A384877 -id:A384877 - cp's OEIS frontend

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.

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]}]

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}]

A384893 Triangle read by rows where T(n,k) is the number of subsets of {1..n} with k maximal anti-runs (increasing by more than 1).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 4, 2, 1, 1, 7, 5, 2, 1, 1, 12, 10, 6, 2, 1, 1, 20, 20, 13, 7, 2, 1, 1, 33, 38, 29, 16, 8, 2, 1, 1, 54, 71, 60, 39, 19, 9, 2, 1, 1, 88, 130, 122, 86, 50, 22, 10, 2, 1, 1, 143, 235, 241, 187, 116, 62, 25, 11, 2, 1, 1, 232, 420, 468, 392, 267, 150, 75, 28, 12, 2, 1

Offset: 0

Gus Wiseman, Jun 21 2025

			The subset {3,6,7,9,11,12} has maximal anti-runs ((3,6),(7,9,11),(12)), so is counted under T(12,3).
The subset {3,6,7,9,10,12} has maximal anti-runs ((3,6),(7,9),(10,12)), so is counted under T(12,3).
Row n = 5 counts the following subsets:
  {}  {1}      {1,2}    {1,2,3}    {1,2,3,4}  {1,2,3,4,5}
      {2}      {2,3}    {2,3,4}    {2,3,4,5}
      {3}      {3,4}    {3,4,5}
      {4}      {4,5}    {1,2,3,5}
      {5}      {1,2,4}  {1,2,4,5}
      {1,3}    {1,2,5}  {1,3,4,5}
      {1,4}    {1,3,4}
      {1,5}    {1,4,5}
      {2,4}    {2,3,5}
      {2,5}    {2,4,5}
      {3,5}
      {1,3,5}
Triangle begins:
   1
   1   1
   1   2   1
   1   4   2   1
   1   7   5   2   1
   1  12  10   6   2   1
   1  20  20  13   7   2   1
   1  33  38  29  16   8   2   1
   1  54  71  60  39  19   9   2   1
   1  88 130 122  86  50  22  10   2   1
   1 143 235 241 187 116  62  25  11   2   1
   1 232 420 468 392 267 150  75  28  12   2   1
   1 376 744 894 806 588 363 188  89  31  13   2   1

Column k = 1 is A000071.
Row sums are A000079.
Column k = 2 is A001629.
For runs instead of anti-runs we have A034839, for strict partitions A116674.
The case containing n is A053538.
For integer partitions instead of subsets we have A268193, strict A384905.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
A384877 gives lengths of maximal anti-runs in binary indices, firsts A384878.
Cf. A010027, A384177, A384879, A384889, A384890.

Table[Length[Select[Subsets[Range[n]],Length[Split[#,#2!=#1+1&]]==k&]],{n,0,10},{k,0,n}]

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}]

A384905 Triangle read by rows where T(n,k) is the number of strict integer partitions of n with k maximal anti-runs (decreasing by more than 1).

Original entry on oeis.org

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

Offset: 0

Gus Wiseman, Jun 21 2025

			The T(10,2) = 3 strict partitions with 2 maximal anti-runs are: (7,2,1), (5,4,1), (5,3,2).
Triangle begins:
  1
  0  1
  0  1  0
  0  1  1  0
  0  2  0  0  0
  0  2  1  0  0  0
  0  3  0  1  0  0  0
  0  3  2  0  0  0  0  0
  0  4  2  0  0  0  0  0  0
  0  5  2  1  0  0  0  0  0  0
  0  6  3  0  1  0  0  0  0  0  0
  0  7  4  1  0  0  0  0  0  0  0  0
  0  9  3  3  0  0  0  0  0  0  0  0  0

Row sums are A000009.
Column k = 1 is A003114.
For subsets instead of strict integer partitions see A053538, A119900, A210034.
For runs instead of anti-runs we have A116674, for subsets A034839.
This is the strict case of A268193.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
A384877 gives lengths of maximal anti-runs in binary indices, firsts A384878.
Cf. A010027, A384177, A384886, A384889, A384890.

Table[Length[Select[IntegerPartitions[n],UnsameQ@@#&&Length[Split[#,#1!=#2+1&]]==k&]],{n,0,10},{k,0,n}]

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&]&]

A053538 Triangle: a(n,m) = ways to place p balls in n slots with m in the rightmost p slots, 0<=p<=n, 0<=m<=n, summed over p, a(n,m)= Sum_{k=0..n} binomial(k,m)*binomial(n-k,k-m), (see program line).

