Self-Similar Sequences and Generalized Wythoff Arrays, Fibonacci Quart. 54 (2016), no. 1, 72-78.
Clark Kimberling, Fractal sequences.
Clark Kimberling, Numeration systems and fractal sequences, Acta Arithmetica 73 (1995) 103-117.
A. J. Macfarlane, On the fibbinary numbers and the Wythoff array, arXiv:2405.18128 [math.CO], 2024. See page 8.
N. J. A. Sloane, Classic Sequences.

Equals A019586(n) + 1. Cf. A003602, A000045, A033192, A035513, A035612, A094967, A265650.

-- according to Kimberling, see formula section.
a003603 n k = a003603_row n !! (k-1)
a003603_row n = a003603_tabl !! (n-1)
a003603_tabl = [1] : [1] : wythoff [2..] [1] [1] where
   wythoff is xs ys = f is xs ys [] where
      f js     []     []     ws = ws : wythoff js ys ws
      f js     []     [v]    ws = f js [] [] (ws ++ [v])
      f (j:js) (u:us) (v:vs) ws
        | u == v = f js us vs (ws ++ [v,j])
        | u /= v = f (j:js) (u:us) vs (ws ++ [v])
-- Reinhard Zumkeller, Jan 26 2012

A003603 := proc(n::posint)
    local r,c,W ;
    for r from 1 do
        for c from 1 do
            W := A035513(r,c) ;
            if W = n then
                return r ;
            elif W > n then
                break ;
            end if;
        end do:
    end do:
end proc:
seq(A003603(n),n=1..100) ; # R. J. Mathar, Aug 13 2021

num[n_, b_] := Last[NestWhile[{Mod[#[[1]], Last[#[[2]]]], Drop[#[[2]], -1], Append[#[[3]], Quotient[#[[1]], Last[#[[2]]]]]} &, {n, b, {}}, #[[2]] =!= {} &]];
left[n_, b_] := If[Last[num[n, b]] == 0, Dot[num[n, b], Rest[Append[Reverse[b], 0]]], n];
fractal[n_, b_] := # - Count[Last[num[Range[#], b]], 0] &@
   FixedPoint[left[#, b] &, n];
Table[fractal[n, Table[Fibonacci[i], {i, 2, 12}]], {n, 30}] (* Birkas Gyorgy, Apr 13 2011 *)
row[1] = row[2] = {1};
row[n_] := row[n] = Module[{ro, pos, lp, ins}, ro = row[n-1]; pos = Position[ro, Alternatives @@ Intersection[ro, row[n-2]]] // Flatten; lp = Length[pos]; ins = Range[lp] + Max[ro]; Do[ro = Insert[ro, ins[[i]], pos[[i]] + i], {i, 1, lp}]; ro];
Array[row, 9] // Flatten (* Jean-François Alcover, Jul 12 2016 *)

Vertical para-budding sequence: says which row of Wythoff array (starting row count at 1) contains n.
If one deletes the first occurrence of 1, 2, 3, ... the sequence is unchanged.
From Clark Kimberling, Oct 29 2009: (Start)
The fractal sequence of the Wythoff array can be constructed without reference to the Wythoff array or Fibonacci numbers. Write initial rows:
Row 1: .... 1
Row 2: .... 1
Row 3: .... 1..2
Row 4: .... 1..3..2
For n>4, to form row n+1, let k be the least positive integer not yet used; write row n, and right after the first number that is also in row n-1, place k; right after the next number that is also in row n-1, place k+1, and continue. A003603 is the concatenation of the rows. (End)
Conjecture: a(n) = abs(floor(n/phi) - floor(n*(1/phi + 1/(-phi)^(A035612(n) + 1)))) where phi = (1+sqrt(5))/2. - Alan Michael Gómez Calderón, Oct 27 2023

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Mar 29 2003

Keyword tabf added by Reinhard Zumkeller, Jan 26 2012

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

Marc LeBrun

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)

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

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.

Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from T. D. Noe)
Joerg Arndt, Matters Computational (The Fxtbook)
A. J. Macfarlane, On the fibbinary numbers and the Wythoff array, arXiv:2405.18128 [math.CO], 2024. See pages 3, 6.
M. Rigo, P. Salimov, and E. Vandomme, Some Properties of Abelian Return Words, Journal of Integer Sequences, Vol. 16 (2013), #13.2.5.
Jeffrey Shallit, The Hurt-Sada Array and Zeckendorf Representations, arXiv:2501.08823 [math.NT], 2025. See p. 2.
N. J. A. Sloane, My favorite integer sequences, in Sequences and their Applications (Proceedings of SETA '98).
N. J. A. Sloane, Classic Sequences
N. J. A. Sloane, Families of Essentially Identical Sequences, Mar 24 2021 (Includes this sequence)
Jiemeng Zhang, Zhixiong Wen, and Wen Wu, Some Properties of the Fibonacci Sequence on an Infinite Alphabet, Electronic Journal of Combinatorics, 24(2) (2017), Article P2.52.

Positions of 0's in A003849.
Complement of A003622.
Cf. A000201, A005206, A035336, A066096, A001950, A062879, A035516, A026274.
Cf. A035612, A094214, A104326, A356749.
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

a022342 n = a022342_list !! (n-1)
a022342_list = filter ((notElem 1) . a035516_row) [0..]
-- Reinhard Zumkeller, Mar 10 2013

[Floor(n*(Sqrt(5)+1)/2)-1: n in [1..100]]; // Vincenzo Librandi, Feb 16 2015

A022342 := proc(n)
      local g;
      g := (1+sqrt(5))/2 ;
    floor(n*g)-1 ;
end proc: # R. J. Mathar, Aug 04 2013

