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

0, 1, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 10, 11, 12, 13, 13, 14, 14, 14, 14, 14, 15

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, A379266, A379297.

Programs

Python

def A379265_list(nterms):
    A = []
    A379266 = []
    for n in range(nterms):
        if n != 0:
            a += (a==A379266[-1])
        else:
            a = 0
        b = sum(1 for x,y in zip(A,reversed(A379266)) if x==y)
        A.append(a)
        A379266.append(b)
    return A

Formula

For n >= 1, a(n) = a(n-1)+1 if a(n-1) = A379266(n-1), otherwise a(n) = a(n-1).