A380188 a(n) is the maximum number of coincidences of the first n terms of this sequence and a cyclic shift of the first n terms of A380189, i.e., the number of equalities a(k) = A380189((s+k) mod n) for 0 <= k < n, maximized over s.

0, 1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 6, 6, 6, 7, 7, 7, 7, 8, 8, 9, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12

Offset: 0

Author

Pontus von Brömssen, Jan 15 2025

Keywords

• nonn

Comments

Is a(n+1)-a(n) always 0 or 1?

Consider a pair of sequences b and c defined in the following manner:

- b(n) is the number of coincidences of the first n terms of sequences b and c,

- c(n) is the number of coincidences of the first n terms of sequence b and the first n terms of sequence c in reverse order.

(The sequences are "self-starting" with no initial values required, because for n = 0 there are obviously no coincidences, so b(0) = c(0) = 0.) For each of b and c, we may or may not allow circular shifts and maximize the number of coincidences over all such shifts, so there are four versions:

- No shifts: (b,c) = (A379265,A379266).

- Shifts in b but not in c: (b,c) = (A380188,A380189).

- Shifts in c and either shifts or no shifts in b: In both these cases, b and c are the following sequences, which are constant from n = 5 and n = 7, respectively:

b: 0, 1, 2, 3, 3, 4, 4, 4, 4, 4, ...

c: 0, 1, 2, 1, 3, 2, 2, 3, 3, 3, ...

Examples

			The first time the shift comes into play is for n = 21. The first 21 terms of this sequence and of A380189 are:
  0, 1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4
  0, 1, 0, 2, 0, 1, 1, 2, 1, 0, 3, 1, 0, 2, 0, 2, 2, 2, 2, 3, 3
  ^  ^     ^                    ^
with only 4 coincidences. But if the second row is shifted 7 steps to the right, we get:
  0, 1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4
  0, 2, 2, 2, 2, 3, 3, 0, 1, 0, 2, 0, 1, 1, 2, 1, 0, 3, 1, 0, 2
  ^     ^  ^     ^  ^
with 5 coincidences. This is the best possible, so a(21) = 5.
		

Links

• Pontus von Brömssen, Table of n, a(n) for n = 0..20000

Crossrefs

Cf. A272727, A276638, A379265, A379266, A380189, A380190.