A348366 -id:A348366 - cp's OEIS frontend

A232559 Sequence (or tree) generated by these rules: 1 is in S, and if x is in S, then x + 1 and 2*x are in S, and duplicates are deleted as they occur.

1, 2, 3, 4, 6, 5, 8, 7, 12, 10, 9, 16, 14, 13, 24, 11, 20, 18, 17, 32, 15, 28, 26, 25, 48, 22, 21, 40, 19, 36, 34, 33, 64, 30, 29, 56, 27, 52, 50, 49, 96, 23, 44, 42, 41, 80, 38, 37, 72, 35, 68, 66, 65, 128, 31, 60, 58, 57, 112, 54, 53, 104, 51, 100, 98, 97

Offset: 1

Clark Kimberling, Nov 26 2013

Let S be the set of numbers defined by these rules:  1 is in S, and if x is in S, then x + 1 and 2*x are in S.  Then S is the set of all positive integers, which arise in generations.  Deleting duplicates as they occur, the generations are given by g(1) = (1), g(2) = (2), g(3) = (3,4), g(4) = (6,5,8), g(5) = (7,12,10,9,16), etc.  Concatenating these gives A232559, a permutation of the positive integers.  The number of numbers in g(n) is A000045(n), the n-th Fibonacci number.  It is helpful to show the results as a tree with the terms of S as nodes and edges from x to x + 1 if x + 1 has not already occurred, and an edge from x to 2*x if 2*x has not already occurred.  The positions of the odd numbers are given by A026352, and of the evens, by A026351.
The previously mentioned tree is an example of a fractal tree; that is, an infinite rooted tree T such that every complete subtree of T contains a subtree isomorphic to T. - Clark Kimberling, Jun 11 2016
The similar sequence S', generated by these rules: 0 is in S', and if x is in S', then 2*x and x+1 are in S', and duplicates are deleted as they occur, appears to equal A048679. - Rémy Sigrist, Aug 05 2017
From Katherine E. Stange and Glen Whitney, Oct 09 2021: (Start)
The beginning of this tree is
                1
                |
                2
               / \
         3..../   \......4
         |              / \
         6         5.../   \...8
        / \        |          / \
      7/   \12     10       9/   \16
This tree contains every positive integer, and one can show that the path from 1 to the integer n is exactly the sequence of intermediate values observed during the Double-And-Add Algorithm AKA Chandra Sutra Method (namely, the algorithm which begins with m = 0, reads the binary representation of n from left to right, and, for each digit 0 read, doubles m, and for each digit 1 read, doubles m and then adds 1 to m; when the algorithm terminates, m = n).
As such, the path between 1 and n is a function of the binary expansion of n.  The elements of the k-th row of the tree (generation g(k)) are all those elements whose binary expansion has k_1 digits and Hamming weight k_2, for some k_1 and k_2 such that k_1 + k_2 = k + 1.
The depth at which integer n appears in this tree is given by A014701(n) = A056792(n)-1.  For example, the depth of 1 is 0, the depth of 2 is 1, and the depths of 3 and 4 are both 2. (End)
Definition need not invoke deletion: Tree is rooted at 1, all even nodes have x+1 as a child, all nodes have 2*x as a child, and any x+1 child precedes its sibling. - Robert Munafo, May 08 2024

			Each x begets x + 1 and 2*x, but if either has already occurred it is deleted.  Thus, 1 begets 2, which begets (3,4); from which 3 begets only 6, and 4 begets (5,8).

Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 1000 terms from Clark Kimberling)
Katherine E. Stange, The Intuition behind the Double-And-Add / Square-And-Multiply Algorithm, YouTube video, 2021.
Dimitri Zucker, I Found a Simple Pattern That Encodes Different Bases, YouTube video, 2024. (Adds 0 above the root of the tree, and shows how to reconstruct the tree backwards from child nodes)
Robert Munafo, Sequences A232559 and A232561, Spanning Trees of N
Index entries for sequences that are permutations of the natural numbers

Cf. A232560 (inverse permutation), A232561, A232563, A226080, A226130.
Cf. A000045, A026352, A026351, A048679.
Cf. A243571 (rows sorted).
Cf. A014701, A056792.

