

A276886


Sumscomplement of the Beatty sequence for 2 + phi.


4



1, 2, 5, 6, 9, 12, 13, 16, 17, 20, 23, 24, 27, 30, 31, 34, 35, 38, 41, 42, 45, 46, 49, 52, 53, 56, 59, 60, 63, 64, 67, 70, 71, 74, 77, 78, 81, 82, 85, 88, 89, 92, 93, 96, 99, 100, 103, 106, 107, 110, 111, 114, 117, 118, 121, 122, 125, 128, 129, 132, 135, 136
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OFFSET

1,2


COMMENTS

See A276871 for a definition of sumscomplement 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)


LINKS

Table of n, a(n) for n=1..62.
Index entries for sequences related to Beatty sequences
Michel Dekking, The Frobenius problem for homomorphic embeddings of languages into the integers, Theoretical Computer Science 732, 7 July 2018, 7379.
J.P. Allouche, F. M. Dekking, Generalized Beatty sequences and complementary triples, arXiv:1809.03424 [math.NT], 2018.


FORMULA

a(n) = 2*floor((n1)*phi)  n + 2, where phi is the golden mean.


EXAMPLE

The Beatty sequence for 2 + phi is 0 followed by A003231, which is (0,3,7,10,14,18,21,...), with difference sequence s = A276867 = (3,4,3,4,4,3,4,3,4,4,3,4,4,3,4,3,4,4,3,...). The sums s(j)+s(j+1)+...+s(k) include (3,4,7,8,10,12,14,15,...), with complement (1,2,5,6,9,12,13,16,...).


MATHEMATICA

z = 500; r = 2 + GoldenRatio; b = Table[Floor[k*r], {k, 0, z}]; (* A003231 *)
t = Differences[b]; (* A276867 *)
c[k_, n_] := Sum[t[[i]], {i, n, n + k  1}];
u[k_] := Union[Table[c[k, n], {n, 1, z  k + 1}]];
w = Flatten[Table[u[k], {k, 1, z}]]; Complement[Range[Max[w]], w]; (* A276886 *)


CROSSREFS

Cf. A003231, A276867, A276871.
Sequence in context: A236072 A055938 A190764 * A047323 A033292 A090500
Adjacent sequences: A276883 A276884 A276885 * A276887 A276888 A276889


KEYWORD

nonn,easy


AUTHOR

Clark Kimberling, Oct 01 2016


STATUS

approved



