A245563 -id:A245563 - cp's OEIS frontend

A385892 In the sequence of run lengths of binary indices of each positive integer (A245563), remove all duplicate rows after the first and take the last term of each remaining row.

1, 2, 1, 3, 1, 2, 4, 1, 1, 2, 3, 5, 1, 1, 1, 2, 2, 3, 4, 6, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 4, 5, 7, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5, 6, 8, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 5, 5, 6, 7

Offset: 1

Gus Wiseman, Jul 18 2025

nonn

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

			The binary indices of 53 are {1,3,5,6}, with maximal runs ((1),(3),(5,6)), with lengths (1,1,2), which is the 16th row of A385817, so a(16) = 2.

In the following references, "before" is short for "before removing duplicate rows".
Positions of firsts appearances appear to be A000071.
Without the removals we have A090996.
For sum instead of last we have A200648, before A000120.
For length instead of last we have A200650+1, before A069010 = A037800+1.
Last term of row n of A385817 (ranks A385818, before A385889), first A083368.
A245563 gives run lengths of binary indices, see A245562, A246029, A328592.
A384877 gives anti-run lengths of binary indices, A385816.
Cf. A001629, A044813, A048793, A164707, A200649, A384878, A384890, A385886.

Last/@DeleteDuplicates[Table[Length/@Split[Join@@Position[Reverse[IntegerDigits[n,2]],1],#2==#1+1&],{n,100}]]

A385890 Positions of first appearances in A245563 = run lengths of binary indices.

Original entry on oeis.org

1, 2, 4, 6, 8, 12, 14, 16, 22, 24, 28, 30, 32, 44, 46, 48, 54, 56, 60, 62, 64, 86, 88, 92, 94, 96, 108, 110, 112, 118, 120, 124, 126, 128, 172, 174, 176, 182, 184, 188, 190, 192, 214, 216, 220, 222, 224, 236, 238, 240, 246, 248, 252, 254, 256, 342, 344, 348

Offset: 1

Gus Wiseman, Jul 18 2025

nonn

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

These are positions of firsts appearances in A245563, ranks A385889, reverse A245562.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
A384877 lists anti-run lengths of binary indices, ranks A385816.
Cf. A000120, A023758, A048793, A069010, A200649, A243815, A246029, A328592, A384890.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
q=Table[Length/@Split[bpe[n],#2==#1+1&],{n,0,1000}];
Select[Range[Length[q]-1],!MemberQ[Take[q,#-1],q[[#]]]&]

A328592 Numbers whose binary expansion has all different lengths of runs of 1's.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 7, 8, 11, 12, 13, 14, 15, 16, 19, 22, 23, 24, 25, 26, 28, 29, 30, 31, 32, 35, 38, 39, 44, 46, 47, 48, 49, 50, 52, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 67, 70, 71, 76, 78, 79, 88, 92, 94, 95, 96, 97, 98, 100, 103, 104, 110, 111, 112, 113, 114

Offset: 1

Gus Wiseman, Oct 20 2019

nonn

Also numbers whose binary indices have different lengths of runs of successive parts. A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

The complement is {5, 9, 10, 17, 18, 20, 21, 27, ...}.

			The sequence of terms together with their binary expansions and binary indices begins:
   0:     0 ~ {}
   1:     1 ~ {1}
   2:    10 ~ {2}
   3:    11 ~ {1,2}
   4:   100 ~ {3}
   6:   110 ~ {2,3}
   7:   111 ~ {1,2,3}
   8:  1000 ~ {4}
  11:  1011 ~ {1,2,4}
  12:  1100 ~ {3,4}
  13:  1101 ~ {1,3,4}
  14:  1110 ~ {2,3,4}
  15:  1111 ~ {1,2,3,4}
  16: 10000 ~ {5}
  19: 10011 ~ {1,2,5}
  22: 10110 ~ {2,3,5}
  23: 10111 ~ {1,2,3,5}
  24: 11000 ~ {4,5}
  25: 11001 ~ {1,4,5}
  26: 11010 ~ {2,4,5}

The version for prime indices is A130091.
The binary expansion of n has A069010(n) runs of 1's.
The lengths of runs of 1's in the binary expansion of n are row n of A245563.
Numbers whose binary expansion has equal lengths of runs of 1's are A164707.
Cf. A000120, A003714, A014081, A098859, A121016, A328593, A328595, A328596.

