A101211 -id:A101211 - cp's OEIS frontend

A066099 Triangle read by rows, in which row n lists the compositions of n in reverse lexicographic order.

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

Offset: 1

Alford Arnold, Dec 30 2001

The representation of the compositions (for fixed n) is as lists of parts, the order between individual compositions (for the same n) is (list-)reversed lexicographic; see the example by Omar E. Pol. - Joerg Arndt, Sep 03 2013
This is the standard ordering for compositions in this database; it is similar to the Mathematica ordering for partitions (A080577). Other composition orderings include A124734 (similar to the Abramowitz & Stegun ordering for partitions, A036036), A108244 (similar to the Maple partition ordering, A080576), etc (see crossrefs).
Factorize each term in A057335; sequence records the values of the resulting exponents. It also runs through all possible permutations of multiset digits.
This can be regarded as a table in two ways: with each composition as a row, or with the compositions of each integer as a row. The first way has A000120 as row lengths and A070939 as row sums; the second has A001792 as row lengths and A001788 as row sums. - Franklin T. Adams-Watters, Nov 06 2006
This sequence includes every finite sequence of positive integers. - Franklin T. Adams-Watters, Nov 06 2006
Compositions (or ordered partitions) are also generated in sequence A101211. - Alford Arnold, Dec 12 2006
The equivalent sequence for partitions is A228531. - Omar E. Pol, Sep 03 2013
The sole partition of zero has no components, not a single component of length one. Hence the first nonempty row is row 1. - Franklin T. Adams-Watters, Apr 02 2014 [Edited by Andrey Zabolotskiy, May 19 2018]
See sequence A261300 for another version where the terms of each composition are concatenated to form one single integer: (0, 1, 2, 11, 3, 21, 12, 111,...). This also shows how the terms can be obtained from the binary numbers A007088, cf. Arnold's first Example. - M. F. Hasler, Aug 29 2015
The k-th composition in the list 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 is described as the standard ordering used in the OEIS, although the sister sequence A228351 is also sometimes considered to be canonical. Both sequences define a bijective correspondence between nonnegative integers and integer compositions. - Gus Wiseman, May 19 2020
First differences of A030303 = positions of bits 1 in the concatenation A030190 (= A030302) of numbers written in binary (A007088). - Indices of record values (= first occurrence of n) are given by A005183: a(A005183(n)) = n, cf. FORMULA for more. - M. F. Hasler, Oct 12 2020
The geometric mean approaches the Somos constant (A112302). - Jwalin Bhatt, Feb 10 2025

			A057335 begins 1 2 4 6 8 12 18 30 16 24 36 ... so we can write
  1 2 1 3 2 1 1 4 3 2 2 1 1 1 1 ...
  . . 1 . 1 2 1 . 1 2 1 3 2 1 1 ...
  . . . . . . 1 . . . 1 . 1 2 1 ...
  . . . . . . . . . . . . . . 1 ...
and the columns here gives the rows of the triangle, which begins
  1
  2; 1 1
  3; 2 1; 1 2; 1 1 1
  4; 3 1; 2 2; 2 1 1; 1 3; 1 2 1; 1 1 2; 1 1 1 1
  ...
From _Omar E. Pol_, Sep 03 2013: (Start)
Illustration of initial terms:
  -----------------------------------
  n  j       Diagram   Composition j
  -----------------------------------
  .               _
  1  1           |_|   1;
  .             _ _
  2  1         |  _|   2,
  2  2         |_|_|   1, 1;
  .           _ _ _
  3  1       |    _|   3,
  3  2       |  _|_|   2, 1,
  3  3       | |  _|   1, 2,
  3  4       |_|_|_|   1, 1, 1;
  .         _ _ _ _
  4  1     |      _|   4,
  4  2     |    _|_|   3, 1,
  4  3     |   |  _|   2, 2,
  4  4     |  _|_|_|   2, 1, 1,
  4  5     | |    _|   1, 3,
  4  6     | |  _|_|   1, 2, 1,
  4  7     | | |  _|   1, 1, 2,
  4  8     |_|_|_|_|   1, 1, 1, 1;
(End)

Franklin T. Adams-Watters, Table of n, a(n) for n = 1..5120 (through compositions of 10)
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
Gus Wiseman, Statistics, classes, and transformations of standard compositions

