A379266 a(n) is the number of coincidences of the first n terms of this sequence and the first n terms of A379265 in reverse order, i.e., the number of equalities a(k) = A379265(n-1-k) for 0 <= k < n.

0, 1, 0, 2, 0, 1, 1, 2, 1, 0, 3, 1, 0, 2, 0, 2, 2, 2, 2, 3, 3, 3, 1, 0, 3, 2, 3, 3, 4, 5, 4, 4, 4, 3, 3, 3, 2, 2, 2, 3, 6, 6, 6, 6, 8, 7, 6, 5, 5, 5, 5, 4, 4, 4, 4, 4, 4, 4, 3, 4, 2, 1, 0, 5, 4, 4, 5, 6, 7, 8, 7, 9, 10, 11, 12, 12, 13, 16, 16, 16, 16, 14, 12

Offset: 0

Author

Pontus von Brömssen, Dec 19 2024

Keywords

• nonn

Comments

a(n) appears to grow roughly like sqrt(n).

Links

• Pontus von Brömssen, Table of n, a(n) for n = 0..9999
• Pontus von Brömssen, Plot of A379265(n), A379266(n), and sqrt(n) for n=0..100000.

Crossrefs

Cf. A272727, A379265, A379297.

Programs

Python

def A379266_list(nterms):
    A = []
    A379265 = []
    for n in range(nterms):
        a = sum(1 for x, y in zip(A, reversed(A379265)) if x==y)
        if n != 0:
            b += (b==A[-1])
        else:
            b = 0
        A.append(a)
        A379265.append(b)
    return A