Select[Range[0,100],UnsameQ@@Length/@Split[Join@@Position[Reverse[IntegerDigits[#,2]],1],#2==#1+1&]&]

A164707 A positive integer n is included if all runs of 1's in binary n are of the same length.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 14, 15, 16, 17, 18, 20, 21, 24, 27, 28, 30, 31, 32, 33, 34, 36, 37, 40, 41, 42, 48, 51, 54, 56, 60, 62, 63, 64, 65, 66, 68, 69, 72, 73, 74, 80, 81, 82, 84, 85, 96, 99, 102, 108, 112, 119, 120, 124, 126, 127, 128, 129, 130, 132, 133, 136

Offset: 1

Leroy Quet, Aug 23 2009

Clarification: A binary number consists of "runs" completely of 1's alternating with runs completely of 0's. No two or more runs all of the same digit are adjacent.

This sequence contains in part positive integers that each contain one run of 1's. For those members of this sequence each with at least two runs of 1's, see A164709.

			From _Gus Wiseman_, Oct 31 2019: (Start)
The sequence of terms together with their binary expansions and binary indices begins:
   1:      1 ~ {1}
   2:     10 ~ {2}
   3:     11 ~ {1,2}
   4:    100 ~ {3}
   5:    101 ~ {1,3}
   6:    110 ~ {2,3}
   7:    111 ~ {1,2,3}
   8:   1000 ~ {4}
   9:   1001 ~ {1,4}
  10:   1010 ~ {2,4}
  12:   1100 ~ {3,4}
  14:   1110 ~ {2,3,4}
  15:   1111 ~ {1,2,3,4}
  16:  10000 ~ {5}
  17:  10001 ~ {1,5}
  18:  10010 ~ {2,5}
  20:  10100 ~ {3,5}
  21:  10101 ~ {1,3,5}
  24:  11000 ~ {4,5}
  27:  11011 ~ {1,2,4,5}
(End)

Ivan Neretin, Table of n, a(n) for n = 1..10000

Cf. A164708, A164709, A164710.
The version for prime indices is A072774.
The binary expansion of n has A069010(n) runs of 1's.
Numbers whose runs are all of different lengths are A328592.
Partitions with equal multiplicities are A047966.
Numbers whose binary expansion is aperiodic are A328594.
Numbers whose reversed binary expansion is a necklace are A328595.
Numbers whose reversed binary expansion is a Lyndon word are A328596.
Cf. A000120, A003714, A014081, A065609, A070939, A121016, A245563, A275692.

isA164707 := proc(n) local bdg,arl,lset ; bdg := convert(n,base,2) ; lset := {} ; arl := -1 ; for p from 1 to nops(bdg) do if op(p,bdg) = 1 then if p = 1 then arl := 1 ; else arl := arl+1 ; end if; else if arl > 0 then lset := lset union {arl} ; end if; arl := 0 ; end if; end do ; if arl > 0 then lset := lset union {arl} ; end if; return (nops(lset) <= 1 ); end proc: for n from 1 to 300 do if isA164707(n) then printf("%d,",n) ; end if; end do; # R. J. Mathar, Feb 27 2010

Select[Range@ 140, SameQ @@ Map[Length, Select[Split@ IntegerDigits[#, 2], First@ # == 1 &]] &] (* Michael De Vlieger, Aug 20 2017 *)

foreach(1..140){
    %runs=();
    $runs{$}++ foreach split /0+/, sprintf("%b",$);
    print "$_, " if 1==keys(%runs);
}
# Ivan Neretin, Nov 09 2015

Extended beyond 42 by R. J. Mathar, Feb 27 2010

A245562 Table read by rows: row n gives list of lengths of runs of 1's in binary expansion of n, starting with high-order bits.