Lists of compositions of integers: this sequence (reverse lexicographic order; minus one gives A108730), A228351 (reverse colexicographic order - every composition is reversed; minus one gives A163510), A228369 (lexicographic), A228525 (colexicographic), A124734 (length, then lexicographic; minus one gives A124735), A296774 (length, then reverse lexicographic), A337243 (length, then colexicographic), A337259 (length, then reverse colexicographic), A296773 (decreasing length, then lexicographic), A296772 (decreasing length, then reverse lexicographic), A337260 (decreasing length, then colexicographic), A108244 (decreasing length, then reverse colexicographic), also A101211 and A227736 (run lengths of bits).
Cf. row length and row sums for different splittings into rows: A000120, A070939, A001792, A001788.
Cf. A228531, A096903, A065120, A057335, A055932, A005811, A261300, A007088.
Cf. lists of partitions of integers, or multisets of integers: A026791 and crosserfs therein, A112798 and crossrefs therein.
See link for additional crossrefs pertaining to standard compositions.
A related ranking of finite sets is A048793/A272020.
Cf. A005183, A030190, A030302, A030303, A035327, A036036, A080576, A080577, A106356, A112302, A238279, A333219.

a066099 = (!!) a066099_list
a066099_list = concat a066099_tabf
a066099_tabf = map a066099_row [1..]
a066099_row n = reverse $ a228351_row n
-- (each composition as a row)
-- Peter Kagey, Aug 25 2016

Table[FactorInteger[Apply[Times, Map[Prime, Accumulate @ IntegerDigits[n, 2]]]][[All, -1]], {n, 41}] // Flatten (* Michael De Vlieger, Jul 11 2017 *)
stc[n_] := Differences[Prepend[Join @@ Position[Reverse[IntegerDigits[n, 2]], 1], 0]] // Reverse;
Table[stc[n], {n, 0, 20}] // Flatten (* Gus Wiseman, May 19 2020 *)
Table[Reverse @ LexicographicSort @ Flatten[Permutations /@ Partitions[n], 1], {n, 10}] // Flatten (* Eric W. Weisstein, Jun 26 2023 *)

arow(n) = {local(v=vector(n),j=0,k=0);
   while(n>0,k++; if(n%2==1,v[j++]=k;k=0);n\=2);
   vector(j,i,v[j-i+1])} \\ returns empty for n=0. - Franklin T. Adams-Watters, Apr 02 2014

from itertools import islice
from itertools import accumulate, count, groupby, islice
def A066099_gen():
    for i in count(1):
        yield [len(list(g)) for _,g in groupby(accumulate(int(b) for b in bin(i)[2:]))]
A066099 = list(islice(A066099_gen(), 120))  # Jwalin Bhatt, Feb 28 2025

def a_row(n): return list(reversed(Compositions(n)))
flatten([a_row(n) for n in range(1,6)]) # Peter Luschny, May 19 2018

From M. F. Hasler, Oct 12 2020: (Start)
a(n) = A030303(n+1) - A030303(n).
a(A005183(n)) = n; a(A005183(n)+1) = n-1 (n>1); a(A005183(n)+2) = 1. (End)

Edited with additional terms by Franklin T. Adams-Watters, Nov 06 2006

0th row removed by Andrey Zabolotskiy, May 19 2018

A294175 a(n) = 2^(n-1) + ((1+(-1)^n)/4)*binomial(n, n/2) - binomial(n, floor(n/2)).

Original entry on oeis.org

0, 0, 1, 1, 5, 6, 22, 29, 93, 130, 386, 562, 1586, 2380, 6476, 9949, 26333, 41226, 106762, 169766, 431910, 695860, 1744436, 2842226, 7036530, 11576916, 28354132, 47050564, 114159428, 190876696, 459312152, 773201629, 1846943453, 3128164186, 7423131482

