cp's OEIS Frontend

This is the unique sequence a satisfying a'(n)=a(a(n))+1 for all n in the set N of natural numbers, where a' denotes the ordered complement (in N) of a. - Clark Kimberling, Feb 17 2003

This sequence and A001950 may be defined as follows. Consider the maps a -> ab, b -> a, starting from a(1) = a; then A000201 gives the indices of a, A001950 gives the indices of b. The sequence of letters in the infinite word begins a, b, a, a, b, a, b, a, a, b, a, ... Setting a = 0, b = 1 gives A003849 (offset 0); setting a = 1, b = 0 gives A005614 (offset 0). - Philippe Deléham, Feb 20 2004

These are the numbers whose lazy Fibonacci representation (see A095791) includes 1; the complementary sequence (the upper Wythoff sequence, A001950) are the numbers whose lazy Fibonacci representation includes 2 but not 1.

a(n) is the unique monotonic sequence satisfying a(1)=1 and the condition "if n is in the sequence then n+(rank of n) is not in the sequence" (e.g. a(4)=6 so 6+4=10 and 10 is not in the sequence) - Benoit Cloitre, Mar 31 2006

Write A for A000201 and B for A001950 (the upper Wythoff sequence, complement of A). Then the composite sequences AA, AB, BA, BB, AAA, AAB,...,BBB,... appear in many complementary equations having solution A000201 (or equivalently, A001950). Typical complementary equations: AB=A+B (=A003623), BB=A+2B (=A101864), BBB=3A+5B (=A134864). - Clark Kimberling, Nov 14 2007

Cumulative sum of A001468 terms. - Eric Angelini, Aug 19 2008

The lower Wythoff sequence also can be constructed by playing the so-called Mancala-game: n piles of total d(n) chips are standing in a row. The piles are numbered from left to right by 1, 2, 3, ... . The number of chips in a pile at the beginning of the game is equal to the number of the pile. One step of the game is described as follows: Distribute the pile on the very left one by one to the piles right of it. If chips are remaining, build piles out of one chip subsequently to the right. After f(n) steps the game ends in a constant row of piles. The lower Wythoff sequence is also given by n -> f(n). - Roland Schroeder (florola(AT)gmx.de), Jun 19 2010

With the exception of the first term, a(n) gives the number of iterations required to reverse the list {1,2,3,...,n} when using the mapping defined as follows: remove the first term of the list, z(1), and add 1 to each of the next z(1) terms (appending 1's if necessary) to get a new list. See A183110 where this mapping is used and other references given. This appears to be essentially the Mancala-type game interpretation given by R. Schroeder above. - John W. Layman, Feb 03 2011

Also row numbers of A213676 starting with an even number of zeros. - Reinhard Zumkeller, Mar 10 2013

From Jianing Song, Aug 18 2022: (Start)

Numbers k such that {k*phi} > phi^(-2), where {} denotes the fractional part.

Proof: Write m = floor(k*phi).

If {k*phi} > phi^(-2), take s = m-k+1. From m < k*phi < m+1 we have k < (m-k+1)*phi < k + phi, so floor(s*phi) = k or k+1. If floor(s*phi) = k+1, then (see A003622) floor((k+1)*phi) = floor(floor(s*phi)*phi) = floor(s*phi^2)-1 = s+floor(s*phi)-1 = m+1, but actually we have (k+1)*phi > m+phi+phi^(-2) = m+2, a contradiction. Hence floor(s*phi) = k.