Original entry on oeis.org

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

Offset: 0

N. J. A. Sloane, Aug 10 2014

A formula for A071053(n) depends on this table.

			Here are the run lengths for the numbers 0 through 21:
0, []
1, [1]
2, [1]
3, [2]
4, [1]
5, [1, 1]
6, [2]
7, [3]
8, [1]
9, [1, 1]
10, [1, 1]
11, [1, 2]
12, [2]
13, [2, 1]
14, [3]
15, [4]
16, [1]
17, [1, 1]
18, [1, 1]
19, [1, 2]
20, [1, 1]
21, [1, 1, 1]

Reinhard Zumkeller, Table of n, a(n) for n = 0..2770
N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168, 2015

Row sums = A000120 (the binary weight).

Cf. A245563, A071053.

for n from 0 to 128 do
lis:=[]; t1:=convert(n,base,2); L1:=nops(t1); out1:=1; c:=0;
for i from 1 to L1 do
   if out1 = 1 and t1[i] = 1 then out1:=0; c:=c+1;
   elif out1 = 0 and t1[i] = 1 then c:=c+1;
   elif out1 = 1 and t1[i] = 0 then c:=c;
   elif out1 = 0 and t1[i] = 0 then lis:=[c,op(lis)]; out1:=1; c:=0;
   fi;
   if i = L1 and c>0 then lis:=[c,op(lis)]; fi;
                   od:
lprint(n,lis);
od:

row(n)=my(v=List(),k); while(n, n>>=valuation(n,2); listput(v, k=valuation(n+1,2)); n>>=k); Vecrev(v) \\ Charles R Greathouse IV, Oct 21 2016

from re import split
A245562_list = [0]
for n in range(1,100):
    A245562_list.extend(len(d) for d in split('0+',bin(n)[2:]) if d != '')
# Chai Wah Wu, Sep 07 2014

A384877 Irregular triangle read by rows where row k lists the lengths of maximal anti-runs (increasing by more than 1) in the binary indices of n.

Original entry on oeis.org

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

Offset: 0

Gus Wiseman, Jun 17 2025

A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.

			The binary indices of 182 are {2,3,5,6,8}, with maximal anti-runs ((2),(3,5),(6,8)) so row 182 is (1,2,2).
Triangle begins:
   0: ()
   1: (1)
   2: (1)
   3: (1,1)
   4: (1)
   5: (2)
   6: (1,1)
   7: (1,1,1)
   8: (1)
   9: (2)
  10: (2)
  11: (1,2)
  12: (1,1)
  13: (2,1)
  14: (1,1,1)
  15: (1,1,1,1)

Row-sums are A000120.
Positions of rows of the form (1,1,...) are A023758.
Positions of first appearances of each distinct row appear to be A052499.
For runs instead of anti-runs we have A245563, reverse A245562.
Row-lengths are A384890.
A355394 counts partitions without a neighborless part, singleton case A355393.
A356606 counts strict partitions without a neighborless part, complement A356607.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
Cf. A044813, A048793, A069010, A164707, A243815, A246029, A328592, A384177, A384877, A384879, A384893.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
Table[Length/@Split[bpe[n],#2!=#1+1&],{n,0,100}]

A384890 Number of maximal anti-runs (increasing by more than 1) in the binary indices of n.

Original entry on oeis.org

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

Offset: 0

Gus Wiseman, Jun 17 2025

nonn

First differs from A272604 at a(51) = 3, A272604(51) = 2.
A binary index of n is any position of a 1 in its reversed binary expansion. The binary indices of n are row n of A048793.
Do all constant runs in this sequence have lengths 1, 2, or 3?

