cp's OEIS Frontend

This is the function named u' in [Carlitz]. - Eric M. Schmidt, Aug 14 2014

See A276871 for a definition of sums-complement and guide to related sequences.

From Michel Dekking, Apr 30 2019: (Start)

This sequence is a generalized Beatty sequence. According to Theorem 3.2 in the paper "The Frobenius problem for homomorphic embeddings of languages into the integers" this sequence (as a subset of the natural numbers) is the complement of the union of the two Beatty sequences

V := A003231 and W = V+1 (as subsets of the natural numbers) given by

V(n):= A(n)+2n = 3,7,10,14,..., W(n):=A(n)+2n+1 = 4,8,11,15,...

Here A = A000201, the lower Wythoff sequence.

Since the sequence Delta A = A014675 of first differences of A is the infinite Fibonacci word on the alphabet {2,1}, the sequence Delta V = (V(n+1)-V(n)) is the infinite Fibonacci word on the alphabet {4,3}. (Delta V equals A276867 shifted by 1.)

Now if for some k, Delta V(k) = 4, then a distance 3 plus a distance 1 are generated between three consecutive numbers in the complement, whereas if Delta V(k) = 3, then only a distance 3 is generated between two consecutive numbers in the complement.

This means that (skipping a(1)=1)

Delta a = (a(n+1)-a(n)) = gamma(Delta V),

where gamma is the morphism

gamma(4) = 31, gamma(3) = 3.

Since the Fibonacci word is a fixed point of the morphism 0->01, 1->0, this implies that Delta a, skipping a(1)=1, is the Fibonacci word on the alphabet {3,1}. It follows that

a(n+1) = 2*A(n) - n + 1.

(End)

The part to the right of the decimal point, reversed, is given by A341722, that is, it is the same as in the Bergman-canonical representation. I asked Jeffrey Shallit to confirm this, and he provided the following verification using the Walnut Theorem-Prover:

[Walnut]$ eval sloane "?msd_fib An,x1,x2,y1,y2 ($saka(n,x1,y1) & $dvl(n,x2,y2)) => $equal(y1,y2)":

(saka(n,x1,y1))&dvl(n,x2,y2))):54 states - 66ms

((saka(n,x1,y1))&dvl(n,x2,y2)))=>equal(y1,y2))):2 states - 25ms

(A n , x1 , x2 , y1 , y2 ((saka(n,x1,y1))&dvl(n,x2,y2)))=>equal(y1,y2)))):1 states - 81ms

Total computation time: 264ms.

TRUE

Numbers k such that A003234(k) equals the image of some x by A000201(A001950()) (see 1.20 p. 339 of Carlitz link). - Michel Marcus, Feb 02 2014

This is the function named t in [Carlitz]. - Eric M. Schmidt, Aug 14 2014

Numbers such that A003231(n) = A003234(n), see Table 1 p. 357 in Carlitz link. - Michel Marcus, Feb 02 2014

From Jeffrey Shallit, Jan 01 2024: (Start)

No integer appears three times or more in this sequence.

If an integer appears twice, it appears as a(n) and a(n-2) for some n.

a(n) = a(n-2) if and only if n belongs to A003231. (observation of Benoit Cloitre)

All these and more properties can be proved using the synchronized Fibonacci automaton for a(n), which has 102 states. (End)

The sequence (a(n+1)-1) = 1,3,7,8,10,... is the union of two generalized Beatty sequences, namely (floor(n*phi)+2*n) = A003231, and the sequence (4*floor(n*phi)+3*n+1), the latter with offset 0. For a proof see my paper "Points of increase...". - Michel Dekking, Apr 01 2020

It appears that the sequences p = A214971, q = A003231, r = A276886 give the positions of 0, 1, 2, respectively. Let t,u,v be the slopes of p, q, r, respectively. Then t = (5+sqrt(5))/2, u = (5+sqrt(5))/2, v = sqrt(5), and 1/t + 1/u + 1/v = 1. If 1 is removed from p (or from r), the resulting three sequences partition the set of positive integers.

From Michel Dekking, Apr 29 2019: (Start)

This sequence is the unique fixed point of the morphism

0->10, 1->2, 2->2210.

To prove this, let phi2 be the square of the Fibonacci morphism given by

phi2(0)=010, phi2(1)=01.

Then xF := A003849 = 0100101001... is the unique fixed point of phi2.

We introduce the morphism beta with fixed point xB := A188432 = 00100101... given by

beta(0) = 001, beta(1) = 01,

and also the morphism psi given by

psi(0) = 010, psi(1) = 10.

CLAIM: psi(xB) = xF.

This claim can be proved by showing with induction that for n>0

psi(beta^n(0)) = phi2^{n+1}(0),

psi(beta^n(01)) = phi2^{n+1}(10).

Why is this claim useful? Well, it implies directly that

(a(n)) = delta(xB),

where delta is the 'decoration' morphism given by

delta(0) = 2, delta(1) = 10.

Now double the 1's in xB: 1->11'. Then beta induces a 'substitution' S

0 -> 0011', 11' -> 011'.

Since 1 is always followed by 1', and 1' always preceded by 1, the action of S is equivalent to the action of the morphism sigma defined by

sigma(0) = 0011', sigma(1) = 0, sigma(1') = 11'.

The decoration morphism delta gives rise to a letter-to-letter map gamma given by

gamma(0) = 2, gamma(1) = 1, gamma(1') = 0.

Now the change of alphabet gamma gives the morphism we have been looking for, since delta(xB) = gamma(xS), where xS is the unique fixed point of sigma.

(End)

This sequence is the {0->2, 1->10}-transform of A188432. - Michel Dekking, Apr 29 2019

Column 0: Nonnegative integers.

Row 0: Upper Wythoff sequence, A001950, with 0 prepended.

Main Diagonal: A003231, with 0 prepended.

Diagonal (2,6,9,13,...) = A054770.

Diagonal (1,4,8,11,...) = A214971.

Diagonal (2,5,9,12,...) = A284624.

Also, the positive integers that do not appear in A118287.