Offset: 0

Enrique Navarrete, Feb 10 2018

nonn

Number of subsets of {1,2,...,n} that contain more even than odd numbers.
Note that A058622 counts the nonempty subsets of {1,2,...,n} that contain more odd than even numbers.
From Gus Wiseman, Jul 22 2021: (Start)
Also the number of integer compositions of n + 1 with alternating sum < 0, where the alternating sum of a sequence (y_1,...,y_k) is Sum_i (-1)^(i-1) y_i. For example, the a(0) = 0 through a(6) = 6 compositions (empty columns indicated by dots) are:
  .  .  (12)  (13)  (14)    (15)
                    (23)    (24)
                    (131)   (141)
                    (1112)  (1113)
                    (1211)  (1212)
                            (1311)
Also the number of integer compositions of n + 1 with reverse-alternating sum < 0. For a bijection, keep the odd-length compositions and reverse the even-length ones.
Also the number of (n+1)-digit binary numbers with more 0's than 1's. For example, the a(0) = 0 through a(5) = 6 binary numbers are:
  .  .  100  1000  10000  100000
                   10001  100001
                   10010  100010
                   10100  100100
                   11000  101000
                          110000
(End)
2*a(n) is the number of all-positive pinnacle sets that are admissible in the group S_{n+1}^B of signed permutations, but not admissible in S_{n+1}. - Bridget Tenner, Jan 06 2023

			For example, for n=5, a(5)=6 and the 6 subsets are {2}, {4}, {2,4}, {1,2,4}, {2,3,4}, {2,4,5}.

Nicolle González, Pamela E. Harris, Gordon Rojas Kirby, Mariana Smit Vega Garcia, and Bridget Eileen Tenner, Pinnacle sets of signed permutations, arXiv:2301.02628 [math.CO] (2023).

The even bisection is A000346.
The odd bisection is A008549.
The following relate to compositions of n + 1 with alternating sum k < 0.
- The k = 1 version is A000984, ranked by A345909/A345911.
- The opposite (k > 0) version is A027306, ranked by A345917/A345918.
- The weak (k <= 0) version A058622, ranked by A345915/A345916.
- The k != 0 version is also A058622, ranked by A345921.
- The complement (k >= 0) is counted by A116406, ranked by A345913/A345914.
- The k = 0 version is A138364, ranked by A344619.
- The unordered version is A344608, ranked by A119899.
- Ranked by A345919 (reverse: A345920).
A097805 counts compositions by alternating (or reverse-alternating) sum.
A101211 lists run-lengths in binary expansion (reverse: A227736).
A103919 counts partitions by sum and alternating sum (reverse: A344612).
A345197 counts compositions by length and alternating sum.
Cf. A000070, A001700, A007318, A025047, A032443, A034871, A106356, A114121, A126869, A163493, A344743, A345908, A289871, A359066, A359067.

f:= gfun:-rectoproc({(8+8*n)*a(n)+(4*n+16)*a(1+n)+(-20-6*n)*a(n+2)+(-5-n)*a(n+3)+(5+n)*a(n+4), a(0) = 0, a(1) = 0, a(2) = 1, a(3) = 1}, a(n), remember):
map(f, [$0..40]); # Robert Israel, Feb 12 2018

f[n_] := 2^(n - 1) + ((1 + (-1)^n)/4) Binomial[n, n/2] - Binomial[n, Floor[n/2]]; Array[f, 38, 0] (* Robert G. Wilson v, Feb 10 2018 *)
Table[Length[Select[Tuples[{0,1},{n+1}],First[#]==1&&Count[#,0]>Count[#,1]&]],{n,0,10}] (* Gus Wiseman, Jul 22 2021 *)

From Robert Israel, Feb 12 2018: (Start)
G.f.: (x+1)*sqrt(1-4*x^2)/(2*x*(4*x^2-1))+(x-1)/(2*(2*x-1)*x).
D-finite with recurrence: (8+8*n)*a(n)+(4*n+16)*a(1+n)+(-20-6*n)*a(n+2)+(-5-n)*a(n+3)+(5+n)*a(n+4) = 0. (End)

A228369 Triangle read by rows in which row n lists the compositions (ordered partitions) of n in lexicographic order.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 3, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 3, 2, 1, 1, 2, 2, 3, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 3, 1, 2, 1, 1, 1, 2, 2, 1, 3, 1, 1, 4, 2, 1, 1, 1, 2, 1, 2, 2, 2, 1, 2, 3, 3, 1, 1, 3, 2, 4, 1, 5, 1, 1, 1, 1, 1, 1

Offset: 1

Omar E. Pol, Aug 28 2013

The representation of the compositions (for fixed n) is as lists of parts, the order between individual compositions (for the same n) is lexicographic. - Joerg Arndt, Sep 02 2013
The equivalent sequence for partitions is A026791.
Row n has length A001792(n-1).
Row sums give A001787, n >= 1.
The m-th composition has length A008687(m+1), m >= 1. - Andrey Zabolotskiy, Jul 19 2017

