%I #7 Jul 15 2021 21:27:00
%S 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,
%T 29,29,29,29,29,29,29,29,29,29,29,29,29,29,29,29,29,46,46,46,46,46,46,
%U 46,46,46,46,46,46,46,46,46,46,46,46,46,46,46,46,46,46
%N a(n) = least k such that the first nblock in A339824 occurs in A339825 beginning at the kth term.
%e 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.
%e Let W denote the infinite Fibonacci word A003849.
%e A339824 = even bisection of W: 001100110001000100011...
%e A339825 = odd bisection of W: 100010001100110011000...
%e 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;
%e block #4 of A339824 is 0011, which first occurs in A339824 beginning at the 8th term, so a(4) = 8.
%t r = (1 + Sqrt[5])/2; z = 3000;
%t f[n_] := 2  Floor[(n + 2) r] + Floor[(n + 1) r]; (*A003849*)
%t u = Table[f[2 n], {n, 0, Floor[z/2]}]; (* A339824 *)
%t v = Table[f[2 n + 1], {n, 0, Floor[z/2]}]; (* A339825 *)
%t a[n_] := Select[Range[z], Take[u, n] == Take[v, {#, # + n  1}] &, 1]
%t Flatten[Table[a[n], {n, 1, 300}]] (* A339826 *)
%Y Cf. A001622, A339051, A339052, A339824, A339825, A339826.
%K nonn
%O 1,1
%A _Clark Kimberling_, Dec 19 2020
