A339826 a(n) = least k such that the first n-block in A339825 occurs in A339824 beginning at the k-th term.

2, 2, 3, 7, 7, 7, 7, 7, 7, 7, 11, 11, 11, 11, 11, 11, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 28, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45

Offset

1,1

Comments

The sequence begins with 2 2's, 1 3, 7 7's, 6 11's, ... Conjecture: the sequence includes infinitely many distinct numbers, in which case, every finite block in A339825 occurs infinitely many times in A339824.

Links

Table of n, a(n) for n=1..67.

Example

Let W denote the infinite Fibonacci word A003849.
A339824 = even bisection of W:  001100110001000100011...
A339825 = odd bisection of W:   100010001100110011000...
Using offset 1 for A339825, block #1 of A339824 is 0, which first occurs in A339825 beginning at the 2nd term, so a(1) = 2;
block #4 of A339824 is 0100, which first occurs in A339825 beginning at the 7th term, so a(4) = 7.

Mathematica

r = (1 + Sqrt[5])/2; z = 3000;
f[n_] := 2 - Floor[(n + 2) r] + Floor[(n + 1) r];  (*A003849*)
u = Table[f[2 n], {n, 0, Floor[z/2]}];      (*A339824 *)
v = Table[f[2 n + 1], {n, 0, Floor[z/2]}];  (*A339825 *)
a[n_] := Select[Range[z], Take[u, n] == Take[v, {#, # + n - 1}] &, 1]
Flatten[Table[a[n], {n, 1, 300}]]       (*A339826 *)

Crossrefs

Cf. A001622, A339051, A339052, A339824, A339825, A339827.

Sequence in context: A002583 A068519 A342848 * A108041 A376348 A259254

Adjacent sequences: A339823 A339824 A339825 * A339827 A339828 A339829

Keyword

nonn

Author

Clark Kimberling, Dec 19 2020

Status

approved