a:= proc() local l, s; l, s:= [1], {1}:
      proc(n) option remember; local i, r; r:= l[1];
        l:= subsop(1=NULL, l);
        for i in [1+r, r+r] do if not i in s then
          l, s:=[l[], i], s union {i} fi
        od; r
      end
    end():
seq(a(n), n=1..100);  # Alois P. Heinz, Aug 06 2017

z = 12; g[1] = {1}; g[2] = {2}; g[n_] := Riffle[g[n - 1] + 1, 2 g[n - 1]]; j[2] = Join[g[1], g[2]]; j[n_] := Join[j[n - 1], g[n]]; g1[n_] := DeleteDuplicates[DeleteCases[g[n], Alternatives @@ j[n - 1]]]; g1[1] = g[1]; g1[2] = g[2]; t = Flatten[Table[g1[n], {n, 1, z}]]  (* this sequence *)
Table[Length[g1[n]], {n, 1, z}] (* Fibonacci numbers *)
t1 = Flatten[Table[Position[t, n], {n, 1, 200}]]  (* A232560 *)

def aupton(terms):
    alst, S, expand = [1, 2], {1, 2}, [2]
    while len(alst) < terms:
        x = expand.pop(0)
        new_elts = [y for y in [x+1, 2*x] if y not in S]
        alst.extend(new_elts); expand.extend(new_elts); S.update(new_elts)
    return alst[:terms]
print(aupton(66)) # Michael S. Branicky, Sep 14 2021

Conjecture: a(n) = A059894(A348366(n)) for n > 0. - Mikhail Kurkov, Jun 14 2022

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)).

A385889 The number k such that the k-th composition in standard order is the 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, 5, 2, 6, 4, 8, 1, 3, 3, 5, 3, 7, 5, 9, 2, 6, 6, 10, 4, 12, 8, 16, 1, 3, 3, 5, 3, 7, 5, 9, 3, 7, 7, 11, 5, 13, 9, 17, 2, 6, 6, 10, 6, 14, 10, 18, 4, 12, 12, 20, 8, 24, 16, 32, 1, 3, 3, 5, 3, 7, 5, 9, 3, 7, 7, 11, 5, 13, 9, 17, 3

Offset: 0

Gus Wiseman, Jul 16 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 27 are {1,2,4,5}, with maximal runs ((1,2),(4,5)), with lengths (2,2), which is the 10th composition in standard order, so a(27) = 10.
The binary indices of 100 are {3,6,7}, with maximal runs ((3),(6,7)), with lengths (1,2), which is the 6th composition in standard order, so a(100) = 6.

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

Sorted positions of firsts appearances appear to be A247648+1.
After removing duplicates we get A385818.
The reverse version is A385887.
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, A023758, A029931, A044813, A048793, A069010, A348366, A384175, A384878, A385817.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
stcinv[q_]:=Total[2^(Accumulate[Reverse[q]])]/2;
Table[stcinv[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}]]

A353654 Numbers whose binary expansion has the same number of trailing 0 bits as other 0 bits.

Original entry on oeis.org

1, 3, 7, 10, 15, 22, 26, 31, 36, 46, 54, 58, 63, 76, 84, 94, 100, 110, 118, 122, 127, 136, 156, 172, 180, 190, 204, 212, 222, 228, 238, 246, 250, 255, 280, 296, 316, 328, 348, 364, 372, 382, 392, 412, 428, 436, 446, 460, 468, 478, 484, 494, 502, 506, 511, 528, 568

Offset: 1

Mikhail Kurkov, Jul 15 2022

Numbers k such that A007814(k) = A086784(k).
To reproduce the sequence through itself, use the following rule: if binary 1xyz is a term then so are 11xyz and 10xyz0 (except for 1 alone where 100 is not a term).
The number of terms with bit length k is equal to Fibonacci(k-1) for k > 1.
Conjecture: 2*A247648(n-1) + 1 with rewrite 1 -> 1, 01 -> 0 applied to binary expansion is the same as a(n) without trailing 0 bits in binary.
Odd terms are positive Mersenne numbers (A000225), because there is no 0 in their binary expansion. - Bernard Schott, Oct 12 2022

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A000045, A000120, A000523, A007814, A010056, A025480, A030109, A054429, A060142, A072649, A084471, A086784, A091892, A132665, A133512, A200650, A213911, A232559, A280514, A343152, A348366.
Cf. A356385 (first differences).
Subsequences with same number k of trailing 0 bits and other 0 bits: A000225 (k=0), 2*A190620 (k=1), 4*A357773 (k=2), 8*A360573 (k=3).