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 3, 3, 1, 1, 5, 5, 4, 1, 1, 8, 10, 7, 5, 1, 1, 13, 18, 16, 9, 6, 1, 1, 21, 33, 31, 23, 11, 7, 1, 1, 34, 59, 62, 47, 31, 13, 8, 1, 1, 55, 105, 119, 101, 66, 40, 15, 9, 1, 1, 89, 185, 227, 205, 151, 88, 50, 17, 10, 1, 1, 144, 324, 426, 414, 321, 213, 113, 61, 19, 11, 1, 1

Offset: 0

Wouter Meeussen, May 23 2001

Riordan array (1/(1-x-x^2), x(1-x)/(1-x-x^2)). Row sums are A000079. Diagonal sums are A006053(n+2). - Paul Barry, Nov 01 2006
Subtriangle of the triangle given by (0, 1, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, -1, 1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 05 2012
Mirror image of triangle in A208342. - Philippe Deléham, Mar 05 2012
A053538 is jointly generated with A076791 as an array of coefficients of polynomials u(n,x):  initially, u(1,x)=v(1,x)=1, for n>1, u(n,x) = x*u(n-1,x) + v(n-1,x) and v(n,x) = u(n-1,x) + v(n-1,x).  See the Mathematica section at A076791. - Clark Kimberling, Mar 08 2012
The matrix inverse starts
    1;
   -1,   1;
   -1,  -1,   1;
    1,  -2,  -1,  1;
    3,   1,  -3, -1,  1;
    1,   6,   1, -4, -1,  1;
   -7,   4,  10,  1, -5, -1,  1;
  -13, -13,   8, 15,  1, -6, -1,  1;
    3, -31, -23, 13, 21,  1, -7, -1, 1; - R. J. Mathar, Mar 15 2013
Also appears to be the number of subsets of {1..n} containing n with k maximal anti-runs of consecutive elements increasing by more than 1. For example, the subset {1,3,6,7,11,12} has maximal anti-runs ((1,3,6),(7,11),(12)) so is counted under a(12,3). For runs instead of anti-runs we get A202064. - Gus Wiseman, Jun 26 2025

			n=4; Table[binomial[k, j]binomial[n-k, k-j], {k, 0, n}, {j, 0, n}] splits {1, 4, 6, 4, 1} into {{1, 0, 0, 0, 0}, {3, 1, 0, 0, 0}, {1, 4, 1, 0, 0}, {0, 0, 3, 1, 0}, {0, 0, 0, 0, 1}} and this gives summed by columns {5, 5, 4, 1, 1}
Triangle begins :
   1;
   1,  1;
   2,  1,  1;
   3,  3,  1, 1;
   5,  5,  4, 1, 1;
   8, 10,  7, 5, 1, 1;
  13, 18, 16, 9, 6, 1, 1;
...
(0, 1, 1, -1, 0, 0, 0, ...) DELTA (1, 0, -1, 1, 0, 0, 0, ...) begins :
  1;
  0,  1;
  0,  1,  1;
  0,  2,  1,  1;
  0,  3,  3,  1, 1;
  0,  5,  5,  4, 1, 1;
  0,  8, 10,  7, 5, 1, 1;
  0, 13, 18, 16, 9, 6, 1, 1;

Alois P. Heinz, Rows n = 0..140, flattened
R. P. Grimaldi, Extraordinary subsets: a generalization, Fib. Quart., 55 (No. 3, 2017), 114-122. See Table 1.

Column k = 1 is A000045.
Row sums are A000079.
Column k = 2 is A010049.
For runs instead of anti-runs we have A202064.
For integer partitions see A268193, strict A384905, runs A116674.
A034839 counts subsets by number of maximal runs.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
A384877 gives lengths of maximal anti-runs in binary indices, firsts A384878.
A384893 counts subsets by number of maximal anti-runs.
Cf. A000071, A001629, A010027, A208342, A384177, A384879, A384889, A384890.