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
```

a(n)=(sqrtint(5*n^2)+n-2)\2 \\ Charles R Greathouse IV, Feb 27 2014

from math import isqrt
def A022342(n): return (n+isqrt(5*n**2)>>1)-1 # Chai Wah Wu, Aug 17 2022

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

Name edited by Peter Munn, Dec 07 2021

A026274 Greatest k such that s(k) = n, where s = A026272.

Original entry on oeis.org

3, 5, 8, 11, 13, 16, 18, 21, 24, 26, 29, 32, 34, 37, 39, 42, 45, 47, 50, 52, 55, 58, 60, 63, 66, 68, 71, 73, 76, 79, 81, 84, 87, 89, 92, 94, 97, 100, 102, 105, 107, 110, 113, 115, 118, 121, 123, 126, 128, 131, 134, 136, 139, 141, 144

Offset: 1

Clark Kimberling

This is the upper s-Wythoff sequence, where s(n)=n+1.
See comments at A026273.
Conjecture: This sequence consists precisely of those numbers without a 1 or 2 in their Zeckendorf representation. [In other words, numbers which are the sum of distinct nonconsecutive Fibonacci numbers greater than 2.] - Charles R Greathouse IV, Jan 28 2015
A Beatty sequence with complement A026273. - Robert G. Wilson v, Jan 30 2015
A035612(a(n)+1) = 1. - Reinhard Zumkeller, Jul 20 2015
From Michel Dekking, Mar 12 2018: (Start)
One has r*r*(n-2*r+3) = n*r^2 -2r^3+3*r^2 = (n+1)*r^2 -2, where r = (1+sqrt(5))/2.
So a(n) = floor((n+1)*r^2)-2, and we see that this sequence is simply the Beatty sequence of the square of the golden ratio, shifted spatially and temporally. In other words, if w = A001950 = 2,5,7,10,13,15,18,20,... is the upper Wythoff sequence, then a(n) = w(n+1) - 2.
(End)
From Michel Dekking, Apr 05 2020: (Start)
Proof of the conjecture by Charles R Greathouse IV.
Let Z(n) = d(L)...d(1)d(0) be the Zeckendorf expansion of n. Well-known is:
      d(0) = 1 if and only if n = floor(k*r^2) - 1
for some integer k (see A003622).
Then the same characterization holds for n with d(1)d(0) = 01, since 11 does not appear in a Zeckendorf expansion. But such an n has predecessor n-1 which always has an expansion with d(1)d(0) = 00. Combined with my comment from March 2018, this proves the conjecture (ignoring n = 0). (End)
It appears that these are the integers m for which A007895(m+1) > A007895(m) where A007895(m) is the number of terms in Zeckendorf representation of m. - Michel Marcus, Oct 30 2020
This follows directly from Theorem 4 in my paper "Points of increase of the sum of digits function of the base phi expansion". - Michel Dekking, Oct 31 2020

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Hung Viet Chu, Difference in the Number of Summands in the Zeckendorf Partitions of Consecutive Integers, arXiv:2010.15592 [math.NT], 2020.
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.

Cf. A184117, A026273, A001950, A003622.

a026274 n = a026274_list !! (n-1)
a026274_list = map (subtract 1) $ tail $ filter ((== 1) . a035612) [1..]
-- Reinhard Zumkeller, Jul 20 2015

r=(1+Sqrt[5])/2;
a[n_]:=Floor[r*r*(n+2r-3)];
Table[a[n],{n,200}]
Table[Floor[GoldenRatio^2 (n+2*GoldenRatio-3)],{n,60}] (* Harvey P. Dale, Dec 23 2022 *)

a(n)=my(w=quadgen(20),phi=(1+w)/2); phi^2*(n+2*phi-3)\1 \\ Charles R Greathouse IV, Nov 10 2021

from math import isqrt
def A026274(n): return (n+1+isqrt(5*(n+1)**2)>>1)+n-1 # Chai Wah Wu, Aug 17 2022

a(n) = floor(r*r*(n+2r-3)), where r = (1+sqrt(5))/2 = A001622. [Corrected by Tom Edgar, Jan 30 2015]
a(n) = 3*n - floor[(n+1)/(1+phi)], phi = (1+sqrt(5))/2. - Joshua Tobin (tobinrj(AT)tcd.ie), May 31 2008
