A385886 -id:A385886 - cp's OEIS frontend

A200649 Number of 1's in the Stolarsky representation of n.

0, 1, 2, 1, 3, 2, 2, 4, 1, 3, 3, 3, 5, 2, 2, 4, 2, 4, 4, 4, 6, 1, 3, 3, 3, 5, 3, 3, 5, 3, 5, 5, 5, 7, 2, 2, 4, 2, 4, 4, 4, 6, 2, 4, 4, 4, 6, 4, 4, 6, 4, 6, 6, 6, 8, 1, 3, 3, 3, 5, 3, 3, 5, 3, 5, 5, 5, 7, 3, 3, 5, 3, 5, 5, 5, 7, 3, 5, 5, 5, 7, 5, 5, 7, 5, 7, 7

Offset: 1

Casey Mongoven, Nov 19 2011

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

Conjecture: a(n) is the length 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 length of each remaining row. For sum instead of length we appear to have A200648. For runs minus 1 instead of anti-runs see A200650. - Gus Wiseman, Jul 21 2025

			The Stolarsky representation of 19 is 11101. This has 4 1's. So a(19) = 4.

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

Cf. A072649, A130312, A135818, A200651.
For length instead of number of 1's we have A200648.
For 0's instead of 1's we have A200650.
Stolarsky representation is listed by A385888, ranks A200714.
A000120 counts 1's in binary expansion.
A384877 lists anti-run lengths of binary indices, duplicates removed A385886.
A384890 counts maximal anti-runs of binary indices, ranked by A385816.
Cf. A023758, A044813, A048793, A069010, A164707, A245562, A245563, A328592, 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_] := Count[stol[n], 1]; 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) = vecsum(stol(n)); \\ Amiram Eldar, Jul 07 2023

a(n) = a(n - A130312(n-1)) + (A072649(n-1) - A072649(n - A130312(n-1) - 1)) mod 2 for n > 2 with a(1) = 0, a(2) = 1. - Mikhail Kurkov, Oct 19 2021 [verification needed]

a(n) = A200648(n) - A200650(n). - Amiram Eldar, Jul 07 2023

More terms from Amiram Eldar, Jul 07 2023

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

A200650 Number of 0's in Stolarsky representation of n.

Original entry on oeis.org

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

Offset: 1

Casey Mongoven, Nov 19 2011

For the Stolarsky representation of n, see the C. Mongoven link.
a(n+1), n >= 1, gives the size of the n-th generation of each of the "[male-female] pair of Fibonacci rabbits" in the Fibonacci rabbits tree read right-to-left by row, the first pair (the root) being the 0th generation. (Cf. OEIS Wiki link below.) - Daniel Forgues, May 07 2015
From Daniel Forgues, May 07 2015: (Start)
Concatenation of:
  0:                      1,
  1:                      0,
  2:                      0,
  3:                   1, 0,
  4:                1, 1, 0,
  5:          2, 1, 1, 1, 0,
  6: 2, 2, 1, 2, 1, 1, 1, 0,
(...),
where row n, n >= 3, is row n-1 prepended by incremented row n-2. (End)
For n >= 3, this algorithm yields the next F_n terms of the sequence, where F_n is the n-th Fibonacci number (A000045). Since it is asymptotic to (phi^n)/sqrt(5), the number of terms thus obtained grows exponentially at each step! - Daniel Forgues, May 22 2015
Conjecture: a(n) is one less than the length of row n-1 of A385817. To obtain it, first take maximal run lengths of binary indices of each nonnegative integer (giving A245563), then remove all duplicate rows (giving A385817), and finally take the length of each remaining row and subtract 1. For sum instead of length we appear to have A200648. For anti-runs instead of runs we appear to have A341259 = A200649-1. - Gus Wiseman, Jul 21 2025
How is this related to A117479? - R. J. Mathar, Aug 10 2025

			The Stolarsky representation of 19 is 11101. This has one 0. So a(19) = 1.

Kenny Lau, Table of n, a(n) for n = 1..20000
Casey Mongoven, Description of Stolarsky Representations.
OEIS Wiki, Fibonacci rabbits per generation.

For length instead of number of 0's we have A200648.
For sum (or number of 1's) instead of number of 0's we have A200649.
Stolarsky representation is listed by A385888, ranks A200714.
A000120 counts 1's in binary expansion, 0's A023416.
A069010 counts maximal runs of binary indices, ranked by A385889.
A245563 lists maximal run lengths of binary indices, duplicates removed A385817.
Cf. A000045, A023758, A135818, A200651, A245562, A328592, A385886.

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, Count[stol[n], 0]]; 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,  my(s = stol(n)); #s - vecsum(s)); \\ Amiram Eldar, Jul 07 2023

a(n) = A200648(n) - A200649(n). - Amiram Eldar, Jul 07 2023

Corrected and extended by Kenny Lau, Jul 04 2016

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}]

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}]]

A083368 a(n) is the position of the highest one in A003754(n).

