A052499 -id:A052499 - cp's OEIS frontend

A003754 Numbers with no adjacent 0's in binary expansion.

0, 1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 21, 22, 23, 26, 27, 29, 30, 31, 42, 43, 45, 46, 47, 53, 54, 55, 58, 59, 61, 62, 63, 85, 86, 87, 90, 91, 93, 94, 95, 106, 107, 109, 110, 111, 117, 118, 119, 122, 123, 125, 126, 127, 170, 171, 173, 174, 175, 181

Offset: 1

N. J. A. Sloane

Theorem (J.-P. Allouche, J. Shallit, G. Skordev): This sequence = A052499 - 1.
Ahnentafel numbers of ancestors contributing the X-chromosome to a female. A280873 gives the male inheritance. - Floris Strijbos, Jan 09 2017 [Equivalence with this sequence pointed out by John Blythe Dobson, May 09 2018]
The k-th composition in standard order (row k of A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions. This sequence lists all numbers k such that the k-th composition in standard order has no parts greater than two. See the corresponding example below. - Gus Wiseman, Apr 04 2020
The binary representation of a(n+1) has the same string of digits as the lazy Fibonacci (also known as dual Zeckendorf) representation of n that uses 0s and 1s. (The "+1" is essentially an adjustment for the offset of this sequence.) - Peter Munn, Sep 06 2022

			21 is in the sequence because 21 = 10101_2. '10101' has no '00' present in it. - _Indranil Ghosh_, Feb 11 2017
From _Gus Wiseman_, Apr 04 2020: (Start)
The terms together with the corresponding compositions begin:
    0: ()            30: (1,1,1,2)         90: (2,1,2,2)
    1: (1)           31: (1,1,1,1,1)       91: (2,1,2,1,1)
    2: (2)           42: (2,2,2)           93: (2,1,1,2,1)
    3: (1,1)         43: (2,2,1,1)         94: (2,1,1,1,2)
    5: (2,1)         45: (2,1,2,1)         95: (2,1,1,1,1,1)
    6: (1,2)         46: (2,1,1,2)        106: (1,2,2,2)
    7: (1,1,1)       47: (2,1,1,1,1)      107: (1,2,2,1,1)
   10: (2,2)         53: (1,2,2,1)        109: (1,2,1,2,1)
   11: (2,1,1)       54: (1,2,1,2)        110: (1,2,1,1,2)
   13: (1,2,1)       55: (1,2,1,1,1)      111: (1,2,1,1,1,1)
   14: (1,1,2)       58: (1,1,2,2)        117: (1,1,2,2,1)
   15: (1,1,1,1)     59: (1,1,2,1,1)      118: (1,1,2,1,2)
   21: (2,2,1)       61: (1,1,1,2,1)      119: (1,1,2,1,1,1)
   22: (2,1,2)       62: (1,1,1,1,2)      122: (1,1,1,2,2)
   23: (2,1,1,1)     63: (1,1,1,1,1,1)    123: (1,1,1,2,1,1)
   26: (1,2,2)       85: (2,2,2,1)        125: (1,1,1,1,2,1)
   27: (1,2,1,1)     86: (2,2,1,2)        126: (1,1,1,1,1,2)
   29: (1,1,2,1)     87: (2,2,1,1,1)      127: (1,1,1,1,1,1,1)
(End)

Indranil Ghosh, Table of n, a(n) for n = 1..50000 (terms 1..1000 from T. D. Noe)
J.-P. Allouche, J. Shallit and G. Skordev, Self-generating sets, integers with missing blocks and substitutions, Discrete Math., Vol. 292, No. 1-3 (2005), pp. 1-15.
Robert Baillie and Thomas Schmelzer, Summing Kempner's Curious (Slowly-Convergent) Series, Mathematica Notebook kempnerSums.nb, Wolfram Library Archive, 2008.
David Garth and Adam Gouge, Affinely Self-Generating Sets and Morphisms, Journal of Integer Sequences, Vol. 10 (2007), Article 07.1.5.
Andreas M. Hinz and Paul K. Stockmeyer, Precious Metal Sequences and Sierpinski-Type Graphs, J. Integer Seq., Vol 25 (2022), Article 22.4.8.
Tomi Kärki, Anne Lacroix, and Michel Rigo, On the recognizability of self-generating sets, Journal of Integer Sequences, Vol. 13 (2010), Article 10.2.2.
Wikipedia, Ahnentafel.
Witzel, Stefan On panel-regular ~A2 lattices Geom. Dedicata 191, 85-135 (2017).
Index entries for 2-automatic sequences.
Index entries for sequences related to binary expansion of n.

