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a(n) = least k such that the first n-block in A339825 occurs in A339824 beginning at the k-th term.
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%I #7 Jun 17 2021 04:38:26

%S 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,

%T 28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,28,45,45,45,45,45,

%U 45,45,45,45,45,45,45,45,45,45,45,45,45,45,45,45,45,45

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

%C 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.

%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 A339825, block #1 of A339824 is 0, which first occurs in A339825 beginning at the 2nd term, so a(1) = 2;

%e block #4 of A339824 is 0100, which first occurs in A339825 beginning at the 7th term, so a(4) = 7.

%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, A339827.

%K nonn

%O 1,1

%A _Clark Kimberling_, Dec 19 2020