			Illustration of initial terms:
-----------------------------------
n  j       Diagram   Composition j
-----------------------------------
.               _
1  1           |_|   1;
.             _ _
2  1         | |_|   1, 1,
2  2         |_ _|   2;
.           _ _ _
3  1       | | |_|   1, 1, 1,
3  2       | |_ _|   1, 2,
3  3       |   |_|   2, 1,
3  4       |_ _ _|   3;
.         _ _ _ _
4  1     | | | |_|   1, 1, 1, 1,
4  2     | | |_ _|   1, 1, 2,
4  3     | |   |_|   1, 2, 1,
4  4     | |_ _ _|   1, 3,
4  5     |   | |_|   2, 1, 1,
4  6     |   |_ _|   2, 2,
4  7     |     |_|   3, 1,
4  8     |_ _ _ _|   4;
.
Triangle begins:
[1];
[1,1],[2];
[1,1,1],[1,2],[2,1],[3];
[1,1,1,1],[1,1,2],[1,2,1],[1,3],[2,1,1],[2,2],[3,1],[4];
[1,1,1,1,1],[1,1,1,2],[1,1,2,1],[1,1,3],[1,2,1,1],[1,2,2],[1,3,1],[1,4],[2,1,1,1],[2,1,2],[2,2,1],[2,3],[3,1,1],[3,2],[4,1],[5];
...

Joerg Arndt, Table of n, a(n) for n = 1..10000

Cf. A001511, A026791, A066099, A101211, A124734, A228351, A228525, A281013.

a228369 n = a228369_list !! (n - 1)
a228369_list = concatMap a228369_row [1..]
a228369_row 0 = []
a228369_row n
  | 2^k == 2 * n + 2 = [k - 1]
  | otherwise        = a228369_row (n `div` 2^k) ++ [k] where
    k = a007814 (n + 1) + 1
-- Peter Kagey, Jun 27 2016

Table[Sort[Join@@Permutations/@IntegerPartitions[n],OrderedQ[PadRight[{#1,#2}]]&],{n,5}] (* Gus Wiseman, Dec 14 2017 *)

gen_comp(n)=
{  /* Generate compositions of n as lists of parts (order is lex): */
    my(ct = 0);
    my(m, z, pt);
    \\ init:
    my( a = vector(n, j, 1) );
    m = n;
    while ( 1,
        ct += 1;
        pt = vector(m, j, a[j]);
        /* for A228369  print composition: */
        for (j=1, m, print1(pt[j],", ") );
\\        /* for A228525 print reversed (order is colex): */
\\        forstep (j=m, 1, -1, print1(pt[j],", ") );
        if ( m<=1,  return(ct) );  \\ current is last
        a[m-1] += 1;
        z = a[m] - 2;
        a[m] = 1;
        m += z;
    );
    return(ct);
}
for(n=1, 12, gen_comp(n) );
\\ Joerg Arndt, Sep 02 2013

a = [[[]], [[1]]]
for s in range(2, 9):
    a.append([])
    for k in range(1, s+1):
        for ss in a[s-k]:
            a[-1].append([k]+ss)
print(a)
# Andrey Zabolotskiy, Jul 19 2017

A043276 a(n) = maximal run length in base-2 representation of n.

Original entry on oeis.org

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

Offset: 1

Clark Kimberling

First occurrence of k is when n=2^k-1 and there is no last occurrence. - Robert G. Wilson v, Dec 14 2008
Sequences A000975, A037969, A037970, A037971 list numbers for which a(n)=1, a(n)=2, a(n)=3, a(n)=4. - M. F. Hasler, Jul 23 2013
a(n) = max(A101211(n,k): k = 1..A005811(n)). - Reinhard Zumkeller, Dec 16 2013

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

Cf. A043277-A043290 for base-3 to base-16 analogs.

a043276 = maximum . a101211_row  -- Reinhard Zumkeller, Dec 16 2013

A043276 := proc(n)
    local a,rl,i ;
    if n > 0 then
        rl := 1 ;
    else
        rl := 0 ;
    end if;
    a := rl ;
    dgs := convert(n,base,2) ;
    for i from 2 to nops(dgs) do
        if op(i,dgs) = op(i-1,dgs) then
            rl := rl+1 ;
            a := max(a,rl) ;
        else
            a := max(a,rl) ;
            rl := 1;
        end if;
    end do:
    a ;
end proc:
seq(A043276(n),n=1...80) ; # R. J. Mathar, Jun 04 2021

f[n_] := Max @@ Length /@ Split@IntegerDigits[n, 2]; Array[f, 105] (* Robert G. Wilson v, Dec 14 2008 *)

