A297771 -id:A297771 - cp's OEIS frontend

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

Clark Kimberling, Jan 26 2018

Every positive integers occurs infinitely many times.
***
Guide to related sequences:
Base b      # runs    # distinct runs
      A005811       A297770
      A043555       A297771
      A043556       A297772
      A043557       A297773
      A043558       A297774
      A043559       A297775
      A043560       A297776
      A043561       A297777
      A043562       A297778
      A043563       A297779
      A043564       A297780
      A043565       A297781
      A043566       A297782
      A043567       A297783
      A043568       A297784

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

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

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

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

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); c}, [0..99]) \\ M. F. Hasler, Jul 13 2024

a(n) = my(s=strjoin(binary(n)), v=vecsort(concat(strsplit(s, "1"), strsplit(s, "0")), , 8)); #v-(v[1]==""); \\ Ruud H.G. van Tol, Aug 05 2024

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

A043555 Number of runs in base-3 representation of n.

Original entry on oeis.org

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

Offset: 0

Clark Kimberling

Every positive integer occurs infinitely many times.  See A297770 for a guide to related sequences.
Having a(0) = 1 (rather than a(0) = 0) is debatable, on the grounds that a(0) = 1 is determined by our culture, rather than the underlying mathematics. See my August 2020 comment in A145204. - Peter Munn, Jul 12 2024
From M. F. Hasler, Jul 13 2024: (Start)
The base-2 version has a(0) = 0, corresponding to the convention that 0 has zero digits, which is the more logical (but maybe less human) convention, such that, e.g., b^n is the least number with n+1 digits in base b (<=> b^n - 1 is the largest number with n digits), valid also for 0. Here and in A043556 (base 4) the convention is that 0 has one digit, '0'.
"Runs" means substrings of consecutive equal digits, here in the base-3 representation of the numbers. See Example for details. (End)

			From _M. F. Hasler_, Jul 13 2024: (Start)
Numbers n = 0, 1, 2, 3, 4, 5, ... are written '0', '1', '2', '10', '11', '12', ... in base 3. The first three have one single digit, so there is just 1 "run" (= subsequence of equal digits), whence a(0) = a(1) = a(2) = 1.
In '10' we have a "run" of '1's of length 1, followed by a run of '0's of length 1, so there are a(3) = 2 runs.
In '11' we have again one single run, here of 2 digits '1', whence a(4) = 1. (End)

Alois P. Heinz, Table of n, a(n) for n = 0..19682 (first 1001 terms from Zak Seidov)

Cf. A005811 (in base 2), A043556 (in base 4), A145204, A297770, A297771 (number of distinct runs).

Cf. A033113.

NRUNS := proc(L::list)
    local a,i;
    a := 1 ;
    for i from 2 to nops(L) do
        if op(i,L) <> op(i-1,L) then
            a := a+1 ;
        end if
    end do:
    a ;
end proc:
A043555 := proc(n)
    convert(n,base,3) ;
    NRUNS(%) ;
end proc:
seq(A043555(n),n=0..80) ; # R. J. Mathar, Jul 12 2024
# second Maple program:
a:= n-> `if`(n<3, 1, a(iquo(n, 3))+`if`(n mod 9 in {0, 4, 8}, 0, 1)):
seq(a(n), n=0..89);  # Alois P. Heinz, Jul 13 2024

b = 3; s[n_] := Length[Split[IntegerDigits[n, b]]];
Table[s[n], {n, 1, 200}]

a(n)=my(d=digits(n,3)); sum(i=2,#d,d[i]!=d[i-1])+1 \\ Charles R Greathouse IV, Jul 20 2014

from itertools import groupby
from sympy.ntheory import digits
def A043555(n): return len(list(groupby(digits(n,3)[1:]))) # Chai Wah Wu, Jul 13 2024

Updated by Clark Kimberling, Feb 03 2018

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

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A043555 Number of runs in base-3 representation of n.

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