A104326(n) = A007088(a(n)); A023416(a(n)) = A087116(a(n)); A107782(a(n)) = 0; A107345(a(n)) = 1; A107359(n) = a(n+1) - a(n); a(A001911(n)) = A000225(n); a(A000071(n+2)) = A000975(n). - Reinhard Zumkeller, May 25 2005
Cf. A003796 (no 000), A004745 (no 001), A004746 (no 010), A004744 (no 011), A004742 (no 101), A004743 (no 110), A003726 (no 111).
Complement of A004753.
Cf. A023705, A196168, A280873.
Positions of numbers <= 2 in A333766 (see this and A066099 for other sequences about compositions in standard order).
Cf. A318928.

a003754 n = a003754_list !! (n-1)
a003754_list = filter f [0..] where
   f x = x == 0 || x `mod` 4 > 0 && f (x `div` 2)
-- Reinhard Zumkeller, Dec 07 2012, Oct 19 2011

isA003754 := proc(n) local bdgs ; bdgs := convert(n,base,2) ; for i from 2 to nops(bdgs) do if op(i,bdgs)=0 and op(i-1,bdgs)= 0 then return false; end if; end do; return true; end proc:
A003754 := proc(n) option remember; if n= 1 then 0; else for a from procname(n-1)+1 do if isA003754(a) then return a; end if; end do: end if; end proc:
# R. J. Mathar, Oct 23 2010

Select[ Range[0, 200], !MatchQ[ IntegerDigits[#, 2], {_, 0, 0, _}]&] (* Jean-François Alcover, Oct 25 2011 *)
Select[Range[0,200],SequenceCount[IntegerDigits[#,2],{0,0}]==0&] (* The program uses the SequenceCount function from Mathematica version 10 *) (* Harvey P. Dale, May 21 2015 *)

is(n)=n=bitor(n,n>>1)+1; n>>=valuation(n,2); n==1 \\ Charles R Greathouse IV, Feb 06 2017

i=0
while i<=500:
    if "00" not in bin(i)[2:]:
        print(str(i), end=',')
    i+=1 # Indranil Ghosh, Feb 11 2017

Sum_{n>=2} 1/a(n) = 4.356588498070498826084131338899394678478395568880140707240875371925764128502... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 12 2022

Removed "2" from the name, because, for example, one could argue that 10001 has 3 adjacent zeros, not 2. - Gus Wiseman, Apr 04 2020

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}]

A384878 Position of first appearance of n in the flattened version of the triangle A384877, whose m-th row lists the lengths of maximal anti-runs in the binary indices of m.

Original entry on oeis.org

1, 6, 34, 178, 882, 4210, 19570, 89202, 400498, 1776754

Offset: 1

Gus Wiseman, Jun 23 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 set of binary indices of each nonnegative integer and its partition into anti-runs begins:
  0: {}      {{}}
  1: {1}     {{1}}
  2: {2}     {{2}}
  3: {1,2}   {{1},{2}}
  4: {3}     {{3}}
  5: {1,3}   {{1,3}}
  6: {2,3}   {{2},{3}}
  7: {1,2,3} {{1},{2},{3}}
The flattened version begins: {}, {1}, {2}, {1}, {2}, {3}, {1,3}, {2}, {3}, {1}, {2}, {3}. Of these sets, the first of length 2 is the sixth (starting with 0), so we have a(2) = 6.

For runs instead of anti-runs we have A001792.
The unflattened version is A052499.
Positions of first appearances in A384877, see A000120, A245562, A245563, A384890.
A023758 lists differences of powers of 2.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
Cf. A044813, A048793, A069010, A384879, A384893.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
mnrm[s_]:=If[Min@@s==1,mnrm[DeleteCases[s-1,0]]+1,0];
q=Join@@Table[Length/@Split[bpe[n],#2!=#1+1&],{n,0,100}];
Table[Position[q,i][[1,1]],{i,mnrm[q]}]

A200648 Length of Stolarsky representation of n.

Original entry on oeis.org

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

Offset: 1

Casey Mongoven, Nov 19 2011

For the Stolarsky representation of n, see the C. Mongoven link.