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 4, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 6, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 7, 2, 1, 4, 1, 3, 2, 1, 6, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 8, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 7, 2, 1, 4, 1, 3, 2, 1, 6, 1, 3, 2, 1, 5, 2, 1, 4, 1, 3, 2, 1, 9, 2, 1, 4, 1, 3, 2, 1, 6, 1, 3, 2, 1, 5, 2

Offset: 1

Gary W. Adamson, Jun 04 2003

Previous name was: A Fibbinary system represents a number as a sum of distinct Fibonacci numbers (instead of distinct powers of two). Using representations without adjacent zeros, a(n) = the highest bit-position which changes going from n-1 to n.
A003754(n), when written in binary, is the representation of n.
Often one uses Fibbinary representations without adjacent ones (the Zeckendorf expansion).
a(A000071(n+3)) = n. - Reinhard Zumkeller, Aug 10 2014
From Gus Wiseman, Jul 24 2025: (Start)
Conjecture: To obtain this sequence, start with A245563 (maximal run lengths of binary indices), then remove empty and duplicate rows (giving A385817), then take the first term of each remaining row. Some variations:
- For sum instead of first term we appear to have A200648.
- For length instead of first term we appear to have A200650+1.
- For last instead of first term we have A385892.
(End)

			27 is represented 110111, 28 is 111010; the fourth position changes, so a(28)=4.

Jay Kappraff, Beyond Measure: A Guided Tour Through Nature, Myth and Number, World Scientific, 2002, page 460.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

A035612 is the analogous sequence for Zeckendorf representations.
A001511 is the analogous sequence for power-of-two representations.
Cf. A003714, A003754.
Cf. A000045, A000071.
Appears to be the first element of each row of A385817, see A083368, A200648, A200650, A385818, A385892.
A000120 counts ones in binary expansion.
A245563 gives run lengths of binary indices, see A089309, A090996, A245562, A246029, A328592.
A384877 gives anti-run lengths of binary indices, ranks A385816.
Cf. A001629, A037800, A044813, A048793, A069010, A200649, A384878, A385886.

a083368 n = a083368_list !! (n-1)
a083368_list = concat $ h $ drop 2 a000071_list where
   h (a:fs@(a':_)) = (map (a035612 . (a' -)) [a .. a' - 1]) : h fs
-- Reinhard Zumkeller, Aug 10 2014

For n = F(a)-1 to F(a+1)-2, a(n) = A035612(F(a+1)-1-n).

a(n) = a(k)+1 if n = ceiling(phi*k) where phi is the golden ratio; otherwise a(n) = 1. - Tom Edgar, Aug 25 2015

Edited by Don Reble, Nov 12 2005

Shorter name from Joerg Arndt, Jul 27 2025

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

nonn

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.

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}]]

A385887 The number k such that the k-th composition in standard order is the reversed sequence of lengths of maximal runs of binary indices of n.

Original entry on oeis.org

0, 1, 1, 2, 1, 3, 2, 4, 1, 3, 3, 6, 2, 5, 4, 8, 1, 3, 3, 6, 3, 7, 6, 12, 2, 5, 5, 10, 4, 9, 8, 16, 1, 3, 3, 6, 3, 7, 6, 12, 3, 7, 7, 14, 6, 13, 12, 24, 2, 5, 5, 10, 5, 11, 10, 20, 4, 9, 9, 18, 8, 17, 16, 32, 1, 3, 3, 6, 3, 7, 6, 12, 3, 7, 7, 14, 6, 13, 12, 24

Offset: 0

Gus Wiseman, Jul 17 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.

			The binary indices of 100 are {3,6,7}, with maximal runs ((3),(6,7)), with reversed lengths (2,1), which is the 5th composition in standard order, so a(100) = 5.

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

Removing duplicates appears to give A232559, see also A348366, A358654, A385818.
Sorted positions of firsts appearances appear to be A247648+1.
The non-reverse version is A385889.
A245563 lists run-lengths of binary indices (ranks A246029), reverse A245562.
A384877 lists anti-run lengths of binary indices (ranks A385816), reverse A209859.
Cf. A000120, A029931, A044813, A048793, A069010, A384175, A385817, A385886.

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}]

A385891 The number k = a(n) whose Stolarsky representation (row k of A385888) is the binary expansion of n (row n of A030302). Inverse Stolarsky representation of binary expansion of n.

Original entry on oeis.org

1, 2, 4, 3, 9, 6, 7, 5, 22, 15, 14, 10, 17, 11, 12, 8, 56, 36, 38, 24, 35, 23, 25, 16, 43, 28, 27, 18, 30, 19, 20, 13, 145, 91, 93, 58, 98, 61, 62, 39, 90, 57, 59, 37, 64, 40, 41, 26, 111, 70, 72, 45, 69, 44, 46, 29, 77, 49, 48, 31, 51, 32, 33, 21, 378, 235, 237

Offset: 0

Gus Wiseman, Jul 24 2025

nonn

A permutation of the positive integers.

Casey Mongoven, Description of Stolarsky Representations.

A000120 counts 1's in binary expansion, zeros A023416.
A030302 gives binary expansion.
A035506 is the Stolarsky array.
A385888 is the Stolarsky representation, lengths A200648, sums A200649, zeros A200650, ranks A200714.
Cf. A000045, A044813, A135818, A200651, A245563, A384890, A385886.

nn=30;
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}]]];
ss=Table[stol[n],{n,nn}];
mnrm[s_]:=If[Min@@s==1,mnrm[DeleteCases[s-1,0]]+1,0];
Table[Position[ss,If[n==0,{},IntegerDigits[n,2]]][[1,1]],{n,0,mnrm[FromDigits[#,2]&/@ss+1]-1}]

A385888(a(n)) = A030302(n).

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.

Original entry on oeis.org

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}]]

cp's OEIS Frontend

A200649 Number of 1's in the Stolarsky representation of n.

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A200648 Length of Stolarsky representation of n.

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A200650 Number of 0's in Stolarsky representation of n.

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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.

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Mathematica

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.

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Mathematica

A083368 a(n) is the position of the highest one in A003754(n).

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Haskell

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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.

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