A339054 a(n) = least k such that the first n-block in A339052 occurs in A339051 beginning at the k-th term.

3, 3, 4, 8, 8, 8, 8, 8, 8, 8, 12, 12, 12, 12, 12, 12, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 29, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46, 46

Offset

1,1

Comments

The sequence begins with 2 3's, 1 4, 7 8's, 6 12's, 28 29's, 47 46's, 117 118's, 116 190's, 494 495's, 493 800's, followed by 2092. Conjecture: the sequence includes infinitely many distinct numbers; if so, then every finite block in A339052 occurs infinitely many times in A339051.

Links

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

Example

Let W denote the infinite Fibonacci word A096270.
A339051 = even bisection of W:  00111011100110011001110 . . .
A339052 = odd bisection of W:   11001100111011101110011 . . .
Using offset 1 for A339051, block #1 of A339052 is 1, which first occurs in A339051 beginning at the 3rd term, so a(1) = 3; block #4 of A339052 is 1100, which first occurs in A339051 beginning at the 8th term, so a(4) = 8.

Mathematica

r = (1 + Sqrt[5])/2; z = 3000;
f[n_] := Floor[(n + 1) r] - Floor[n r] - 1   (* A096270 *)
u = Table [f[2 n], {n, 0, Floor[z/2]}];      (* A339051 *)
v = Table [f[2 n + 1], {n, 0, Floor[z/2]}];  (* A339052 *)
a[n_] := Select[Range[z], Take[v, n] == Take[u, {#, # + n - 1}] &, 1]
Flatten[Table[a[n], {n, 1, 300}]]  (* A339053 *)

Crossrefs

Cf. A001622, A096270, A339051, A339052, A339053.

Sequence in context: A360846 A340429 A147679 * A137417 A137418 A357306

Adjacent sequences: A339051 A339052 A339053 * A339055 A339056 A339057

Keyword

nonn

Author

Clark Kimberling, Dec 08 2020

Status

approved