			The binary indices of 51 are {1,2,5,6}, with maximal anti-runs ((1),(2,5),(6)), so a(51) = 3.

For runs instead of anti-runs we have A069010 = run-lengths of A245563 (reverse A245562).
Row-lengths of A384877, firsts A384878.
For prime indices instead of binary indices we have A384906.
A000120 counts binary indices.
A356606 counts strict partitions without a neighborless part, complement A356607.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
Cf. A044813, A048793, A052499, A164707, A243815, A300820, A328592, A384177, A384879, A384893.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
Table[Length[Split[bpe[n],#2!=#1+1&]],{n,0,100}]

A039004 Numbers whose base-4 representation has the same number of 1's and 2's.

Original entry on oeis.org

0, 3, 6, 9, 12, 15, 18, 24, 27, 30, 33, 36, 39, 45, 48, 51, 54, 57, 60, 63, 66, 72, 75, 78, 90, 96, 99, 102, 105, 108, 111, 114, 120, 123, 126, 129, 132, 135, 141, 144, 147, 150, 153, 156, 159, 165, 177, 180, 183, 189, 192, 195, 198, 201, 204, 207, 210, 216, 219

Offset: 1

Olivier Gérard

Numbers such that sum (-1)^k*b(k) = 0 where b(k)=k-th binary digit of n (see A065359). - Benoit Cloitre, Nov 18 2003
Conjecture: a(C(2n,n)-1) = 4^n - 1. (A000984 is C(2n,n)). - Gerald McGarvey, Nov 18 2007
From Russell Jay Hendel, Jun 23 2015: (Start)
We prove the McGarvey conjecture (A) a(e(n,n)-1) = 4^n-1, with e(n,m) = A034870(n,m) = binomial(2n,m), the even rows of Pascal's triangle. By the comment from Hendel in A034870, we have the function s(n,k) = #{n-digit, base-4 numbers with n-k more 1-digits than 2-digits}. As shown in A034870, (B) #s(n,k)= e(n,k) with # indicating cardinality, that is, e(n,k) = binomial(2n,k) gives the number of n-digit, base-4 numbers with n-k more 1-digits than 2-digits.
We now show that (B) implies (A). By definition, s(n,n) contains the e(n,n) = binomial(2n,n) numbers with an equal number of 1-digits and 2-digits. The biggest n-digit, base-4 number is 333...3 (n copies of 3). Since 333...33 has zero 1-digits and zero 2-digits it follows that 333...333 is a member of s(n,n) and hence it is the biggest member of s(n,n). But 333...333 (n copies of 3) in base 4 has value 4^n-1. Since A039004 starts with index 0 (that is, 0 is the 0th member of A039004), it immediately follows that 4^n-1 is the (e(n,n)-1)st member of A039004, proving the McGarvey conjecture. (End)
Also numbers whose alternating sum of binary expansion is 0, i.e., positions of zeros in A345927. These are numbers whose binary expansion has the same number of 1's at even positions as at odd positions. - Gus Wiseman, Jul 28 2021

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A139370, A139371, A139372, A139373.
Cf. A139351, A139352, A139353, A139354, A139355.
A subset of A001969 (evil numbers).
A base-2 version is A031443 (digitally balanced numbers).
Positions of 0's in A065359 and A345927.
Positions of first appearances are A086893.
The version for standard compositions is A344619.
A000120 and A080791 count binary digits, with difference A145037.
A003714 lists numbers with no successive binary indices.
A011782 counts compositions.
A030190 gives the binary expansion of each nonnegative integer.
A070939 gives the length of an integer's binary expansion.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A101211 lists run-lengths in binary expansion:
- row-lengths: A069010
- reverse: A227736
- ones only: A245563
A138364 counts compositions with alternating sum 0:
- bisection: A001700/A088218
- complement: A058622
A328594 lists numbers whose binary expansion is aperiodic.
A345197 counts compositions by length and alternating sum.
Cf. A000302, A000346, A003242, A029837, A053738, A053754, A071321, A103919, A191232, A328596.