a(n) = A003622(n+1) - 1 for n>1 (conjectured). - Michel Marcus, Oct 30 2020
This conjectured formula follows directly from the formula a(n) = floor((n+1)*r^2)-2 in my Mar 12 2018 comment above. - Michel Dekking, Oct 31 2020

Extended by Clark Kimberling, Jan 14 2011

A035614 Horizontal para-Fibonacci sequence: says which column of Wythoff array (starting column count at 0) contains n+1.

Original entry on oeis.org

0, 1, 2, 0, 3, 0, 1, 4, 0, 1, 2, 0, 5, 0, 1, 2, 0, 3, 0, 1, 6, 0, 1, 2, 0, 3, 0, 1, 4, 0, 1, 2, 0, 7, 0, 1, 2, 0, 3, 0, 1, 4, 0, 1, 2, 0, 5, 0, 1, 2, 0, 3, 0, 1, 8, 0, 1, 2, 0, 3, 0, 1, 4, 0, 1, 2, 0, 5, 0, 1, 2, 0, 3, 0, 1, 6, 0, 1, 2, 0, 3

Offset: 0

J. H. Conway and N. J. A. Sloane

This is probably the same as the "Fibonacci ruler function" mentioned by Knuth. - N. J. A. Sloane, Aug 03 2012
From Amiram Eldar, Mar 10 2021: (Start)
a(n) is the number of the trailing zeros in the Zeckendorf representation of (n+1) (A014417).
The asymptotic density of the occurrences of k is 1/phi^(k+2), where phi is the golden ratio (A001622).
The asymptotic mean of this sequence is phi. (End)

D. E. Knuth, The Art of Computer Programming, Vol. 4A, Section 7.1.3, p. 82, solution to Problem 179.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
Casey Mongoven, Sonification of multiple Fibonacci-related sequences, Annales Mathematicae et Informaticae, 41 (2013) pp. 175-192.
N. J. A. Sloane, Classic Sequences

Cf. A000045, A001622, A014417, A019586, A035513, A035612, A122840, A139764.

a035614 = a122840 . a014417 . (+ 1)  -- Reinhard Zumkeller, Mar 10 2013

max = 81; wy = Table[(n-k)*Fibonacci[k] + Fibonacci[k+1]*Floor[ GoldenRatio*(n - k + 1)], {n, 1, max}, {k, 1, n}]; a[n_] := Position[wy, n][[1, 2]]-1; Table[a[n], {n, 1, max}] (* Jean-François Alcover, Nov 02 2011 *)

from sympy import fibonacci
def a122840(n): return len(str(n)) - len(str(int(str(n)[::-1])))
def a014417(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 x
def a(n): return a122840(a014417(n + 1)) # Indranil Ghosh, Jun 09 2017, after Haskell code by Reinhard Zumkeller

The segment between the first M and the first M+1 is given by the segment before the first M-1.

a(n) = A122840(A014417(n + 1)). - Indranil Ghosh, Jun 09 2017

A083368 a(n) is the position of the highest one in A003754(n).

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 4, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 6, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 7, 2, 1, 4, 1, 3, 2, 1, 6, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 8, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 7, 2, 1, 4, 1, 3, 2, 1, 6, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 9, 2, 1, 4, 1, 3, 2, 1, 6, 1, 3, 2, 1, 5, 2

Offset: 1

Gary W. Adamson, Jun 04 2003

Previous name was: A Fibbinary system represents a number as a sum of distinct Fibonacci numbers (instead of distinct powers of two). Using representations without adjacent zeros, a(n) = the highest bit-position which changes going from n-1 to n.
A003754(n), when written in binary, is the representation of n.
Often one uses Fibbinary representations without adjacent ones (the Zeckendorf expansion).
a(A000071(n+3)) = n. - Reinhard Zumkeller, Aug 10 2014
From Gus Wiseman, Jul 24 2025: (Start)