A043276(n,b=2)={my(m,c=1);while(n>0,n%b==(n\=b)%b && c++ && next;m=max(m,c);c=1);m} \\ M. F. Hasler, Jul 23 2013

a(n)=my(r,t); while(n, t=valuation(n,2); if(t>r, r=t); n>>=t; t=valuation(n+1,2); if(t>r, r=t); n>>=t); r \\ Charles R Greathouse IV, Nov 02 2016

from itertools import groupby
def A043276(n): return max(len(list(g)) for k, g in groupby(bin(n)[2:])) # Chai Wah Wu, Mar 09 2023

More terms from Robert G. Wilson v, Dec 14 2008

A296774 Triangle read by rows in which row n lists the compositions of n ordered first by length and then reverse-lexicographically.

Original entry on oeis.org

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

Offset: 1

Gus Wiseman, Dec 20 2017

			Triangle of compositions begins:
(1),
(2),(11),
(3),(21),(12),(111),
(4),(31),(22),(13),(211),(121),(112),(1111),
(5),(41),(32),(23),(14),(311),(221),(212),(131),(122),(113),(2111),(1211),(1121),(1112),(11111).

Cf. A066099, A101211, A108730, A124734, A228369, A281013, A294859, A296302, A296656, A296772, A296773.

Table[Sort[Join@@Permutations/@IntegerPartitions[n],Or[Length[#1]

A039004 Numbers whose base-4 representation has the same number of 1's and 2's.

Original entry on oeis.org

0, 3, 6, 9, 12, 15, 18, 24, 27, 30, 33, 36, 39, 45, 48, 51, 54, 57, 60, 63, 66, 72, 75, 78, 90, 96, 99, 102, 105, 108, 111, 114, 120, 123, 126, 129, 132, 135, 141, 144, 147, 150, 153, 156, 159, 165, 177, 180, 183, 189, 192, 195, 198, 201, 204, 207, 210, 216, 219

Offset: 1

Olivier Gérard

Numbers such that sum (-1)^k*b(k) = 0 where b(k)=k-th binary digit of n (see A065359). - Benoit Cloitre, Nov 18 2003
Conjecture: a(C(2n,n)-1) = 4^n - 1. (A000984 is C(2n,n)). - Gerald McGarvey, Nov 18 2007
From Russell Jay Hendel, Jun 23 2015: (Start)
We prove the McGarvey conjecture (A) a(e(n,n)-1) = 4^n-1, with e(n,m) = A034870(n,m) = binomial(2n,m), the even rows of Pascal's triangle. By the comment from Hendel in A034870, we have the function s(n,k) = #{n-digit, base-4 numbers with n-k more 1-digits than 2-digits}. As shown in A034870, (B) #s(n,k)= e(n,k) with # indicating cardinality, that is, e(n,k) = binomial(2n,k) gives the number of n-digit, base-4 numbers with n-k more 1-digits than 2-digits.
We now show that (B) implies (A). By definition, s(n,n) contains the e(n,n) = binomial(2n,n) numbers with an equal number of 1-digits and 2-digits. The biggest n-digit, base-4 number is 333...3 (n copies of 3). Since 333...33 has zero 1-digits and zero 2-digits it follows that 333...333 is a member of s(n,n) and hence it is the biggest member of s(n,n). But 333...333 (n copies of 3) in base 4 has value 4^n-1. Since A039004 starts with index 0 (that is, 0 is the 0th member of A039004), it immediately follows that 4^n-1 is the (e(n,n)-1)st member of A039004, proving the McGarvey conjecture. (End)
Also numbers whose alternating sum of binary expansion is 0, i.e., positions of zeros in A345927. These are numbers whose binary expansion has the same number of 1's at even positions as at odd positions. - Gus Wiseman, Jul 28 2021

