A348556 Binary expansion contains 4 adjacent 1's.

15, 30, 31, 47, 60, 61, 62, 63, 79, 94, 95, 111, 120, 121, 122, 123, 124, 125, 126, 127, 143, 158, 159, 175, 188, 189, 190, 191, 207, 222, 223, 239, 240, 241, 242, 243, 244, 245, 246, 247, 248, 249, 250, 251, 252, 253, 254, 255, 271, 286, 287, 303, 316, 317, 318

Offset: 1

Author

Charles R Greathouse IV, Oct 22 2021

Keywords

• nonn
• base
• easy

Comments

For k > 0, each term m = 2^(k+3) - 1 is the end of a run of A083593(k-1) consecutive terms. For k = 4, from a(13) = 120 up to a(20) = 2^7-1 = 127, there are A083593(3) = 8 consecutive terms corresponding to 1111000, 1111001, 1111010, 1111011, 1111100, 1111101, 111110 and 1111111. - Bernard Schott, Feb 20 2022

Links

• Michel Marcus, Table of n, a(n) for n = 1..10000
• Index entries for 2-automatic sequences.

Crossrefs

Binary expansion contains k adjacent 1s: A000027 (1), A004780 (2), A004781 (3), this sequence (4).

Cf. A006520, A083593.

Subsequences: A110286, A195744.

Programs

Maple

q:= n-> verify([1$4], Bits[Split](n), 'sublist'):
select(q, [$0..400])[];  # Alois P. Heinz, Oct 22 2021

Mathematica

Select[Range[300], StringContainsQ[IntegerString[#, 2], "1111"] &] (* Amiram Eldar, Oct 22 2021 *)

PARI

is(n)=n=bitand(n,n<<2); !!bitand(n,n<<1);

Python

def ok(n): return "1111" in bin(n)
print([k for k in range(319) if ok(k)]) # Michael S. Branicky, Oct 22 2021

Formula

a(n) ~ n.

a(n+1) <= a(n) + 16.