Conjecture: To obtain this sequence, start with A245563 (maximal run lengths of binary indices), then remove empty and duplicate rows (giving A385817), then take the first term of each remaining row. Some variations:
- For sum instead of first term we appear to have A200648.
- For length instead of first term we appear to have A200650+1.
- For last instead of first term we have A385892.
(End)

			27 is represented 110111, 28 is 111010; the fourth position changes, so a(28)=4.

Jay Kappraff, Beyond Measure: A Guided Tour Through Nature, Myth and Number, World Scientific, 2002, page 460.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

A035612 is the analogous sequence for Zeckendorf representations.
A001511 is the analogous sequence for power-of-two representations.
Cf. A003714, A003754.
Cf. A000045, A000071.
Appears to be the first element of each row of A385817, see A083368, A200648, A200650, A385818, A385892.
A000120 counts ones in binary expansion.
A245563 gives run lengths of binary indices, see A089309, A090996, A245562, A246029, A328592.
A384877 gives anti-run lengths of binary indices, ranks A385816.
Cf. A001629, A037800, A044813, A048793, A069010, A200649, A384878, A385886.

a083368 n = a083368_list !! (n-1)
a083368_list = concat $ h $ drop 2 a000071_list where
   h (a:fs@(a':_)) = (map (a035612 . (a' -)) [a .. a' - 1]) : h fs
-- Reinhard Zumkeller, Aug 10 2014

For n = F(a)-1 to F(a+1)-2, a(n) = A035612(F(a+1)-1-n).

a(n) = a(k)+1 if n = ceiling(phi*k) where phi is the golden ratio; otherwise a(n) = 1. - Tom Edgar, Aug 25 2015

Edited by Don Reble, Nov 12 2005

Shorter name from Joerg Arndt, Jul 27 2025

A316628 a(1)=1, a(2)=2, a(3)=2, a(4)=3; a(n) = a(n-a(n-1))+a(n-1-a(n-2)-a(n-2-a(n-2))) for n > 4.

Original entry on oeis.org

1, 2, 2, 3, 3, 3, 4, 5, 5, 5, 5, 6, 7, 7, 8, 8, 8, 8, 8, 9, 10, 10, 11, 11, 11, 12, 13, 13, 13, 13, 13, 13, 14, 15, 15, 16, 16, 16, 17, 18, 18, 18, 18, 19, 20, 20, 21, 21, 21, 21, 21, 21, 21, 22, 23, 23, 24, 24, 24, 25, 26, 26, 26, 26, 27, 28, 28, 29, 29, 29, 29, 29, 30, 31, 31

Offset: 1

Nathan Fox, Jul 08 2018

nonn

This sequence increases slowly.
k occurs A035612(k) times.
Each Fibonacci number occurs more times than any number before it.

Nathan Fox, Table of n, a(n) for n = 1..10000
Nathan Fox, Trees, Fibonacci Numbers, and Nested Recurrences, Rutgers University Experimental Math Seminar, Mar 07, 2019

Cf. A000045, A005185, A035612, A046699.

a:=[1,2,2,3];; for n in [5..80] do a[n]:=a[n-a[n-1]]+a[n-1-a[n-2]-a[n-2-a[n-2]]]; od; a; # Muniru A Asiru, Jul 09 2018

I:=[1,2,2,3]; [n le 4 select I[n] else Self(n-Self(n-1))+Self(n-1-Self(n-2)-Self(n-2-Self(n-2))): n in [1..100]]; // Vincenzo Librandi, Jul 09 2018

A316628:=proc(n) option remember: if n <= 0 then 0: elif n = 1 then 1: elif n = 2 then 2: elif n = 3 then 2: elif n = 4 then 3: else A316628(n-A316628(n-1)) + A316628(n-1-A316628(n-2)-A316628(n-2-A316628(n-2))): fi: end:

a(n+1)-a(n)=1 or 0.

a(n)/n -> C=(sqrt(5)-1)/(sqrt(5)+1).

A255671 Number of the column of the Wythoff array (A035513) that contains U(n), where U = A001950, the upper Wythoff sequence.

Original entry on oeis.org

2, 4, 2, 2, 6, 2, 4, 2, 2, 4, 2, 2, 8, 2, 4, 2, 2, 6, 2, 4, 2, 2, 4, 2, 2, 6, 2, 4, 2, 2, 4, 2, 2, 10, 2, 4, 2, 2, 6, 2, 4, 2, 2, 4, 2, 2, 8, 2, 4, 2, 2, 6, 2, 4, 2, 2, 4, 2, 2, 6, 2, 4, 2, 2, 4, 2, 2, 8, 2, 4, 2, 2, 6, 2, 4, 2, 2, 4, 2, 2, 6, 2, 4, 2, 2, 4

Offset: 1

Clark Kimberling, Mar 03 2015