If floor(s*phi) = k, suppose otherwise that k*phi - m <= phi^(-2), then m < (k+1)*phi <= m+2, so floor((k+1)*phi) = m+1. Suppose that A035513(p,q) = k for p,q >= 1, then A035513(p,q+1) = floor((k+1)*phi) - 1 = m = A035513(s,1). But it is impossible for one number (m) to occur twice in A035513. (End)

The formula from Jianing Song above is a direct consequence of an old result by Carlitz et al. (1972). Their Theorem 11 states that (a(n)) consists of the numbers k such that {k*phi^(-2)} < phi^(-1). One has {k*phi^(-2)} = {k*(2-phi)} = {-k*phi}. Using that 1-phi^(-1) = phi^(-2), the Jianing Song formula follows. - Michel Dekking, Oct 14 2023

In the Fokkink-Joshi paper, this sequence is the Cloitre (1,1,2,1)-hiccup sequence, i.e., a(1) = 1; for m < n, a(n) = a(n-1)+2 if a(m) = n-1, else a(n) = a(n-1)+1. - Michael De Vlieger, Jul 28 2025

Suppose that s(n) is a nondecreasing sequence of positive integers. The lower and upper s(n)-Wythoff sequences, a and b, are introduced here. Define

a(1) = 1; b(1) = s(1) + a(1); and for n>=2,

a(n) = least positive integer not in {a(1),...,a(n-1),b(1),...,b(n-1)},

b(n) = s(n) + a(n).

Clearly, a and b are complementary. If s(n)=n, then

a=A000201, the lower Wythoff sequence, and

b=A001950, the upper Wythoff sequence.

A184117 is chosen to represent the class of s-Wythoff sequences for which s is an arithmetic sequence given by s(n) = kn - r. Such sequences (lower and upper) are indexed in the OEIS as shown here:

n+1....A026273...A026274

n......A000201...A001950 (the classical Wythoff sequences)

2n+1...A184117...A184118

2n.....A001951...A001952

2n-1...A136119...A184119

3n+1...A184478...A184479

3n.....A184480...A001956

3n-1...A184482...A184483

3n-2...A184484...A184485

4n+1...A184486...A184487

4n.....A001961...A001962

4n-1...A184514...A184515

The pattern continues for A184516 to A184531.

s-Wythoff sequences for choices of s other than arithmetic sequences include these:

A184419 and A184420 (s = lower Wythoff sequence)

A184421 and A184422 (s = upper Wythoff sequence)

A184425 and A184426 (s = triangular numbers)

A184427 and A184428 (s = squares)

A036554 and A003159 (invariant and limiting sequences).

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)

Ordinal transform of A003603. Removing all 1's from this sequence and decrementing the remaining numbers generates the original sequence. - Franklin T. Adams-Watters, Aug 10 2012

It can be shown that a(n) is the index of the smallest Fibonacci number used in the Zeckendorf representation of n, where f(0)=f(1)=1. - Rachel Chaiser, Aug 18 2017

The asymptotic density of the occurrences of k = 1, 2, ..., is (2-phi)/phi^(k-1), where phi is the golden ratio (A001622). The asymptotic mean of this sequence is 1 + phi (A104457). - Amiram Eldar, Nov 02 2023

Let n = Sum_{i >= 2} eps(i) Fib_i and k = Sum_{j >= 2} eps(j) Fib_j be the Zeckendorf expansions of n and k, respectively (cf. A035517, A014417). (The eps(i) are 0 or 1 and no two consecutive eps(i) are both 1.) Then the Fibonacci (or circle) product of n and k is n o k = Sum_{i,j} eps(i)*eps(j) Fib_{i+j} (= T(n,k)).

The Zeckendorf expansion can be written n = Sum_{i=1..k} F(a_i), where a_{i+1} >= a_i + 2. In this formulation, the product becomes: if n = Sum_{i=1..k} F(a_i) and m = Sum_{j=1..l} F(b_j) then n o m = Sum_{i=1..k} Sum_{j=1..l} F(a_i + b_j).

Knuth shows that this multiplication is associative. This is not true if we change the product to n X k = Sum_{i,j} eps(i)*eps(j) Fib_{i+j-2}, see A101646. Of course 1 is not a multiplicative identity here, whereas it is in A101646.

The papers by Arnoux, Grabner et al. and Messaoudi discuss this sequence and generalizations.

