A368842 a(n) gives the number of triples of equally spaced equal digits in the binary expansion of n (without leading zeros).

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

Offset: 0

Author

Rémy Sigrist, Jan 07 2024

Keywords

• nonn
• base
• easy

Comments

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

A000225 \ {1, 3} corresponds to indices of records.

Examples

			For n = 277:
- the binary expansion of 277 is "100010101",
- we have the following triples:  1   1   1
                                   000
                                   0 0 0
                                     0 0 0
                                      1 1 1
- so a(277) = 5.
		

Links

• Index entries for sequences related to binary expansion of n

Crossrefs

Cf. A000225, A002620, A368841, A368843, A368844.

Programs

PARI

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

Python

def A368842(n):
    l = len(s:=bin(n)[2:])
    return sum(1 for i in range(l-2) for j in range(1,l-i+1>>1) if s[i:i+(j<<1)+1:j] in {'000','111'}) # Chai Wah Wu, Jan 10 2024

Formula

a(2^k) = A002620(k - 1) for any k > 0.

a(2^k - 1) = A002620(k - 1) for any k > 0.

a(n) = A368843(n) + A368844(n).

a(floor(n/2)) <= a(n).