Conjecture: a(n) is the sum of row n-1 of A385886. To obtain it, first take maximal anti-run lengths of binary indices of each nonnegative integer (giving A384877), then remove all duplicate rows (giving A385886), and finally take the sum of each remaining row. For length instead of sum we appear to have A200649. - Gus Wiseman, Jul 21 2025

			The Stolarsky representation of 19 is 11101. This is of length 5. So a(19) = 5.

Amiram Eldar, Table of n, a(n) for n = 1..10000
Casey Mongoven, Description of Stolarsky Representations.

Cf. A135817, A200651.
Counting just ones gives A200649.
Counting just zeros gives A200650.
Stolarsky representation is listed by A385888, ranks A200714.
A000120 counts 1's in binary expansion.
A384890 counts maximal anti-runs of binary indices, ranks A385816.
A385886 lists maximal anti-run lengths of binary indices.
Cf. A023758, A048793, A052499, A069010, A072649, A083368, A135818, A245562, A245563, A348366, A384877, A384893.

stol[n_] := stol[n] = If[n == 1, {}, If[n != Round[Round[n/GoldenRatio]*GoldenRatio], Join[stol[Floor[n/GoldenRatio^2] + 1], {0}], Join[stol[Round[n/GoldenRatio]], {1}]]];
a[n_] := If[n == 1, 1, Length[stol[n]]]; Array[a, 100] (* Amiram Eldar, Jul 07 2023 *)

stol(n) = {my(phi=quadgen(5)); if(n==1, [], if(n != round(round(n/phi)*phi), concat(stol(floor(n/phi^2) + 1), [0]), concat(stol(round(n/phi)), [1])));}
a(n) = if(n == 1, 1, #stol(n)); \\ Amiram Eldar, Jul 07 2023

a(n) = A200649(n) + A200650(n). - Michel Marcus, Mar 14 2023

More terms from Amiram Eldar, Jul 07 2023

A385816 The number k such that the k-th composition in standard order lists the maximal anti-run lengths of the binary indices of n. Standard composition number of row n of A384877.

Original entry on oeis.org

0, 1, 1, 3, 1, 2, 3, 7, 1, 2, 2, 6, 3, 5, 7, 15, 1, 2, 2, 6, 2, 4, 6, 14, 3, 5, 5, 13, 7, 11, 15, 31, 1, 2, 2, 6, 2, 4, 6, 14, 2, 4, 4, 12, 6, 10, 14, 30, 3, 5, 5, 13, 5, 9, 13, 29, 7, 11, 11, 27, 15, 23, 31, 63, 1, 2, 2, 6, 2, 4, 6, 14, 2, 4, 4, 12, 6, 10, 14

Offset: 0

Gus Wiseman, Jul 15 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 k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.
If the k-th composition in standard order is y, then the standard composition number of y is defined to be k.

			The binary indices of 181 are {1,3,5,6,8}, with maximal anti-runs ((1,3,5),(6,8)), with lengths (3,2), which is the 18th composition in standard order, so a(181) = 18.

Gus Wiseman, Statistics, classes, and transformations of standard compositions

The reverse version is A209859.
Sorted positions of first appearances are A247648.
These are standard composition numbers of rows of A384877 (duplicates removed A385886).
For runs instead of anti-runs the reverse is A385887 (duplicates removed A232559).
For runs instead of anti-runs we have A385889 (duplicates removed A385818).
A245563 lists run lengths of binary indices (ranks A246029), reverse A245562.
Cf. A000120, A029931, A044813, A048793, A052499, A069010, A348366, A384175, A384878, A384879, A384893.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
stcinv[q_]:=Total[2^(Accumulate[Reverse[q]])]/2;
stcinv/@Table[Length/@Split[bpe[n],#2!=#1+1&],{n,0,100}]

A385886 Irregular triangle read by rows listing the lengths of maximal anti-runs (sequences of distinct consecutive elements increasing by more than 1) of binary indices, duplicate rows removed.

Original entry on oeis.org

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

Offset: 0

Gus Wiseman, Jul 14 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.

This is the triangle A384877, except all duplicates after the first instance of each composition are removed. It lists all compositions in order of their first appearance as a row of A384877.