All the terms are even, and every even positive integer occurs infinitely many times.
From Michel Dekking, Dec 09 2024 and Ad van Loon: (Start)
This sequence has a self-similarity property:
      a(U(n)) = a(n) + 2 for all n.
Proof: it is known that the columns C_h of the Wythoff array are compound Wythoff sequences. For example: C_1 = L^2, C_2 = UL.
In general column C_h is equal to LU^{(h-1)/2} if h is odd, and to U^{h/2}L if h is even (see Theorem 10 in Kimberling’s 2008 paper in JIS).
Now if h is odd then the elements of column C_h are a subsequence of L, so no U(m) can occur in such a column.
If h is even then the elements of column C_h form a subsequence of U, and so many U(m) occur. Suppose that a(m) = h. Then U(U(m)) is an element of column UU^{h/2}L = U^{(h+2)/2}L. This implies a(U(m)) = a(m) +2. (End)

			Corner of the Wythoff array:
  1   2   3   5   8   13
  4   7   11  18  29  47
  6   10  16  26  42  68
  9   15  24  39  63  102
L = (1,3,4,6,8,9,11,...); U = (2,5,7,10,13,15,18,...), so that
A255670 = (1,3,1,1,5,...) and A255671 = (2,4,2,2,6,...).

Ad van Loon, The structure of the expansions, See Section 5.

Cf. A255670, A035612, A000201, A001950.

z = 13; r = GoldenRatio; f[1] = {1}; f[2] = {1, 2};
f[n_] := f[n] = Join[f[n - 1], Most[f[n - 2]], {n}]; f[z];
g[n_] := g[n] = f[z][[n]]; Table[g[n], {n, 1, 100}]  (* A035612 *)
Table[g[Floor[n*r]], {n, 1, (1/r) Length[f[z]]}]     (* A255670 *)
Table[g[Floor[n*r^2]], {n, 1, (1/r^2) Length[f[z]]}] (* A255671 *)

a(n) = 2 if and only if n = L(j) for some j; otherwise, n = U(k) for some k.

a(n) = A255670(n) + 1 = A035612(A001950(n)).

A352538 Primes whose position in the Wythoff array is immediately followed by another prime in the next column.

Original entry on oeis.org

2, 3, 7, 19, 23, 29, 67, 97, 103, 107, 149, 181, 227, 271, 311, 353, 379, 433, 449, 563, 631, 719, 761, 883, 919, 941, 971, 997, 1049, 1087, 1223, 1291, 1297, 1427, 1447, 1453, 1531, 1601, 1627, 1699, 1753, 1831, 1861, 1877, 2039, 2207, 2213, 2239, 2269, 2281, 2287

Offset: 1

Michel Marcus, Mar 20 2022

nonn

			The Wythoff array begins:
   1    2    3    5    8 ...
   4    7   11   18   29 ...
   6   10   16   26   42 ...
   ...
So 2, 3 and 7 are terms, since they are horizontally followed by 3, 5 and 11.

Cf. A003603, A022342, A035612, A035513 (Wythoff array).

Cf. A352537 (next row and column), A352539 (next row).

T(n,k) = (n+sqrtint(5*n^2))\2*fibonacci(k+1) + (n-1)*fibonacci(k); \\ A035513
cell(n) = for (r=1, oo, for (c=1, oo, if (T(r,c) == n, return([r, c])); if (T(r,c) > n, break);););
isokh(m) = {my(pos = cell(prime(m))); isprime (T(pos[1], pos[2]+1))};
lista(nn) = for (n=1, nn, if (isokh(n), print1(prime(n), ", ")));

right(n) = n++; (sqrtint(5*n^2)+n-2)\2; \\ see A022342
isokh(n) = isprime(right(n));
lista(nn) = for (n=1, nn, my(p=prime(n)); if (isokh(p), print1(p, ", ")));

A255670 Number of the column of the Wythoff array (A035513) that contains L(n), where L = A000201, the lower Wythoff sequence.

Original entry on oeis.org

1, 3, 1, 1, 5, 1, 3, 1, 1, 3, 1, 1, 7, 1, 3, 1, 1, 5, 1, 3, 1, 1, 3, 1, 1, 5, 1, 3, 1, 1, 3, 1, 1, 9, 1, 3, 1, 1, 5, 1, 3, 1, 1, 3, 1, 1, 7, 1, 3, 1, 1, 5, 1, 3, 1, 1, 3, 1, 1, 5, 1, 3, 1, 1, 3, 1, 1, 7, 1, 3, 1, 1, 5, 1, 3, 1, 1, 3, 1, 1, 5, 1, 3, 1, 1, 3

Offset: 1

Clark Kimberling, Mar 03 2015

All the terms are odd, and every odd positive integer occurs infinitely many times.

