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.

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A256654 Least Fibonacci number not less than n.

Original entry on oeis.org

1, 2, 3, 5, 5, 8, 8, 8, 13, 13, 13, 13, 13, 21, 21, 21, 21, 21, 21, 21, 21, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 34, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 55, 89, 89, 89, 89, 89, 89, 89, 89, 89, 89, 89, 89
Offset: 1

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Author

Clark Kimberling, Apr 08 2015

Keywords

Comments

This sequence plays a role in the definition of minimal alternating Fibonacci representations, introduced at A256655.
Essentially the same as A246104. - R. J. Mathar, Apr 14 2015

Crossrefs

Programs

  • Mathematica
    h[0] = {1}; h[n_] := Join[h[n - 1], Table[Fibonacci[n + 2], {k, 1, Fibonacci[n]}]]; h[10]

Formula

Concatenate these numbers: F(2), F(3), F(4), then F(3) F(5)'s, then F(4) F(6)'s, then F(5) F(7)'s, ... F(n+2) F(n)'s, ..., where F = A000045, the Fibonacci numbers.
Sum_{n>=1} 1/a(n)^2 = 1 + Sum_{n>=1} F(n)/F(n+2)^2 = 1.5651369873... . - Amiram Eldar, Aug 16 2022

A246105 Least m > 0 for which (s(m),...,s(n+m-1)) = (s(n),...,s(0)), the reverse of the first n+1 terms of the infinite Fibonacci word A003849.

Original entry on oeis.org

2, 1, 3, 2, 1, 5, 4, 3, 2, 1, 8, 7, 6, 5, 4, 3, 2, 1, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 34, 33, 32, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13
Offset: 0

Views

Author

Clark Kimberling, Aug 14 2014

Keywords

Examples

			reverse(s(0),...,s(6)) = reverse(0,1,0,0,1,0,1) = (1,0,1,0,0,1,0)
first repeats in A003849 at (s(4),...,s(10)), so that a(6) = 4.
		

Crossrefs

Programs

  • Mathematica
    s = Flatten[Nest[{#, #[[1]]} &, {0, 1}, 12]]; b[m_, n_] := b[m, n] = Take[s, {m, n}]; t = -1 + Flatten[Table[Select[Range[2, 1600], b[#, n + # - 1] == Reverse[b[1, n]] &, 1], {n, 1, 120}]]

Formula

Concatenation of blocks (F(k), F(k - 1), ..., F(3), F(2)) beginning with k = 3, where F = A000045 (Fibonacci numbers).
Showing 1-2 of 2 results.