A245562 -id:A245562 - cp's OEIS frontend

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

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}]

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)).

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]}]

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

Original entry on oeis.org

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

A384879 Numbers whose binary indices have all distinct lengths of maximal anti-runs (increasing by more than 1).

Original entry on oeis.org

1, 2, 4, 5, 8, 9, 10, 11, 13, 16, 17, 18, 19, 20, 21, 22, 25, 26, 32, 33, 34, 35, 36, 37, 38, 40, 41, 42, 43, 44, 49, 50, 52, 53, 64, 65, 66, 67, 68, 69, 70, 72, 73, 74, 75, 76, 80, 81, 82, 83, 84, 85, 86, 88, 97, 98, 100, 101, 104, 105, 106, 128, 129, 130

Offset: 1

Gus Wiseman, Jun 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 binary indices of 813 are {1,3,4,6,9,10}, with maximal anti-runs ((1,3),(4,6,9),(10)), with lengths (2,3,1), so 813 is in the sequence.
The terms together with their binary expansions and binary indices begin:
    1:       1 ~ {1}
    2:      10 ~ {2}
    4:     100 ~ {3}
    5:     101 ~ {1,3}
    8:    1000 ~ {4}
    9:    1001 ~ {1,4}
   10:    1010 ~ {2,4}
   11:    1011 ~ {1,2,4}
   13:    1101 ~ {1,3,4}
   16:   10000 ~ {5}
   17:   10001 ~ {1,5}
   18:   10010 ~ {2,5}
   19:   10011 ~ {1,2,5}
   20:   10100 ~ {3,5}
   21:   10101 ~ {1,3,5}
   22:   10110 ~ {2,3,5}
   25:   11001 ~ {1,4,5}
   26:   11010 ~ {2,4,5}

Subsets of this type are counted by A384177, for runs A384175 (complement A384176).
These are the indices of strict rows in A384877, see A384878, A245563, A245562, A246029.
A000120 counts binary indices.
A098859 counts Wilf partitions (distinct multiplicities), complement A336866.
A356606 counts strict partitions without a neighborless part, complement A356607.
A384890 counts maximal anti-runs in binary indices, runs A069010.
Cf. A023758, A044813, A048793, A164707, A242882, A243815, A325325, A328592, A384879, A384884, A384886, A384893.

bpe[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
Select[Range[100],UnsameQ@@Length/@Split[bpe[#],#2!=#1+1&]&]

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

A245180 A160239(n)/8.

Original entry on oeis.org

1, 1, 3, 1, 8, 3, 14, 1, 8, 8, 24, 3, 24, 14, 52, 1, 8, 8, 24, 8, 64, 24, 112, 3, 24, 24, 72, 14, 112, 52, 216, 1, 8, 8, 24, 8, 64, 24, 112, 8, 64, 64, 192, 24, 192, 112, 416, 3, 24, 24, 72, 24, 192, 72, 336, 14, 112, 112, 336, 52, 416, 216, 848, 1, 8, 8, 24, 8, 64, 24, 112, 8, 64, 64, 192

Offset: 1

N. J. A. Sloane, Jul 16 2014

			The entries may be arranged into blocks of sizes 1,2,4,8,...:
B_0: 1,
B_1: 1, 3,
B_2: 1, 8, 3, 14,
B_3: 1, 8, 8, 24, 3, 24, 14, 52,
B_4: 1, 8, 8, 24, 8, 64, 24, 112, 3, 24, 24, 72, 14, 112, 52, 216,
B_5: 1, 8, 8, 24, 8, 64, 24, 112, 8, 64, 64, 192, 24, 192, 112, 416, 3, 24, 24, 72, 24, 192, 72, 336, 14, 112, 112, 336, 52, 416, 216, 848,
...
The first half of each block is equal to 1 followed by 8 times an initial segment of the sequence itself.
The next quarter of each block consists of 3 times (1 followed by 8 times an initial segment of the sequence itself).
The next one-eighth of each block consists of 14 times (1 followed by 8 times an initial segment of the sequence itself).
And so on, the successive multipliers 1,3,14,52,... being given by A083424.
Also, the final quarter of any block consists of the twice the last half of the previous block added to eight times the full block before that.
Consider for example the 4th block,
[1, 8, 8, 24, 8, 64, 24, 112; 3, 24, 24, 72; 14, 112, 52, 216].
This is [1 8*(1,1,3,1,8,3,14); 3*(1 8*(1,1,3)); 2*(3,24,14,52)+8*(1,8,3,14)].
The final entries in the blocks give A083424.
See also the formula section.
.
From _Omar E. Pol_, Mar 18 2015: (Start)
Also, the sequence can be written as an irregular tetrahedron T(s,r,k) as shown below:
1;
..
1;
3;
........
1,    8;
3;
14;
................
1,    8,  8, 24;
3,   24;
14;
52;
..................................
1,    8,  8, 24,  8,  64, 24, 112;
3,   24, 24, 72;
14, 112;
52;
216;
.....................................................................
1,    8,  8, 24,  8,  64, 24, 112, 8, 64, 64, 192, 24, 192, 112, 416;
3,   24, 24, 72, 24, 192, 72, 336;
14, 112,112,336;
52, 416;
216;
848;
...
Note that T(s,r,k) = T(s+1,r,k).
(End)

