cp's OEIS Frontend

Positions of 2's is given by the positive Fibonacci numbers: 1, 2, 3, 5, 8, 13, 21, ..., that is, A000045(n) from n >= 2 onward.

Positions of 3's is given by Lucas numbers larger than 3: 4, 7, 11, 18, ..., that is, A000032(n) from n >= 3 onward.

Sequence allots a distinct value for each distinct multiset formed from the lengths of 1-runs in the binary representation of A048679(n). Compare to the scatter plot of A286622.

Let u = A000201 = (lower Wythoff sequence) and v = A001950 = (upper Wythoff sequence). Conjecture: This sequence is the sequence p given by p(1) = 0 and p(u(k)) = p(k); p(v(k)) = 1-p(k). - Clark Kimberling, Apr 15 2011

[base 2] 0.111010010001100... = 0.9105334708635617... - Joerg Arndt, May 13 2011

From Michel Dekking, Nov 28 2019: (Start)

This sequence is a morphic sequence.

Let the morphism sigma be given by

sigma(1) = 12, sigma(2) = 4, sigma(3) = 1, sigma(4) = 43,

and let the letter-to-letter map lambda be given by

lambda(1) = 0, lambda(2) = 1, lambda(3) = 0, lambda(4) = 1.

Then a(n) = lambda(x(n)), where x(0)x(1)... = 1244343... is the fixed point of sigma starting with 1.

See footnote number 4 in the paper by Emmanuel Ferrand. (End)

From Michel Dekking, Nov 29 2019: (Start)

Proof of Kimberling's conjecture, by virtue of the four symbol morphism sigma in the previous comment.

We first show that the fixed point x = 1244343143112... of sigma has the remarkable property that the letters 1 and 4 exclusively occur at positions u(k), k=1,2,..., and the letters 2 and 3 exclusively occur at positions v(k) k=1,2,....

To see this, let alpha be the Fibonacci morphism on the alphabet {a,b}:

alpha(a) = ab, alpha(b) = a.

It is well known that the lower Wythoff sequence u gives the positions of a in the fixed point abaababaab... of alpha, and that the upper Wythoff sequence v gives the positions of b in this infinite Fibonacci word.

Let pi be the projection from {1,2,3,4} to {a,b} given by

pi(1) = pi(4) = a, pi(2) = pi(3) = b.

One easily checks that (composition of morphisms)

pi sigma = alpha pi.

This implies the remarkable property stated above.

What remains to be proved is that lambda(x(u(k)) = a(k) and lambda(x(v(k)) = a(k), where lambda is the letter-to-letter map in the previous comment. We tackle this problem by an extensive analysis of the return words of the letter 1 in the sequence x= 1244343143112...

These are the words occurring in x which start with 1 and have no other occurrences of 1 in them.

One easily finds that they are given by

A:=1, B:=12, C:=124, D:=143, E:=1243, F:=12443, G:=1244343.

These words partition x. Moreover, sigma induces a morphism rho on the alphabet {A,B,C,D,E,F,G} given by

rho(A) = B, rho(B) =C, rho(C) = F, rho(D) = EA,

rho(E) = FA, rho(F) = GA, rho(G) = GDA.

CLAIM 1: Let mu be the letter-to-letter map given by

mu(A)=mu(B) = 1, mu(C)=mu(D)=mu(E) = 2, mu(F) = 3, mu(G) = 4,

then mu(R) = (s(n+1)-s(n)), where R = GDAEABFAB... is the unique fixed point of rho and s = 1,5,7,8,10,... is the sequence of positions of 1 in (x(u(k)).

Proof of CLAIM 1: The 1's in x occur exclusively at positions u(k), so if we want the differences s(n+1)-s(n), we can strip the 2's and 3's from the return words of 1, and record how long it takes to the next 1. This is the length of the stripped words A~=1, B~=1, C~=14, ... which is given by mu.