Fortran
```
c See link in A139351.
```

N:= 1000: # to get all terms up to N, which should be divisible by 4
B:= Array(0..N-1):
d:= ceil(log[4](N));
S:= Array(0..N-1,[seq(op([0,1,-1,0]),i=1..N/4)]):
for i from 1 to d do
  B:= B + S;
  S:= Array(0..N-1,i-> S[floor(i/4)]);
od:
select(t -> B[t]=0, [$0..N-1]); # Robert Israel, Jun 24 2015

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Select[Range[0,100],ats[IntegerDigits[#,2]]==0&] (* Gus Wiseman, Jul 28 2021 *)

for(n=0,219,if(sum(i=1,length(binary(n)),(-1)^i*component(binary(n),i))==0,print1(n,",")))

Conjecture: there is a constant c around 5 such that a(n) is asymptotic to c*n. - Benoit Cloitre, Nov 24 2002

That conjecture is false. The number of members of the sequence from 0 to 4^d-1 is binomial(2d,d) which by Stirling's formula is asymptotic to 4^d/sqrt(Pi*d). If Cloitre's conjecture were true we would have 4^d-1 asymptotic to c*4^d/sqrt(Pi*d), a contradiction. - Robert Israel, Jun 24 2015

A227736 Irregular table read by rows: the first entry of n-th row is length of run of rightmost identical bits (either 0 or 1, equal to n mod 2), followed by length of the next run of bits, etc., in the binary representation of n, when scanned from the least significant to the most significant end.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 2, 3, 3, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 1, 3, 4, 4, 1, 1, 3, 1, 1, 1, 2, 1, 2, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 3, 2, 1, 2, 2, 1, 1, 1, 2, 2, 1, 2, 2, 3, 1, 1, 3, 1, 4, 5, 5, 1, 1, 4, 1, 1, 1, 3, 1

Offset: 1

Antti Karttunen, Jul 25 2013

Row n has A005811(n) terms. In rows 2^(k-1)..2^k-1 we have all the compositions (ordered partitions) of k. Other orderings of compositions: A101211 (same with rows reversed), A066099, A108244 and A124734.

Each row n >= 1 contains the initial A005811(n) nonzero terms from the beginning of row n of A227186. A070939(n) gives the sum of terms on row n, while A167489(n) gives the product of its terms. A136480 gives the first column. A101211 lists the terms of each row in reverse order.