Numbers whose Zeckendorf representation ends with 000. - Benoit Cloitre, Jan 11 2014

The asymptotic density of this sequence is sqrt(5)-2. - Amiram Eldar, Mar 21 2022

This is the lower s-Wythoff sequence, where s(n)=n+1.

See A184117 for the definition of lower and upper s-Wythoff sequences. The first few terms of a and its complement, b=A026274, are obtained generated as follows:

s=(2,3,4,5,6,...);

a=(1,2,4,6,7,...)=A026273;

b=(3,5,8,11,13,...)=A026274.

Briefly: b=s+a, and a=mex="least missing".

From Michel Dekking, Mar 12 2018: (Start)

One has r*(n-2*r+3) = n*r-2r^2+3*r = (n+1)*r-2.

So a(n) = (n+1)*r-2, and we see that this sequence is simply the Beatty sequence of the golden ratio, shifted spatially and temporally. In other words: if w = A000201 = 1,3,4,6,8,9,11,12,14,... is the lower Wythoff sequence, then a(n) = w(n+2) - 2.

(N.B. As so often, there is the 'offset 0 vs 1 argument', w = A000201 has offset 1; it would have been better to give (a(n)) offset 1, too).

This observation also gives an answer to Lenormand's question, and a simple proof of Mathar's conjecture in A059426.

(End)

Let n = Sum_{i >= 2} eps(i) Fib_i and k = Sum_{j >= 2} eps(j) Fib_j be the Zeckendorf expansions of n and k, respectively (cf. A035517, A014417). The product of n and k is defined here to be n x k = Sum_{i,j} eps(i)*eps(j) Fib_{i+j-2} (= T(n,k)). [Comment corrected by R. J. Mathar, Aug 07 2007]

Although now 1 is the multiplicative identity, in contrast to A101330, this multiplication is not associative. For example, as pointed out by Grabner et al., we have (4 x 7 ) x 9 = 25 x 9 = 198 but 4 x (7 x 9 ) = 4 x 54 = 195.

From Jianing Song, Aug 21 2022: (Start)

Numbers k > 0 such that floor((k^2+1)*phi) - 1 is a square, phi = A001622.

Suppose that k is a term and that floor((k^2+1)*phi) = m^2+1, then (m^2+1)/(k^2+1) < phi < (m^2+2)/(k^2+1), so |sqrt(phi) - m/k| < max{m/k - sqrt((m^2+1)/(k^2+1)), sqrt((m^2+2)/(k^2+1)) - m/k} = m/k - sqrt((m^2+1)/(k^2+1)) <= sqrt((k^2+1)*phi-1)/k - sqrt(phi) < 1/(2*sqrt(phi)*k^2). According to the Mathematics Stack Exchange link, m/k is a convergent to sqrt(phi), so this is a subsequence of A225205. The terms are b(3), b(5), b(11), b(15), b(19), b(20), ... for b = A225205.

For k = A225205(r), m = A225204(r), we have |sqrt(phi) - m/k| < 1/(k*A225205(r+1)) (by Theorem 5 of the Wikipedia link), so k = A225205(r) is a term if 1/(k*A225205(r+1)) < min{m/k - sqrt((m^2+1)/(k^2+1)), sqrt((m^2+2)/(k^2+1)) - m/k} = sqrt((m^2+2)/(k^2+1)) - m/k, or A225205(r+1) > (k*sqrt((m^2+2)/(k^2+1)) - m)^(-1).

If k = A225205(r) is a term with even r, then k is also in A354549, since m^2 < k^2*phi < k^2*(m^2+2)/(k^2+1) < m^2+phi^(-2) for m = A225204(r), so floor(k^2*phi) = m^2. Furthermore we have {k^2*phi} < phi^(-2), where {} denotes the fractional part. Conversely, if k is in A354549 and {k^2*phi} < phi^(-2), then k is in this sequence since floor((k^2+1)*phi) = floor(k^2*phi)+1 in this case. (End)