A368857 a(n) gives the maximum number of equally spaced equal digits in the binary expansion of n (without leading zeros).

0, 1, 1, 2, 2, 2, 2, 3, 3, 2, 2, 2, 2, 2, 3, 4, 4, 3, 2, 2, 2, 3, 2, 3, 3, 2, 2, 2, 3, 3, 4, 5, 5, 4, 3, 3, 2, 2, 2, 3, 3, 2, 3, 3, 2, 2, 3, 4, 4, 3, 2, 2, 2, 3, 2, 3, 3, 3, 3, 3, 4, 4, 5, 6, 6, 5, 4, 4, 3, 3, 3, 3, 3, 3, 3, 3, 2, 3, 3, 4, 4, 3, 2, 2, 3, 4, 3

Offset: 0

Author

Rémy Sigrist, Jan 08 2024

Keywords

• nonn
• base

Comments

This sequence diverges to infinity by Van der Waerden's theorem.

Links

• Index entries for sequences related to binary expansion of n

Crossrefs

Cf. A368841.

Programs

PARI

a(n, base = 2) = { my (b = digits(n, base), v = if (n, 1, 0)); for (i = 1, #b-1, for (j = i+1, #b, if (b[i]==b[j], my (d = j-i, k = j); while (k + d <= #b && b[k + d]==b[i], k += d;); v = max(v, 1 + (k-i) / d);););); return (v); }

Python

def A368857(n):
    if n == 0: return 0
    l = len(s:=bin(n)[2:])
    return 1+max((k-1-i)//j for i in range(l) for j in range(1,l-i+3>>1) for k in range(i+1,l+1,j) if len(set(s[i:k:j]))==1) # Chai Wah Wu, Jan 10 2024

Formula

a(2^k) = k for any k > 0.

a(2^k - 1) = k for any k >= 0.

a(2*n) >= a(n).