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

0,4

Comments

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

Links

Table of n, a(n) for n=0..86.

Index entries for sequences related to binary expansion of n

Formula

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

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

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

Prog

(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

Crossrefs

Cf. A368841.

Sequence in context: A292137 A292138 A322665 * A273632 A347387 A196046

Adjacent sequences: A368854 A368855 A368856 * A368858 A368859 A368860

Keyword

nonn,base

Author

Rémy Sigrist, Jan 08 2024

Status

approved