Alois P. Heinz, Table of n, a(n) for n = 1..10000

Cf. A139370, A139371, A139372, A139373.
Cf. A139351, A139352, A139353, A139354, A139355.
A subset of A001969 (evil numbers).
A base-2 version is A031443 (digitally balanced numbers).
Positions of 0's in A065359 and A345927.
Positions of first appearances are A086893.
The version for standard compositions is A344619.
A000120 and A080791 count binary digits, with difference A145037.
A003714 lists numbers with no successive binary indices.
A011782 counts compositions.
A030190 gives the binary expansion of each nonnegative integer.
A070939 gives the length of an integer's binary expansion.
A097805 counts compositions by alternating (or reverse-alternating) sum.
A101211 lists run-lengths in binary expansion:
- row-lengths: A069010
- reverse: A227736
- ones only: A245563
A138364 counts compositions with alternating sum 0:
- bisection: A001700/A088218
- complement: A058622
A328594 lists numbers whose binary expansion is aperiodic.
A345197 counts compositions by length and alternating sum.
Cf. A000302, A000346, A003242, A029837, A053738, A053754, A071321, A103919, A191232, A328596.

Fortran
```
c See link in A139351.
```

N:= 1000: # to get all terms up to N, which should be divisible by 4
B:= Array(0..N-1):
d:= ceil(log[4](N));
S:= Array(0..N-1,[seq(op([0,1,-1,0]),i=1..N/4)]):
for i from 1 to d do
  B:= B + S;
  S:= Array(0..N-1,i-> S[floor(i/4)]);
od:
select(t -> B[t]=0, [$0..N-1]); # Robert Israel, Jun 24 2015

ats[y_]:=Sum[(-1)^(i-1)*y[[i]],{i,Length[y]}];
Select[Range[0,100],ats[IntegerDigits[#,2]]==0&] (* Gus Wiseman, Jul 28 2021 *)

for(n=0,219,if(sum(i=1,length(binary(n)),(-1)^i*component(binary(n),i))==0,print1(n,",")))

Conjecture: there is a constant c around 5 such that a(n) is asymptotic to c*n. - Benoit Cloitre, Nov 24 2002

That conjecture is false. The number of members of the sequence from 0 to 4^d-1 is binomial(2d,d) which by Stirling's formula is asymptotic to 4^d/sqrt(Pi*d). If Cloitre's conjecture were true we would have 4^d-1 asymptotic to c*4^d/sqrt(Pi*d), a contradiction. - Robert Israel, Jun 24 2015

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)).

A125106 Enumeration of partitions by binary representation: each 1 is a part; the part size is 1 more than the number of 0's in the rest of the number.

Original entry on oeis.org

1, 2, 1, 1, 3, 2, 1, 2, 2, 1, 1, 1, 4, 3, 1, 3, 2, 2, 1, 1, 3, 3, 2, 2, 1, 2, 2, 2, 1, 1, 1, 1, 5, 4, 1, 4, 2, 3, 1, 1, 4, 3, 3, 2, 1, 3, 2, 2, 2, 1, 1, 1, 4, 4, 3, 3, 1, 3, 3, 2, 2, 2, 1, 1, 3, 3, 3, 2, 2, 2, 1, 2, 2, 2, 2, 1, 1, 1, 1, 1

Offset: 1

Alford Arnold, Dec 10 2006

Another way to describe this: starting with the binary representation and a counter set at one, count the 0's from right to left. Write a term equal to the counter for each "1" encountered.
A101211 is a similar sequence, with A005811 elements per row which maps natural numbers to compositions (ordered partitions).
There are two ways to consider this as a table: taking each partition as a row, or taking the partitions generated by 2^(n-1) through 2^n-1 as a row.
Taking the n-th row as multiple partitions, it consists of those partitions with the first hook size (largest part plus number of parts minus 1) equal to n. The number of integers in this n-th row is A001792(n-1), and the row sum is A049611.
Taking each partition as a separate row, the row lengths are A000120, and the row sums are A161511.
Heinz numbers of the rows are A005940. - Gus Wiseman, Jan 17 2023

			Row 4:
1000 [4]
1001 [3,1]
1010 [3,2]
1011 [2,1,1]
1100 [3,3]
1101 [2,2,1]
1110 [2,2,2]
1111 [1,1,1,1]

Alois P. Heinz, Rows n = 1..12, flattened

Cf. A000041, A005811, A037016, A101211, A001792, A049611, A126411.
Each partition as row: A000120 (row widths), A161511 (row sums), A243499 (row products).
Cf. A005940. - Franklin T. Adams-Watters, Mar 06 2010
Lasts are A001511.
Firsts are A008687.
Cf. A019565, A029837, A029931, A048793, A059893, A066099, A070939, A242628.

