A297770 Number of distinct runs in base-2 digits of n.

1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 3, 2, 3, 2, 1, 2, 2, 3, 3, 3, 2, 3, 3, 2, 3, 3, 2, 2, 3, 2, 1, 2, 2, 3, 3, 2, 3, 4, 3, 3, 3, 2, 3, 4, 3, 3, 3, 2, 3, 4, 2, 4, 3, 2, 3, 2, 3, 3, 3, 2, 3, 2, 1, 2, 2, 3, 3, 3, 3, 4, 3, 3, 2, 3, 4, 3, 4, 4, 3, 3, 3, 3, 4, 3, 2, 3

Offset

1,2

Comments

Every positive integers occurs infinitely many times.

***

Guide to related sequences:

Base b # runs # distinct runs

2 A005811 A297770

3 A043555 A297771

4 A043556 A297772

5 A043557 A297773

6 A043558 A297774

7 A043559 A297775

8 A043560 A297776

9 A043561 A297777

10 A043562 A297778

11 A043563 A297779

12 A043564 A297780

13 A043565 A297781

14 A043566 A297782

15 A043567 A297783

16 A043568 A297784

Links

Clark Kimberling, Table of n, a(n) for n = 1..10000

Example

27 in base-2: 1,1,0,1,1; three runs, of which 2 are distinct:  0 and 11, so that a(27) = 2.

Mathematica

b = 2; s[n_] := Length[Union[Split[IntegerDigits[n, b]]]]
Table[s[n], {n, 1, 200}]

Prog

(Python)
from itertools import groupby
def A297770(n): return len(set(map(lambda x:tuple(x[1]), groupby(bin(n)[2:])))) # Chai Wah Wu, Jul 13 2024
(PARI) apply( {A297770(n)=my(r=[0, 0], c); while(n, my(d=bitand(n, 1), L=valuation(n+d, 2)); !bittest(r[1+d], L) && c++ && r[1+d] += 1<<L; n>>=L); c}, [0..99]) \\ M. F. Hasler, Jul 13 2024

Crossrefs

Cf. A005811 (number of runs, not necessarily distinct).

Sequence in context: A160520 A235708 A353929 * A330617 A343240 A145866

Adjacent sequences: A297767 A297768 A297769 * A297771 A297772 A297773

Keyword

nonn,base,easy,changed

Author

Clark Kimberling, Jan 26 2018

Status

approved