login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A276886
Sums-complement 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
OFFSET
1,2
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)
LINKS
FORMULA
a(n) = 2*floor((n-1)*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
Sequence in context: A236072 A055938 A190764 * A047323 A343076 A033292
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Oct 01 2016
STATUS
approved