cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A276886 Sums-complement of the Beatty sequence for 2 + phi.

Original entry on oeis.org

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
Offset: 1

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Author

Clark Kimberling, Oct 01 2016

Keywords

Comments

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)

Examples

			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,...).
		

Crossrefs

Programs

  • 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 *)

Formula

a(n) = 2*floor((n-1)*phi) - n + 2, where phi is the golden mean.