Flat(List([0..12], n-> List([0..n], k-> Sum([0..n], j->  Binomial(j,k)*Binomial(n-j,j-k)) ))); # G. C. Greubel, May 16 2019

[[(&+[Binomial(j,k)*Binomial(n-j,j-k): j in [0..n]]): k in [0..n]]: n in [0..12]]; // G. C. Greubel, May 16 2019

a:= (n, m)-> add(binomial(k, m)*binomial(n-k, k-m), k=0..n):
seq(seq(a(n,m), m=0..n), n=0..12);  # Alois P. Heinz, Sep 19 2013

Table[Sum[Binomial[k, m]*Binomial[n-k, k-m], {k,0,n}], {n,0,12}, {m,0,n}]

{T(n,k) = sum(j=0,n, binomial(j,k)*binomial(n-j,j-k))}; \\ G. C. Greubel, May 16 2019

[[sum(binomial(j,k)*binomial(n-j,j-k) for j in (0..n)) for k in (0..n)] for n in (0..12)] # G. C. Greubel, May 16 2019

From Philippe Deléham, Mar 05 2012: (Start)
T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k) - T(n-2,k-1), T(0,0) = T(1,0) = T(1,1) = 1 and T(n,k) = 0 if k<0 or if k>n.
G.f.: 1/(1-(1+y)*x-(1-y)*x^2).
Sum_{k, 0<=k<=n} T(n,k)*x^k = A077957(n), A000045(n+1), A000079(n), A001906(n+1), A007070(n), A116415(n), A084326(n+1), A190974(n+1), A190978(n+1), A190984(n+1), A190990(n+1), A190872(n+1) for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 respectively. (End)

A210034 Triangle of coefficients of polynomials v(n,x) jointly generated with A210033; see the Formula section.

Original entry on oeis.org

1, 2, 1, 4, 2, 1, 7, 5, 2, 1, 12, 10, 6, 2, 1, 20, 20, 13, 7, 2, 1, 33, 38, 29, 16, 8, 2, 1, 54, 71, 60, 39, 19, 9, 2, 1, 88, 130, 122, 86, 50, 22, 10, 2, 1, 143, 235, 241, 187, 116, 62, 25, 11, 2, 1, 232, 420, 468, 392, 267, 150, 75, 28, 12, 2, 1, 376, 744, 894, 806

Offset: 1

Clark Kimberling, Mar 16 2012

For a discussion and guide to related arrays, see A208510.
From Gus Wiseman, Jun 29 2025: (Start)
This appears to be the number of subsets of {1..n} with k>0 maximal anti-runs (sequences of consecutive elements increasing by more than 1). For example, the subset {1,2,4,5} has maximal anti-runs ((1),(2,4),(5)) so is counted under T(5,3). Row n = 5 counts the following:
  {1}      {1,2}    {1,2,3}    {1,2,3,4}  {1,2,3,4,5}
  {2}      {2,3}    {2,3,4}    {2,3,4,5}
  {3}      {3,4}    {3,4,5}
  {4}      {4,5}    {1,2,3,5}
  {5}      {1,2,4}  {1,2,4,5}
  {1,3}    {1,2,5}  {1,3,4,5}
  {1,4}    {1,3,4}
  {1,5}    {1,4,5}
  {2,4}    {2,3,5}
  {2,5}    {2,4,5}
  {3,5}
  {1,3,5}
For runs instead of anti-runs we have A034839, with n A202064. For reversed partitions instead of subsets we have A268193. (End)

			First five rows:
  1
  2    1
  4    2    1
  7    5    2   1
  12   10   6   2   1
First three polynomials v(n,x): 1, 2 + x, 4 + 2*x + x^2.

Cf. A210033, A208510.
Column k = 1 is A000071.
Row sums are A000225.
Column k = 2 is A001629.
Column k = 3 is A055243.
The version including k = 0 is A384893.
A034839 counts subsets by number of maximal runs, see also A202023, A202064.
A384175 counts subsets with all distinct lengths of maximal runs, complement A384176.
A384877 gives lengths of maximal anti-runs of binary indices, firsts A384878.
Cf. A053538, A116674, A119900, A268193, A384177, A384879, A384889, A384905.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + v[n - 1, x] + 1;
v[n_, x_] := u[n - 1, x] + x*v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A210033 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A210034 *)

u(n,x)=u(n-1,x)+v(n-1,x)+1,
v(n,x)=u(n-1,x)+x*v(n-1,x)+1,
where u(1,x)=1, v(1,x)=1.

cp's OEIS Frontend

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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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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A384893 Triangle read by rows where T(n,k) is the number of subsets of {1..n} with k maximal anti-runs (increasing by more than 1).

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

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A384905 Triangle read by rows where T(n,k) is the number of strict integer partitions of n with k maximal anti-runs (decreasing by more than 1).

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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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