A339827 a(n) = least k such that the first n-block in A339824 occurs in A339825 beginning at the k-th term.

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

Links

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

Example

The sequence begins with one 3, two 4's, six 8's, six 12's, ...  Conjecture: the sequence includes infinitely many distinct numbers, in which case, every finite block in A339824 occurs infinitely many times in A339825.
Let W denote the infinite Fibonacci word A003849.
A339824 = even bisection of W:  001100110001000100011...
A339825 =  odd bisection of W:  100010001100110011000...
Using offset 1 for A339824, block #1 of A339825 is 1, which first occurs in A339824 beginning at the 3rd term, so a(1) = 3;
block #4 of A339824 is 0011, which first occurs in A339824 beginning at the 8th term, so a(4) = 8.

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, A339826.

Sequence in context: A269714 A146944 A294113 * A240875 A127735 A330249

Adjacent sequences: A339824 A339825 A339826 * A339828 A339829 A339830

Keyword

nonn

Author

Clark Kimberling, Dec 19 2020

Status

approved