b:= proc(n) local c, l, m; l:=[][]; m:= n; c:=1;
      while m>0 do if irem(m, 2, 'm')=0 then c:= c+1
         else l:= c, l fi
      od; l
    end:
T:= n-> seq(b(i), i=2^(n-1)..2^n-1):
seq(T(n), n=1..7);  # Alois P. Heinz, Sep 25 2015

f[k_] := (bits = IntegerDigits[k, 2]; zerosCount = Reverse[ Accumulate[ 1-Reverse[bits] ] ] + 1; Select[ Transpose[ {bits, zerosCount} ], First[#] == 1 & ][[All, 2]]); row[n_] := Table[ f[k], {k, 2^(n-1), 2^n-1}]; Flatten[ Table[ row[n], {n, 1, 5}]] (* Jean-François Alcover, Jan 24 2012 *)
scc[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
Table[Reverse[scc[n]-Range[Length[scc[n]]]+1],{n,0,20}] (* Gus Wiseman, Jan 17 2023 *)

Partition 2n is partition n with every part size increased by 1; partition 2n+1 is partition n with an additional part of size 1.

T(n,k) = A272020(n,k) - A000120(n) + k. - Gus Wiseman, Jan 17 2023

Edited by Franklin T. Adams-Watters, Jun 11 2009

A167489 Product of run lengths in binary representation of n.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 2, 3, 3, 2, 1, 2, 4, 2, 3, 4, 4, 3, 2, 4, 2, 1, 2, 3, 6, 4, 2, 4, 6, 3, 4, 5, 5, 4, 3, 6, 4, 2, 4, 6, 3, 2, 1, 2, 4, 2, 3, 4, 8, 6, 4, 8, 4, 2, 4, 6, 9, 6, 3, 6, 8, 4, 5, 6, 6, 5, 4, 8, 6, 3, 6, 9, 6, 4, 2, 4, 8, 4, 6, 8, 4, 3, 2, 4, 2, 1, 2, 3, 6, 4, 2, 4, 6, 3, 4, 5, 10, 8, 6, 12, 8, 4, 8

Offset: 0

Andrew Weimholt, Nov 05 2009

			a(56) = 9, because 56 in binary is written 111000 giving the run lengths 3,3 and 3x3 = 9.
a(99) = 12, because 99 in binary is written 1100011 giving the run lengths 2,3,2, and 2x3x2 = 12.

Antti Karttunen, Table of n, a(n) for n = 0..8192
Index entries for sequences related to binary expansion of n

Row products of A101211 and A227736 (for n > 0).
Cf. A167490 (smallest number with binary run length product = n).
Cf. A167491 (members of A167490 sorted in ascending order).
Cf. A227355, A227184, A227190.
Cf. A227349, A227350, A227352, A246588, A284558, A284559, A284582, A284583.
Differs from similar A284579 for the first time at n=56, where a(56) = 9, while A284579(56) = 5.

import Data.List (group)
a167489 = product . map length . group . a030308_row
-- Reinhard Zumkeller, Jul 05 2013

Table[ Times @@ (Length /@ Split[IntegerDigits[n, 2]]), {n, 0, 100}](* Olivier Gérard, Jul 05 2013 *)

a(n) = {my(p=1, b=n%2, i=0); while(n!=0, n=n>>1; i=i+1; if((n%2)!=b, p=p*i; i=0; b=n%2)); p} \\ Indranil Ghosh, Apr 17 2017, after the Python Program by Antti Karttunen

def A167489(n):
  '''Product of run lengths in binary representation of n.'''
  p = 1
  b = n%2
  i = 0
  while (n != 0):
    n >>= 1
    i += 1
    if ((n%2) != b):
      p *= i
      i = 0
      b = n%2
  return(p)
# Antti Karttunen, Jul 24 2013 (Cf. Python program for A227184).

(define (A167489 n) (apply * (binexp->runcount1list n)))
(define (binexp->runcount1list n) (if (zero? n) (list) (let loop ((n n) (rc (list)) (count 0) (prev-bit (modulo n 2))) (if (zero? n) (cons count rc) (if (eq? (modulo n 2) prev-bit) (loop (floor->exact (/ n 2)) rc (1+ count) (modulo n 2)) (loop (floor->exact (/ n 2)) (cons count rc) 1 (modulo n 2)))))))
;; Antti Karttunen, Jul 05 2013

a(n) = A227349(n) * A227350(n) = A227355(A227352(2n+1)). - Antti Karttunen, Jul 25 2013

a(n) = A284558(n) * A284559(n) = A284582(n) * A284583(n). - Antti Karttunen, Apr 16 2017

A140690 A positive integer n is included if n written in binary can be subdivided into a number of runs all of equal-length, the first run from the left consisting of all 1's, the next run consisting of all 0's, the next run consisting of all 1's, the next run consisting of all 0's, etc.