			Table begins as:
  Row  n in    Terms on
   n   binary  that row
   1      1    1;
   2     10    1,1;
   3     11    2;
   4    100    2,1;
   5    101    1,1,1;
   6    110    1,2;
   7    111    3;
   8   1000    3,1;
   9   1001    1,2,1;
  10   1010    1,1,1,1;
  11   1011    2,1,1;
  12   1100    2,2;
  13   1101    1,1,2;
  14   1110    1,3;
  15   1111    4;
  16  10000    4,1;
etc. with the terms of row n appearing in reverse order compared how the runs of the same length appear in the binary expansion of n (Cf. A101211).
From _Omar E. Pol_, Sep 08 2013: (Start)
Illustration of initial terms:
  ---------------------------------------
  k   m     Diagram        Composition
  ---------------------------------------
  .          _
  1   1     |_|_           1;
  2   1     |_| |          1, 1,
  2   2     |_ _|_         2;
  3   1     |_  | |        2, 1,
  3   2     |_|_| |        1, 1, 1,
  3   3     |_|   |        1, 2,
  3   4     |_ _ _|_       3;
  4   1     |_    | |      3, 1,
  4   2     |_|_  | |      1, 2, 1,
  4   3     |_| | | |      1, 1, 1, 1,
  4   4     |_ _|_| |      2, 1, 1,
  4   5     |_  |   |      2, 2,
  4   6     |_|_|   |      1, 1, 2,
  4   7     |_|     |      1, 3,
  4   8     |_ _ _ _|_     4;
  5   1     |_      | |    4, 1,
  5   2     |_|_    | |    1, 3, 1,
  5   3     |_| |   | |    1, 1, 2, 1,
  5   4     |_ _|_  | |    2, 2, 1,
  5   5     |_  | | | |    2, 1, 1, 1,
  5   6     |_|_| | | |    1, 1, 1, 1, 1,
  5   7     |_|   | | |    1, 2, 1, 1,
  5   8     |_ _ _|_| |    3, 1, 1,
  5   9     |_    |   |    3, 2,
  5  10     |_|_  |   |    1, 2, 2,
  5  11     |_| | |   |    1, 1, 1, 2,
  5  12     |_ _|_|   |    2, 1, 2,
  5  13     |_  |     |    2, 3,
  5  14     |_|_|     |    1, 1, 3,
  5  15     |_|       |    1, 4,
  5  16     |_ _ _ _ _|    5;
.
Also irregular triangle read by rows in which row k lists the compositions of k, k >= 1.
Triangle begins:
 [1];
 [1,1], [2];
 [2,1], [1,1,1], [1,2],[3];
 [3,1], [1,2,1], [1,1,1,1], [2,1,1], [2,2], [1,1,2], [1,3], [4];
 [4,1], [1,3,1], [1,1,2,1], [2,2,1], [2,1,1,1], [1,1,1,1,1], [1,2,1,1], [3,1,1], [3,2], [1,2,2], [1,1,1,2], [2,1,2], [2,3], [1,1,3], [1,4], [5];
Row k has length A001792(k-1).
Row sums give A001787(k), k >= 1.
(End)

Antti Karttunen, The rows 1..1023 of the table, flattened
Mikhail Kurkov, Comments on A227736.
Claude Lenormand, Deux transformations sur les mots, Preprint, 5 pages, Nov 17 2003. Apparently unpublished. This is a scanned copy of the version that the author sent to me in 2003. - _N. J. A. Sloane_, Sep 09 2018. See Procedure 1.

Cf. A227738 and also A227739 for similar table for unordered partitions.
Cf. A030308, A245562, A245563.
Cf. A101211 (rows in reversed order).

import Data.List (group)
a227736 n k = a227736_tabf !! (n-1) !! (k-1)
a227736_row n = a227736_tabf !! (n-1)
a227736_tabf = map (map length . group) $ tail a030308_tabf
-- Reinhard Zumkeller, Aug 11 2014

Array[Length /@ Reverse@ Split@ IntegerDigits[#, 2] &, 34] // Flatten (* Michael De Vlieger, Dec 11 2020 *)

apply( {A227736_row(n, r=[1], b=n%2)=while(n\=2, n%2==b && r[#r]++ || [b=1-b, r=concat(r,1)]); r}, [1..22]) \\ M. F. Hasler, Mar 11 2025

def A227736_row(n): return[len(list(g))for _,g in groupby(bin(n)[:1:-1])]
from itertools import groupby # M. F. Hasler, Mar 11 2025

(define (A227736 n) (A227186bi (A227737 n) (A227740 n))) ;; The Scheme-function for A227186bi has been given in A227186.

a(n) = A227186(A227737(n), A227740(n)).

a(n) = A101211(A227741(n)).

A319411 Triangle read by rows: T(n,k) = number of binary vectors of length n with runs-resistance k (1 <= k <= n).

