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.

Showing 1-3 of 3 results.

A189377 a(n) = n + floor(ns/r) + floor(nt/r) with r=2, s=(-1+sqrt(5))/2, t=(1+sqrt(5))/2.

Original entry on oeis.org

1, 3, 5, 8, 10, 11, 14, 16, 18, 21, 22, 24, 27, 29, 31, 32, 35, 37, 39, 42, 43, 45, 48, 50, 52, 55, 56, 58, 60, 63, 65, 66, 69, 71, 73, 76, 77, 79, 82, 84, 86, 87, 90, 92, 94, 97, 99, 100, 103, 105, 107, 110, 111, 113, 115, 118, 120, 121, 124, 126, 128, 131, 132, 134, 137, 139, 141, 144, 145, 147, 149, 152, 154, 155, 158, 160, 162, 165, 166, 168, 171, 173, 175, 176
Offset: 1

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Author

Clark Kimberling, Apr 20 2011

Keywords

Comments

This is one of three sequences that partition the positive integers. In general, suppose that r, s, t are positive real numbers for which the sets {i/r: i>=1}, {j/s: j>=1}, {k/t: k>=1} are pairwise disjoint. Let a(n) be the rank of n/r when all the numbers in the three sets are jointly ranked. Define b(n) and c(n) as the ranks of n/s and n/t. It is easy to prove that
a(n) = n+[ns/r]+[nt/r],
b(n) = n+[nr/s]+[nt/s],
c(n) = n+[nr/t]+[ns/t], where []=floor.
With r=2, s=(-1+sqrt(5))/2, t=(1+sqrt(5))/2 gives a=A189377, b=A189378, c=A189379.
(Conjecture) These are the numbers n such that (n+1)-sections of the Fibonacci word contain "000" (the commoner bit) but not "111" (the rarer bit). - Don Reble, Apr 07 2021
Conjecture proved April 8 2021, using the Walnut theorem prover. - Jeffrey Shallit, Apr 09 2021

Links

Crossrefs

Cf. A189378, A189379.

Programs

  • Mathematica
    r=2; s=(-1+5^(1/2))/2; t=(1+5^(1/2))/2;
a[n_] := n + Floor[n*s/r] + Floor[n*t/r];
b[n_] := n + Floor[n*r/s] + Floor[n*t/s];
c[n_] := n + Floor[n*r/t] + Floor[n*s/t]
Table[a[n], {n, 1, 120}]  (*A189377*)
Table[b[n], {n, 1, 120}]  (*A189378*)
Table[c[n], {n, 1, 120}]  (*A189379*)

A189378 a(n) = n + [nr/s] + [nt/s]; r=2, s=(-1+sqrt(5))/2, t=(1+sqrt(5))/2.

Original entry on oeis.org

6, 13, 19, 26, 34, 40, 47, 53, 61, 68, 74, 81, 89, 95, 102, 108, 116, 123, 129, 136, 142, 150, 157, 163, 170, 178, 184, 191, 197, 205, 212, 218, 225, 233, 239, 246, 252, 259, 267, 273, 280, 286, 294, 301, 307, 314, 322, 328, 335, 341, 349, 356, 362, 369, 375, 383, 390, 396, 403, 411, 417, 424, 430, 438, 445, 451, 458, 466, 472, 479, 485, 492, 500, 506, 513, 519, 527, 534, 540, 547, 555, 561
Offset: 1

Views

Author

Clark Kimberling, Apr 20 2011

Keywords

Comments

Theorem: These are the numbers k such that (k+1)-sections of the Fibonacci word contain neither "000" nor "111". Proved by J. Shallit and "Walnut", Apr 06 2021. - Don Reble, Apr 06 2021

Links

Crossrefs

Cf. A189377, A189379.
Cf. A001622, A094214.
See also A339950.

Programs

A339950 Numbers k such that all k-sections of the infinite Fibonacci word A014675 have just two different run-lengths.

Original entry on oeis.org

1, 7, 14, 20, 27, 35, 41, 48, 54, 62, 69, 75, 82, 90, 96, 103, 109, 117, 124, 130, 137, 143, 151, 158, 164, 171, 179, 185, 192, 198, 206, 213, 219, 226, 234, 240, 247, 253, 260, 268, 274, 281, 287, 295, 302, 308, 315, 323, 329, 336, 342, 350, 357, 363, 370, 376, 384, 391, 397, 404
Offset: 1

Views

Author

Clark Kimberling, Dec 26 2020

Keywords

Comments

Equivalent definition: these are the numbers n such that all n-sections of the infinite Fibonacci word A003849 have just two run-lengths.
The distinct terms of the difference sequence of the first 40 terms are 6, 7, and 8.
Conjecture: a(n) = A189378(n-1)+1 for n >= 2. - Don Reble, Apr 06 2021.
"All n-sections" means all subsequences S(k) = (A014675(n*i+k); i = 0, 1, 2, ...), for k = 0, ..., n-1. "Run-lengths" means the numbers of consecutive equal terms in the sequence: see examples. - M. F. Hasler, Apr 07 2021

Examples

			Let W = A014675, so that as a word, W = 21221212212212122121221221212212212122121221221...
The unique 1-section of W is W itself, which is a concatenation of runs 1, 2, and 22, so that a(1) = 2. The sequence A339949 shows that a(n) > 2 for n = 2,3,4,5,6. For n = 7, the n-section of W that starts with its first letter, 2, is 221221221221221221221221221221221221121..., in which the runs are 22, 1, 11, supporting the conjecture that a(2) = 7.
Some run-lengths may appear quite late. For example, when n = 68, the third run-length appears in the n-section S(k=0) only with the 2829th element, corresponding to the 192372-th element of the original sequence. - _M. F. Hasler_, Apr 07 2021

Links

Crossrefs

Cf. A001622, A003849, A014675, A339949.
See also A189377, A189378, A189379.

Programs

  • Mathematica
    r = (1 + Sqrt[5])/2; z = 80000;
f[n_] := Floor[(n + 1) r] - Floor[n r];  (* A014675 *)
t = Table[Max[Map[Length,
     Union[Split[Table [f[n d], {n, 0, Floor[z/d]}]]]]], {d, 1,
    400}, {n, 1, d}];
u = Map[Max, t]
Flatten[Position[u, 2]]  (* A339950 *)

Extensions

More terms from Don Reble, Apr 13 2021
Showing 1-3 of 3 results.

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