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

1,1

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.

Formula

a(n) ~ n.

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

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 *)

Prog

(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

Crossrefs

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

Cf. A006520, A083593.

Subsequences: A110286, A195744.

Sequence in context: A297285 A115776 A095783 * A067400 A115801 A343343

Adjacent sequences: A348553 A348554 A348555 * A348557 A348558 A348559

Keyword

nonn,base,easy

Author

Charles R Greathouse IV, Oct 22 2021

Status

approved