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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)

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)