Original entry on oeis.org

1, 2, 3, 5, 7, 10, 12, 15, 21, 31, 42, 51, 56, 63, 85, 127, 170, 204, 240, 255, 341, 455, 511, 682, 819, 992, 1023, 1365, 2047, 2730, 3276, 3640, 3855, 4032, 4095, 5461, 8191, 10922, 13107, 16256, 16383, 21845, 29127, 31775, 32767, 43690, 52428, 61680

Offset: 1

Leroy Quet, Jul 11 2008

Also: numbers of the form (2^s-1)*[4^{s*(k+1)}-1]/(4^s-1) or 2^s(2^s-1)*[4^{s*(k+1)}-1]/(4^s-1), s>=1, k>=0. Subsequences are, with the possible exception of terms at n=0, A002450(n), A043291(n), A015565(2n), A093134(2n+1), A000225(n), A020522(n). [R. J. Mathar, Aug 04 2008]
From Emeric Deutsch, Jan 25 2018: (Start)
Also the indices of the compositions having equal parts.
We define the index of a composition to be the positive integer whose binary form has run-lengths (i.e. runs of 1's, runs of 0's, etc., from left to right) equal to the parts of the composition. Example: the composition [1,1,3,1] has index 46 since the binary form of 46 is 101110. The integer 992 is in the sequence since its binary form is 1111100000 and the composition [5,5] has equal parts. The integer 100 is not in the sequence since its binary form is 1100100 and the composition [2,2,1,2] does not have equal parts.
The command c(n) from the Maple program yields the composition having index n. (End)

			819 in binary is 1100110011. The runs of 0's and 1's are (11)(00)(11)(00)(11). Each run (alternating 1's and 0's) is the same length. So 819 is in the sequence.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for sequences related to binary expansion of n

Cf. A101211, A298644.

import Data.Set (singleton, deleteFindMin, insert)
a140690 n = a140690_list !! (n-1)
a140690_list = f $ singleton (1, 1, 2) where
   f s | k == 1 = m : f (insert (2*b-1, 1, 2*b) $ insert (b*m, k+1, b) s')
       | even k    = m : f (insert (b*m+b-1, k+1, b) s')
       | otherwise = m : f (insert (b*m, k+1, b) s')
       where ((m, k, b), s') = deleteFindMin s
-- Reinhard Zumkeller, Feb 21 2014

Runs := proc (L) local j, r, i, k: j := 1: r[j] := L[1]: for i from 2 to nops(L) do if L[i] = L[i-1] then r[j] := r[j], L[i] else j := j+1: r[j] := L[i] end if end do: [seq([r[k]], k = 1 .. j)] end proc: RunLengths := proc (L) map(nops, Runs(L)) end proc: c := proc (n) ListTools:-Reverse(convert(n, base, 2)): RunLengths(%) end proc: A := {}: for n to 62000 do if nops(convert(c(n), set)) = 1 then A := `union`(A, {n}) else  end if end do: A; # most of the Maple program is due to W. Edwin Clark. - Emeric Deutsch, Jan 25 2018

Select[Range[62000],Length[Union[Length/@Split[IntegerDigits[#,2]]]]==1&] (* Harvey P. Dale, Mar 22 2012 *)

Terms beyond 42 from R. J. Mathar, Aug 04 2008

cp's OEIS Frontend

A066099 Triangle read by rows, in which row n lists the compositions of n in reverse lexicographic order.

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A294175 a(n) = 2^(n-1) + ((1+(-1)^n)/4)*binomial(n, n/2) - binomial(n, floor(n/2)).

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A228369 Triangle read by rows in which row n lists the compositions (ordered partitions) of n in lexicographic order.

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A043276 a(n) = maximal run length in base-2 representation of n.

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A296774 Triangle read by rows in which row n lists the compositions of n ordered first by length and then reverse-lexicographically.

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A039004 Numbers whose base-4 representation has the same number of 1's and 2's.

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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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