Original entry on oeis.org

2, 2, 2, 2, 2, 4, 2, 4, 6, 4, 2, 2, 12, 12, 4, 2, 6, 30, 18, 8, 0, 2, 2, 44, 44, 32, 4, 0, 2, 6, 82, 76, 74, 16, 0, 0, 2, 4, 144, 138, 172, 52, 0, 0, 0, 2, 6, 258, 248, 350, 156, 4, 0, 0, 0, 2, 2, 426, 452, 734, 404, 28, 0, 0, 0, 0, 2, 10, 790, 752, 1500, 938, 104, 0, 0, 0, 0, 0

Offset: 1

N. J. A. Sloane, Sep 20 2018

"Runs-resistance" is defined in A318928.
Row sums are 2,4,8,16,... (the binary vectors may begin with 0 or 1).
This is similar to A329767 but without the k = 0 column and with a different row n = 1. - Gus Wiseman, Nov 25 2019

			Triangle begins:
2,
2, 2,
2, 2, 4,
2, 4, 6, 4,
2, 2, 12, 12, 4,
2, 6, 30, 18, 8, 0,
2, 2, 44, 44, 32, 4, 0,
2, 6, 82, 76, 74, 16, 0, 0,
2, 4, 144, 138, 172, 52, 0, 0, 0,
2, 6, 258, 248, 350, 156, 4, 0, 0, 0,
2, 2, 426, 452, 734, 404, 28, 0, 0, 0, 0,
2, 10, 790, 752, 1500, 938, 104, 0, 0, 0, 0, 0,
...
Lenormand gives the first 20 rows.
The calculation of row 4 is as follows.
We may assume the first bit is a 0, and then double the answers.
vector / runs / steps to reach a single number:
0000 / 4 / 1
0001 / 31 -> 11 -> 2 / 3
0010 / 211 -> 12 -> 11 -> 2 / 4
0011 / 22 -> 2 / 2
0100 / 112 -> 21 -> 11 -> 2 / 4
0101 / 1111 -> 4 / 2
0110 / 121 -> 111 -> 3 / 3
0111 / 13 -> 11 -> 2 / 3
and we get 1 (once), 2 (twice), 3 (three times) and 4 (twice).
So row 4 is: 2,4,6,4.

Hiroaki Yamanouchi, Table of n, a(n) for n = 1..1830
Claude Lenormand, Deux transformations sur les mots, Preprint, 5 pages, Nov 17 2003. Apparently unpublished. This is a scanned copy of the version that the author sent to me in 2003.

Row sums are A000079.
Column k = 2 is 2 * A032741 = A319410.
Column k = 3 is 2 * A329745 (because runs-resistance 2 for compositions corresponds to runs-resistance 3 for binary words).
The version for compositions is A329744.
The version for partitions is A329746.
The number of nonzero entries in row n > 0 is A319412(n).
The runs-resistance of the binary expansion of n is A318928.
Cf. A096365, A225485, A245563, A319413, A319414, A319415, A325280, A329747, A329750, A329767, A329870.

runsresist[q_]:=If[Length[q]==1,1,Length[NestWhileList[Length/@Split[#]&,q,Length[#]>1&]]-1];
Table[Length[Select[Tuples[{0,1},n],runsresist[#]==k&]],{n,10},{k,n}] (* Gus Wiseman, Nov 25 2019 *)

cp's OEIS Frontend

A385892 In the sequence of run lengths of binary indices of each positive integer (A245563), remove all duplicate rows after the first and take the last term of each remaining row.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A385890 Positions of first appearances in A245563 = run lengths of binary indices.

Views

Author

Keywords

Comments

Crossrefs

Programs

Mathematica

A328592 Numbers whose binary expansion has all different lengths of runs of 1's.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A164707 A positive integer n is included if all runs of 1's in binary n are of the same length.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

Mathematica

Perl

Extensions

A245562 Table read by rows: row n gives list of lengths of runs of 1's in binary expansion of n, starting with high-order bits.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Maple

PARI

Python

A384877 Irregular triangle read by rows where row k lists the lengths of maximal anti-runs (increasing by more than 1) in the binary indices of n.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A384890 Number of maximal anti-runs (increasing by more than 1) in the binary indices of n.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A039004 Numbers whose base-4 representation has the same number of 1's and 2's.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Fortran

Maple