N. J. A. Sloane, Table of n, a(n) for n = 1..16383
David Applegate, Omar E. Pol and N. J. A. Sloane, The Toothpick Sequence and Other Sequences from Cellular Automata, Congressus Numerantium, Vol. 206 (2010), 157-191. [There is a typo in Theorem 6: (13) should read u(n) = 4.3^(wt(n-1)-1) for n >= 2.], which is also available at arXiv:1004.3036v2

Cf. A160239, A083424, A245190, A245195, A245196, A245562, A245540 (partial sums).

See A245181 for the numbers that appear.

a245180 = flip div 8 . a160239  -- Reinhard Zumkeller, Feb 13 2015

R:=proc(n) option remember;
if n=1 then 1
elif (n mod 2) = 0 then R(n/2)
elif (n mod 4) = 3 then 2*R((n-1)/2)+R(n-2)
else 8*R((n-1)/4); fi; end;
[seq(R(n),n=1..200)];

R[n_] := R[n] = Which[n == 1, 1, Mod[n, 2] == 0, R[n/2], Mod[n, 4] == 3, 2*R[(n - 1)/2] + R[n - 2], True, 8*R[(n - 1)/4] ];
Array[R, 200] (* Jean-François Alcover, Nov 16 2017, translated from Maple *)

The following is a fairly simple explicit formula for a(n) as a function of n: a(n) = 8^(r-1) * Product_{lengths i of runs of 1 in binary expansion of n} R(i), where r is the number of runs of 1 in the binary expansion of n and R(i) = A083424(i-1) = (5*4^(i-1)+(-2)^(i-1))/6. Note that row i of the table in A245562 lists the lengths of runs of 1 in binary expansion of i. Example: n = 27 = 11011 in binary, there are two runs each of length 2, so r=1, R(2) = A083424(1) = 3, and so a(27) = 8^1*3*3 = 72. - N. J. A. Sloane, Aug 10 2014
Many 2-D cellular automata studied in the Toothpick paper (Applegate et al.) have a recursive formula for the general term in a typical block of 2^k terms (see Equations 2, 4, 5, 9, 10 12, 38, 39 of that paper). An analogous formula for the present sequence is the following.
Consider the block B_{k-1} containing terms a(2^(k-1)), a(2^(k-1)+1), ..., a(2^k-1). It is convenient to index the terms working backwards from the next, 2^k-th, term. For n in the range 2^(k-1) <= n < 2^k, write n = 2^k-2^r+j, with 0 <= r <= k-1 and 0 <= j < 2^(r-1), and j=0 if r=0. Then
(if j=0) a(2^k-2^r) = f(k-r-1),
(if j>0) a(2^k-2^r+j) = 8*f(k-r-1)*a(j),
where f(i) = A083424(i) = (5*4^i+(-2)^i)/6.
For example, here is block B_4, consisting of terms a(16)=a(31), so k=5:
n:   16 17 18 19 20 21 22 23 24 25 26 27 28  29 30 31
a(n): 1  8  8 24  8 64 24 112 3 24 24 72 14 112 52 216
r:    4  4  4  4  4  4  4  4  3  3  3  3  2   2  1  0
j:    0  1  2  3  4  5  6  7  0  1  2  3  0   1  0  0
Then we have a(24) = a(32-8) = f(5-3-1) = f(1) = 3, illustrating the first equation, and a(21) = a(32-16+5) = 8*f(0)*a(5) = 8*1*8 = 64, illustrating the second equation.
See A245196 for a list of sequences produced by this type of recurrence.

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}]]

cp's OEIS Frontend

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

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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.

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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.

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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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A384879 Numbers whose binary indices have all distinct lengths of maximal anti-runs (increasing by more than 1).

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

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