A260311 Difference sequence of A260317.

1, 1, 1, 1, 1, 2, 2, 1, 2, 1, 2, 3, 2, 3, 2, 3, 3, 2, 3, 3, 2, 3, 5, 3, 2, 3, 5, 3, 2, 3, 5, 3, 5, 3, 2, 3, 5, 3, 5, 3, 2, 3, 5, 3, 5, 5, 3, 5, 3, 2, 3, 5, 3, 5, 5, 3, 5, 3, 2, 3, 5, 3, 5, 5, 3, 5, 3, 5, 5, 3, 5, 3, 2, 3, 5, 3, 5, 5, 3, 5, 3, 5, 5, 3, 5, 3

Offset: 1

Author

Clark Kimberling, Jul 31 2015

Keywords

• nonn
• easy

Comments

Conjecture: a(n) is a Fibonacci number (A000045) for every n.

In fact, a(n) is in {1,2,3,5}; proved with the Walnut theorem-prover. - Jeffrey Shallit, Oct 12 2022

Comment from Jonathan F. P. Grube and Benjamin Mason, Oct 10 2024 [Corrected by Jonathan F. P. Grube, Nov 05 2024]: (Start)

With an offset of 41, the sequence is of the form 2[c_{1,1},...,c_{1,n_1}]2[c_{2,1},...,c_{2,n_2}]2[c_{3,1},...,c_{3,n_3}]2..., where [c_{i,1},...,c_{i,n_i}] is the word 35(355)^{c_{i,1}}35(355)^{c_{i,2}}35...35(355)^{c_{i,n_i}}353 over the alphabet {3,5} for some nonnegative integers c_{i,j}. Furthermore c_{i,j} is in {1,2}. Proved with Walnut, the theorem-prover.

Conjectures: n_{2k-1} = n_{2k} and c_{2k-1, j} = c_{2k, j} for all positive k and 0 0} is the Fibonacci sequence. (End)

Links

• Wieb Bosma, Rene Bruin, Robbert Fokkink, Jonathan Grube, Anniek Reuijl, and Thian Tromp, Using Walnut to solve problems from the OEIS, arXiv:2503.04122 [math.NT], 2025.
• The Walnut code at https://cs.uwaterloo.ca/~shallit/oeis-walnut.txt proves the conjecture. Walnut itself can be downloaded from https://cs.uwaterloo.ca/~shallit/walnut.html.

Crossrefs

Cf. A259556, A259600, A259601, A260317, A000045.

Programs

Mathematica

r = GoldenRatio; z = 1060;
u[n_] := u[n] = Floor[n*r]; v[n_] := v[n] = Floor[n*r^2];
s[m_, n_] := v[m] + v[n];
t = Table[s[m, n], {n, 2, z}, {m, 1, n - 1}]; (* A259601 *)
w = Flatten[Table[Count[Flatten[t], n], {n, 1, z}]];
p0 = Flatten[Position[w, 0]]  (* A260317 *)
d = Differences[p0] (* A260311 *)