			The binary indices of 27 are {1,2,4,5}, with maximal anti-runs ((1),(2,4),(5)), with lengths (1,2,1). After removing duplicates, this is our row 10.
The binary indices of 53 are {1,3,5,6}, with maximal anti-runs ((1,3,5),(6)), with lengths (3,1). After removing duplicates, this is our row 16.
Triangle begins:
   0: .
   1: 1
   2: 1 1
   3: 2
   4: 1 1 1
   5: 1 2
   6: 2 1
   7: 1 1 1 1
   8: 3
   9: 1 1 2
  10: 1 2 1
  11: 2 1 1
  12: 1 1 1 1 1
  13: 1 3
  14: 2 2
  15: 1 1 1 2
  16: 3 1
  17: 1 1 2 1
  18: 1 2 1 1
  19: 2 1 1 1
  20: 1 1 1 1 1 1

In the following references, "before" is short for "before removing duplicate rows".
Positions of singleton rows appear to be A001906 = A055588 - 1.
Positions of rows of the form (1,1,...) appear to be A001911-2, before A023758.
Row sums appear to be A200648, before A000120.
Row lengths appear to be A200649, before A384890.
Standard composition numbers of each row appear to be A348366.
Before we had A384877, ranks A385816, firsts A052499.
For runs instead of anti-runs we have A385817, see A245563, A245562, A246029.
Cf. A044813, A048793, A069010, A164707, A242882, A328592, A384175, A384177, A384878, A384879, A384893.

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

A209859 Rewrite the binary expansion of n from the most significant end, 1 -> 1, 0+1 (one or more zeros followed by one) -> 0, drop the trailing zeros of the original n.

Original entry on oeis.org

0, 1, 1, 3, 1, 2, 3, 7, 1, 2, 2, 5, 3, 6, 7, 15, 1, 2, 2, 5, 2, 4, 5, 11, 3, 6, 6, 13, 7, 14, 15, 31, 1, 2, 2, 5, 2, 4, 5, 11, 2, 4, 4, 9, 5, 10, 11, 23, 3, 6, 6, 13, 6, 12, 13, 27, 7, 14, 14, 29, 15, 30, 31, 63, 1, 2, 2, 5, 2, 4, 5, 11, 2, 4, 4, 9, 5, 10, 11, 23, 2, 4, 4, 9, 4, 8, 9, 19, 5, 10, 10, 21, 11, 22, 23, 47, 3, 6, 6, 13, 6, 12, 13, 27, 6, 12, 12, 25, 13

Offset: 0

Antti Karttunen, Mar 24 2012

This is the number k such that the k-th composition in standard order is the reversed sequence of lengths of the maximal anti-runs of the binary indices of n. Here, the binary indices of n are row n of A048793, and the k-th composition in standard order is row k of A066099. For example, the binary indices of 100 are {3,6,7}, with maximal anti-runs ((3,6),(7)), with reversed lengths (1,2), which is the 6th composition in standard order, so a(100) = 6. - Gus Wiseman, Jul 27 2025

			102 in binary is 1100110, we rewrite it from the left so that first two 1's stay same ("11"), then "001" is rewritten to "0", the last 1 to "1", and we ignore the last 0, thus getting 1101, which is binary expansion of 13, thus a(102) = 13.

Gus Wiseman, Statistics, classes, and transformations of standard compositions

This is an "inverse" of A071162, i.e. a(A071162(n)) = n for all n. Bisection: A209639. Used to construct permutation A209862.
Removing duplicates appears to give A358654.
Sorted positions of firsts appearances appear to be A247648+1.
A245563 lists run-lengths of binary indices (ranks A246029), reverse A245562.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
Cf. A000120, A023758, A029931, A044813, A048793, A052499, A069010, A348366, A384878, A384879, A384893.
Cf. A384877, A385816, A385887, A385889.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
stcinv[q_]:=Total[2^(Accumulate[Reverse[q]])]/2;
Table[stcinv[Reverse[Length/@Split[bpe[n],#2!=#1+1&]]],{n,0,100}] (* Gus Wiseman, Jul 25 2025 *)

import re
def a(n): return int(re.sub("0+1", "0", bin(n)[2:].rstrip("0")), 2) if n else 0
print([a(n) for n in range(109)])  # Michael S. Branicky, Jul 25 2025

(define (A209859 n) (let loop ((n n) (s 0) (i (A053644 n))) (cond ((zero? n) s) ((> i n) (if (> (/ i 2) n) (loop n s (/ i 2)) (loop (- n (/ i 2)) (* 2 s) (/ i 4)))) (else (loop (- n i) (+ (* 2 s) 1) (/ i 2))))))

a(n) = a(A000265(n)).

