cp's OEIS Frontend

The sums-complement of a sequence s(1), s(2), ... of positive integers is introduced here as the set of numbers c(1), c(2), ... such that no c(n) is a sum s(j)+s(j+1)+...+s(k) for any j and k satisfying 1 <= j <= k. If this set is not empty, the term "sums-complement" also applies to the (possibly finite) sequence of numbers c(n) arranged in increasing order. In particular, the difference sequence D(r) of a Beatty sequence B(r) of an irrational number r > 2 has an infinite sums-complement, abbreviated as SC(r) in the following table:

r B(r) D(r) SC(r)

----------------------------------------------------

sqrt(5) A022839 A081427 A276871

sqrt(6) A022840 A276856 A276872

sqrt(7) A022841 A276857 A276873

sqrt(8) A022842 A276858 A276874

e A022843 A276859 A276875

2*e A276853 A276860 A276876

Pi A022844 A063438 A276877

2*Pi A028130 A276861 A276878

1+sqrt(2) A003151 A276862 A276879

1+sqrt(3) A054088 A007538 A276880

1+sqrt(5) A276854 A276863 A276881

2+sqrt(2) A001952 A276864 A276882

2+sqrt(3) A003512 A276865 A276883

2+sqrt(5) A004976 A276866 A276884

1+tau A001950 2 + A003849 A276885

2+tau A003231 A276867 A276886

3+tau A276855 A276868 A276887

2+sqrt(1/2) A182769 A276869 A276888

sqrt(2)+sqrt(3) A110117 A276870 A276889

From Jeffrey Shallit, Aug 15 2023: (Start)

Simpler description: this sequence represents those positive integers that CANNOT be expressed as a difference of two elements of A022839.

There is a 20-state Fibonacci automaton for the terms of this sequence (see a276871.pdf). It takes as input the Zeckendorf representation of n and accepts iff n is a member of A276871. (End)

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