CLAIM 2: Let delta be the 'decoration' morphism given by

delta(A) = 1, delta(B) = 2, delta(C) = 3, delta(D) = 21,

delta(E) = 31, delta(F) = 41, delta(G) = 421,

then delta(R) = (t(n+1-t(n)), where rho(R) = R, and t is the sequence of positions of 1 or 3 in the sequence x = 1244343143112....

Proof of CLAIM 2: We have to record the differences in the occurrences of 1 and 3 in the return words of 1. These are given by delta. For example: F = 12443, where 1 and 3 occur at position 1 and 5; and the next 1 will occur at the beginning of any of the seven words A,...,G.

If we combine CLAIM 1 and CLAIM 2 with lambda(1) = lambda(3) = 0, we obtain the first half of Kimberling's conjecture, simply because delta = mu rho, and rho(R) = R.

The second half of the conjecture is obtained in a similar way. (End)

From Michel Dekking, Mar 10 2020: (Start)

The fact that this sequence is a morphic sequence can be easily deduced from the morphism tau given in A007895:

tau((j,0)) = (j,0) (j+1,1),

tau((j,1)) = (j,0).

The sum of digits of a number n in Zeckendorf expansion is given by projecting the n-th letter in the fixed point of tau starting with (0,0) on its first coordinate: (j,i)->j for i=0,1.

If we consider all the j's modulo 2, we obtain a morphism sigma' on 4 letters:

sigma'((0,0)) = (0,0) (1,1),

sigma'((0,1)) = (0,0),

sigma'((1,0)) = (1,0)(0,1),

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

The change of alphabet (0,0)->1, (0,1)->3, (1,0)->4, (1,1)->2, turns sigma' into sigma given in my Nov 28, 2019 comment. The projection map turns into the map 1->0, 2->1, 3->0, 4->1, as in my Nov 28 2019 comment.

(End)

Another proof that this sequence is a morphic sequence has been given in the monograph by Allouche and Shallit. They demonstrate on page 239 that (a(n)) is a letter to letter projection of a fixed point of a morphism on an alphabet of six letters. However, it can be seen easily that two pairs of letters can be merged, yielding a morphism on an alphabet of four letters, which after a change of alphabet equals the morphism from my previous comments. - Michel Dekking, Jun 25 2020

Union of A003754 and A003714, complement of A107911;

a(A023548(n+2)) = A052940(n+1) for n>0;

a(A001924(n)) = A000225(n) = 2^n - 1;

a(A000126(n)) = A000079(n) = 2^n for n>0;

A107910(n) = a(n+1) - a(n).

Let F_{-n} be the negative Fibonacci numbers (as defined in the first comment in A039834): F_{-1}=1, F_{-2}=-1, F_{-3}=2, F_{-4}=-3, F_{-5}=5, ..., F_{-n}=(-1)^(n-1)F_n.

Every integer has a unique representation as n = Sum_{k=1..r} c_k F_{-k} for some r, where the c_k are 0 or 1 and no two adjacent c's are 1.

Then a(n) = c_r ... c_3 c_2 c_1.

The Run Length Transform of a sequence {S(n), n>=0} is defined to be the sequence {T(n), n>=0} given by T(n) = Product_i S(i), where i runs through the lengths of runs of 1's in the binary expansion of n. E.g., 19 is 10011 in binary, which has two runs of 1's, of lengths 1 and 2. So T(19) = S(1)*S(2). T(0)=1 (the empty product).

The Run Length Transform of a sequence {S(n), n>=0} is defined to be the sequence {T(n), n>=0} given by T(n) = Product_i S(i), where i runs through the lengths of runs of 1's in the binary expansion of n. E.g. 19 is 10011 in binary, which has two runs of 1's, of lengths 1 and 2. So T(19) = S(1)*S(2). T(0)=1 (the empty product).

a(n) can be also computed by replacing all consecutive runs of zeros in the binary expansion of n with * (multiplication sign), and then performing that multiplication, still in binary, after which the result is converted into decimal. See the example below.