A385817 Irregular triangle read by rows listing the lengths of maximal runs (sequences of consecutive elements increasing by 1) of binary indices, duplicate rows removed.

Original entry on oeis.org

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

Offset: 0

Gus Wiseman, Jul 14 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.

This is the triangle A245563, except all duplicates after the first instance of each composition are removed. It lists all compositions in order of their first appearance as a row of A245563.

			The binary indices of 53 are {1,3,5,6}, with maximal runs ((1),(3),(5,6)), with lengths (1,1,2). After removing duplicates, this is our row 16.
Triangle begins:
   0: .
   1: 1
   2: 2
   3: 1 1
   4: 3
   5: 2 1
   6: 1 2
   7: 4
   8: 1 1 1
   9: 3 1
  10: 2 2
  11: 1 3
  12: 5
  13: 2 1 1
  14: 1 2 1
  15: 4 1
  16: 1 1 2
  17: 3 2
  18: 2 3
  19: 1 4
  20: 6
  21: 1 1 1 1

In the following references, "before" is short for "before removing duplicate rows".
Positions of singleton rows appear to be A000071 = A000045-1, before A023758.
Positions of firsts appearances appear to be A001629.
Positions of rows of the form (1,1,...) appear to be A055588 = A001906+1.
First term of each row appears to be A083368.
Row sums appear to be A200648, before A000120.
Row lengths after the first row appear to be A200650+1, before A069010 = A037800+1.
Before the removals we had A245563 (except first term), see A245562, A246029, A328592.
For anti-run ranks we have A385816, before A348366, firsts A052499.
Standard composition numbers of rows are A385818, before A385889.
For anti-runs we have A385886, before A384877, firsts A384878.
Cf. A044813, A048793, A164707, A200649, A242882, A243815, A384175, A384890.

DeleteDuplicates[Table[Length/@Split[Join@@Position[Reverse[IntegerDigits[n,2]],1],#2==#1+1&],{n,0,100}]]

A385818 The number k such that the k-th composition in standard order lists the maximal run lengths of each nonnegative integer's binary indices, with duplicates removed.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 8, 7, 9, 10, 12, 16, 11, 13, 17, 14, 18, 20, 24, 32, 15, 19, 21, 25, 33, 22, 26, 34, 28, 36, 40, 48, 64, 23, 27, 35, 29, 37, 41, 49, 65, 30, 38, 42, 50, 66, 44, 52, 68, 56, 72, 80, 96, 128, 31, 39, 43, 51, 67, 45, 53, 69, 57, 73, 81, 97

Offset: 0

Gus Wiseman, Jul 18 2025

A permutation of the nonnegative integers.
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 k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions.

			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 14th composition in standard order, so A385889(53) = 14, and after removing duplicate rows a(16) = 14.

John Tyler Rascoe, Table of n, a(n) for n = 0..8192

For anti-runs instead of runs we appear to have A348366.
See also A385816 (standard compositions of rows of A384877), reverse A209859.
The compositions themselves are listed by A385817.
Before removing duplicates we had A385889.
A245563 lists run lengths of binary indices (ranks A246029), rev A245562, strict A328592.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
Cf. A000120, A029931, A048793, A052499, A069010, A200648, A200649, A200650, A384890, A385886, A385887.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
stcinv[q_]:=Total[2^(Accumulate[Reverse[q]])]/2;
stcinv/@DeleteDuplicates[Table[Length/@Split[bpe[n],#2==#1+1&],{n,0,100}]]

cp's OEIS Frontend

A003754 Numbers with no adjacent 0's in binary expansion.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Haskell

Maple

Mathematica

PARI

Python

Formula

Extensions

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

A384878 Position of first appearance of n in the flattened version of the triangle A384877, whose m-th row lists the lengths of maximal anti-runs in the binary indices of m.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A200648 Length of Stolarsky representation of n.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

PARI

Formula

Extensions

A385816 The number k such that the k-th composition in standard order lists the maximal anti-run lengths of the binary indices of n. Standard composition number of row n of A384877.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

A385886 Irregular triangle read by rows listing the lengths of maximal anti-runs (sequences of distinct consecutive elements increasing by more than 1) of binary indices, duplicate rows removed.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A209859 Rewrite the binary expansion of n from the most significant end, 1 -> 1, 0+1 (one or more zeros followed by one) -> 0, drop the trailing zeros of the original n.

Views

Author

Keywords