A374176 a(n) is the maximum number of consecutive bit changes in the binary representation of n.

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

Offset

1,5

Links

Hugo Pfoertner, Table of n, a(n) for n = 1..8192

Formula

a(n) = A038374(A038554(n)), with A038374(0) = 0. - Michael S. Branicky, Jul 06 2024

Example

a(1117) = 3:
  1117_2 = [1 0 0 0 1 0 1 1 1 0 1]
             ^     ^ ^ ^     ^ ^
             1       3        2
            consecutive changes

Maple

b:= n-> `if`(n<2, [0$2], (f-> (t-> [t, max(t, f[2])])(
        `if`(n mod 4 in {0, 3}, 0, f[1]+1)))(b(iquo(n, 2)))):
a:= n-> b(n)[2]:
seq(a(n), n=1..105);  # Alois P. Heinz, Jul 07 2024

Prog

(PARI) a(n) = {my(b=digits(n, 2), d=#b, m=0, j=b[1], c=0); for(k=2, d, if(b[k]!=j, c++; m=max(m, c), c=0); j=b[k]); m}
(Python)
def a(n):
    b, c, m = bin(n)[2:], 0, 0
    for i in range(len(b)-1):
        if b[i] != b[i+1]: c += 1
        else: m = max(m, c); c = 0
    return max(m, c)
print([a(n) for n in range(1, 89)]) # Michael S. Branicky, Jul 06 2024
(Python) # using formula
from itertools import groupby
def a(n): return max((len(list(g)) for k, g in groupby(bin(n^(n>>1))[3:]) if k=="1"), default=0)
print([a(n) for n in range(1, 89)]) # Michael S. Branicky, Jul 06 2024

Crossrefs

Cf. A005811, A037834, A038554, A038374.

Cf. A000975 (index of first occurrence of n).

Sequence in context: A221857 A133607 A103631 * A263191 A192517 A309896

Adjacent sequences: A374173 A374174 A374175 * A374177 A374178 A374179

Keyword

nonn,base

Author

Hugo Pfoertner, Jul 05 2024

Status

approved