			Corner of the Wythoff array:
  1   2   3   5   8   13
  4   7   11  18  29  47
  6   10  16  26  42  68
  9   15  24  39  63  102
L = (1,3,4,6,8,9,11,...); U = (2,5,7,10,13,15,18,...), so that
this sequence = (1,3,1,1,5,...) and A255671 = (2,4,2,2,6,...).

Nicholas John Bizzell-Browning, LIE scales: Composing with scales of linear intervallic expansion, Ph. D. Thesis, Brunel Univ. (UK, 2024). See p. 39.

Cf. A255671, A035612, A000201, A001950.

z = 13; r = GoldenRatio; f[1] = {1}; f[2] = {1, 2};
f[n_] := f[n] = Join[f[n - 1], Most[f[n - 2]], {n}]; f[z];
g[n_] := g[n] = f[z][[n]]; Table[g[n], {n, 1, 100}]  (* A035612 *)
Table[g[Floor[n*r]], {n, 1, (1/r) Length[f[z]]}]     (* A255670 *)
Table[g[Floor[n*r^2]], {n, 1, (1/r^2) Length[f[z]]}] (* A255671 *)

a(n) = A255671(n) - 1 = A035612(A000201(n)).

a(n) = 1 if and only if n = L(j) for some j; otherwise, n = U(k) for some k.

A268034 A268032 with repeated 1's removed.

Original entry on oeis.org

3, 5, 11, 3, 21, 3, 5, 43, 3, 5, 11, 3, 85, 3, 5, 11, 3, 21, 3, 5, 171, 3, 5, 11, 3, 21, 3, 5, 43, 3, 5, 11, 3, 341, 3, 5, 11, 3, 21, 3, 5, 43, 3, 5, 11, 3, 85, 3, 5, 11, 3, 21, 3, 5, 683, 3, 5, 11, 3, 21, 3, 5, 43, 3, 5

Offset: 1

Jeremy Gardiner, Jan 24 2016

nonn

Records appear to be given by A001045 Jacobsthal numbers.
(a(n)-1)/2 appears to be A085358.
The terms between the A001045(n+3) are:
3
5
11
      3,
21
      3, 5,
43
      3, 5, 11, 3,
85
      3, 5, 11, 3, 21, 3, 5,
171
      3, 5, 11, 3, 21, 3, 5, 43, 3, 5, 11, 3,
341
      3, 5, 11, 3, 21, 3, 5, 43, 3, 5, 11, 3, 85, 3, 5, 11, 3, 21, 3, 5,
683
This gives the same sequence. Every column has the same number.
By rows, there are 0, 0, 1, 2, 4, 7, 12, 20, ... apparently = Fibonacci(n+1) - 1 = A000071 terms.
From Paul Curtz, Jan 26 2016: (Start)
a(n) is also in
  0,  1,  1  0,  3, 0,  1,  5, 0, ...  equivalent to A035614(n)
  1,  1,  3, 1,  5, 1,  1, 11, 1, ...  equivalent to A035612(n)
  1,  3,  5, 1, 11, 1,  3, 21, 1, ... (compare to A268032)
  3,  5, 11, 3, 21, 3,  5, 43, 3, ...  a(n) (equivalent to a3(n) in A035612)
  5, 11, 21, 5, 43, 5, 11, 85, 5, ...
etc.
Every vertical comes from A001045 (*).
Second row: first one removing all 0's.
Third row: second one removing a part of 1's respecting (*)
Fourth row: third one removing all 1's.
etc.
The offset 0 is homogeneous to these sequences. (End)

			A268032 begins 1, 1, 1, 1, 1, 3, 1, 1, 1, 5, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 3, 1, 1, 1, 21, ... hence this sequence begins 3, 5, 11, 3, 21, ...

Cf. A000071, A001045, A035612, A035614, A085358, A268032.

cp's OEIS Frontend

A003603 Fractal sequence obtained from Fibonacci numbers (or Wythoff array).

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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.

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A026274 Greatest k such that s(k) = n, where s = A026272.

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A035614 Horizontal para-Fibonacci sequence: says which column of Wythoff array (starting column count at 0) contains n+1.

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A083368 a(n) is the position of the highest one in A003754(n).

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A316628 a(1)=1, a(2)=2, a(3)=2, a(4)=3; a(n) = a(n-a(n-1))+a(n-1-a(n-2)-a(n-2-a(n-2))) for n > 4.

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