A339053 a(n) = least k such that the first n-block in A339051 occurs in A339052 beginning at the k-th term.

3, 3, 3, 3, 7, 7, 7, 7, 7, 7, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 45, 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 4 3's, 6 7's, 62 45's, 116 117's, 1104 799's, followed by 2091. Conjecture: the sequence includes infinitely many distinct numbers; if so, then every finite block in A339051 occurs infinitely many times in A339052.

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 A339052, the first 1-block of A339051 is 0, which first occurs in A339052 beginning at the 3rd term, so a(1) = 3; the first 5-block of A339051 is 00111, which first occurs in A339052 beginning at the 7th term, so a(5) = 7.

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[u, n] == Take[v, {#, # + n - 1}] &, 1]
Flatten[Table[a[n], {n, 1, 300}]]  (* A339053 *)

Crossrefs

Cf. A001622, A096270, A339051, A339052, A339054.

Sequence in context: A381041 A079084 A092323 * A244584 A369233 A092531

Adjacent sequences: A339050 A339051 A339052 * A339054 A339055 A339056

Keyword

nonn

Author

Clark Kimberling, Dec 08 2020

Status

approved