N:= 10: # for terms <= 2^N
S:= {1};
for d from 1 to N do
  for k from 0 to d/2-1 do
    B:= combinat:-choose([$k+1..d-2],k);
    S:= S union convert(map(proc(t) local s; 2^d - 2^k - add(2^(s),s=t) end proc,B),set);
od od:
sort(convert(S,list)); # Robert Israel, Sep 21 2023

Join[{1}, Select[Range[2, 600], IntegerExponent[#, 2] == Floor[Log2[# - 1]] - DigitCount[# - 1, 2, 1] &]] (* Amiram Eldar, Jul 16 2022 *)

isok(k) = if (k==1, 1, (logint(k-1, 2)-hammingweight(k-1) == valuation(k, 2))); \\ Michel Marcus, Jul 16 2022

from itertools import islice, count
def A353654_gen(startvalue=1): # generator of terms >= startvalue
    return filter(lambda n:(m:=(~n & n-1).bit_length()) == bin(n>>m)[2:].count('0'),count(max(startvalue,1)))
A353654_list = list(islice(A353654_gen(),30)) # Chai Wah Wu, Oct 14 2022

a(n) = a(n-1) + A356385(n-1) for n > 1 with a(1) = 1.
Conjectured formulas: (Start)
a(n) = 2^g(n-1)*(h(n-1) + 2^A000523(h(n-1))*(2 - g(n-1))) for n > 2 with a(1) = 1, a(2) = 3 where f(n) = n - A130312(n), g(n) = [n > 2*f(n)] and where h(n) = a(f(n) + 1).
a(n) = 1 + 2^r(n-1) + Sum_{k=1..r(n-1)} (1 - g(s(n-1, k)))*2^(r(n-1) - k) for n > 1 with a(1) = 1 where r(n) = A072649(n) and where s(n, k) = f(s(n, k-1)) for n > 0, k > 1 with s(n, 1) = n.
a(n) = 2*(2 + Sum_{k=1..n-2} 2^(A213911(A280514(k)-1) + 1)) - 2^A200650(n) for n > 1 with a(1) = 1.
A025480(a(n)-1) = A348366(A343152(n-1)) for n > 1.
a(A000045(n)) = 2^(n-1) - 1 for n > 1. (End)

A358654 a(n) = A025480(A353654(n+1) - 1).

Original entry on oeis.org

0, 1, 3, 2, 7, 5, 6, 15, 4, 11, 13, 14, 31, 9, 10, 23, 12, 27, 29, 30, 63, 8, 19, 21, 22, 47, 25, 26, 55, 28, 59, 61, 62, 127, 17, 18, 39, 20, 43, 45, 46, 95, 24, 51, 53, 54, 111, 57, 58, 119, 60, 123, 125, 126, 255, 16, 35, 37, 38, 79, 41, 42, 87, 44, 91, 93

Offset: 0

Mikhail Kurkov, Nov 25 2022

Permutation of the nonnegative integers.

Conjecture: A247648(n) with rewrite 1 -> 1, 01 -> 0 applied to binary expansion is the same as a(n).

Cf. A025480, A048679, A247648, A343152, A348366, A353654, A355489.

Conjecture: a(n) = A348366(A343152(n)) for n > 0 with a(0) = 1.

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

A232559 Sequence (or tree) generated by these rules: 1 is in S, and if x is in S, then x + 1 and 2*x are in S, and duplicates are deleted as they occur.

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A200648 Length of 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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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.

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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.

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A385889 The number k such that the k-th composition in standard order is the sequence of lengths of maximal runs of binary indices of n.

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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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A353654 Numbers whose binary expansion has the same number of trailing 0 bits as other 0 bits.

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