A227737 -id:A227737 - cp's OEIS frontend

A101211 Triangle read by rows: n-th row is length of run of leftmost 1's, followed by length of run of 0's, followed by length of run of 1's, etc., in the binary representation of n.

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

Offset: 1

Leroy Quet, Dec 13 2004

Row n has A005811(n) elements. In rows 2^(k-1)..2^k-1 we have all the compositions (ordered partitions) of k. Other orderings of compositions: A066099, A108244, and A124734. - Jason Kimberley, Feb 09 2013
A043276(n) = largest term in n-th row. - Reinhard Zumkeller, Dec 16 2013
From the first comment it follows that we have a bijection between the positive integers and the set of all compositions. - Emeric Deutsch, Jul 11 2017
From Robert Israel, Jan 23 2018: (Start)
If n is even, row 2*n is row n with its last element incremented by 1, and row 2*n+1 is row n with 1 appended.
If n is odd, row 2*n+1 is row n with its last element incremented by 1, and row 2*n is row n with 1 appended. (End)

			Since 9 is 1001 in binary, the 9th row is 1,2,1.
Since 11 is 1011 in binary, the 11th row is 1,1,2.
Triangle begins:
  1;
  1,1;
  2;
  1,2;
  1,1,1;
  2,1;
  3;
  1,3;

Antti Karttunen, The rows 1..1023 of the table, flattened

A070939(n) gives the sum of terms in row n, while A167489(n) gives the product of its terms. A090996 gives the first column. A227736 lists the terms of each row in reverse order.
Cf. also A227186.
Cf. A030308, A175911.
Cf. A318927 (concatenation of each row), A318926 (concatenations of reversed rows).
Cf. A382255 (Heinz numbers of the rows: Product_k prime(T(n,k))).

import Data.List (group)
a101211 n k = a101211_tabf !! (n-1) !! (k-1)
a101211_row n = a101211_tabf !! (n-1)
a101211_tabf = map (reverse . map length . group) $ tail a030308_tabf
-- Reinhard Zumkeller, Dec 16 2013

# Maple program due to W. Edwin Clark:
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: # Row n is obtained with the command c(n). - Emeric Deutsch, Jul 03 2017
# Maple program due to W. Edwin Clark, yielding the integer ind corresponding to a given composition (the index of the composition):
ind := proc (x) local X, j, i: X := NULL: for j to nops(x) do if type(j, odd) then X := X, seq(1, i = 1 .. x[j]) end if: if type(j, even) then X := X, seq(0, i = 1 .. x[j]) end if end do: X := [X]: add(X[i]*2^(nops(X)-i), i = 1 .. nops(X)) end proc; # Clearly, ind(c(n))= n. - Emeric Deutsch, Jan 23 2018

Table[Length /@ Split@ IntegerDigits[n, 2], {n, 38}] // Flatten (* Michael De Vlieger, Jul 11 2017 *)

apply( {A101211_row(n)=Vecrev((n=vecextract([-1..exponent(n)], bitxor(2*n, bitor(n,1))))[^1]-n[^-1])}, [1..19]) \\ replacing older code by M. F. Hasler, Mar 24 2025

from itertools import groupby
def arow(n): return [len(list(g)) for k, g in groupby(bin(n)[2:])]
def auptorow(rows):
    alst = []
    for i in range(1, rows+1): alst.extend(arow(i))
    return alst
print(auptorow(38)) # Michael S. Branicky, Oct 02 2021

a(n) = A227736(A227741(n)) = A227186(A056539(A227737(n)),A227740(n)) - Antti Karttunen, Jul 27 2013

More terms from Emeric Deutsch, Apr 12 2005

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)).

A227739 Irregular table where row n lists in nondecreasing order the parts of unordered partition encoded in the runlengths of binary expansion of n; nonzero terms of A227189.

Original entry on oeis.org

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

Offset: 1

Antti Karttunen, Jul 25 2013

Row n has A005811(n) elements. Each row contains a unique (unordered) partition of some integer, and all possible partitions of finite natural numbers eventually occur. The first partition that sums to k occurs at row A227368(k) and the last at row A000225(k).

Other similar tables of unordered partitions: A036036, A036037, A080576, A080577 and A112798.

			Rows are constructed as:
  Row    n in   Runlengths  With one     Partial sums   The row sums
   n    binary  collected   subtracted   of which give  to, i.e. is
                from lsb-   from all     terms on       a partition of
                to msb-end  except 1st   that row       of A227183(n)
   1       "1"        [1]        [1]     1;             1
   2      "10"      [1,1]      [1,0]     1, 1;          2
   3      "11"        [2]        [2]     2;             2
   4     "100"      [2,1]      [2,0]     2, 2;          4
   5     "101"    [1,1,1]    [1,0,0]     1, 1, 1;       3
   6     "110"      [1,2]      [1,1]     1, 2;          3
   7     "111"        [3]        [3]     3;             3
   8    "1000"      [3,1]      [3,0]     3, 3;          6
   9    "1001"    [1,2,1]    [1,1,0]     1, 2, 2;       5
  10    "1010"  [1,1,1,1]  [1,0,0,0]     1, 1, 1, 1;    4
  11    "1011"    [2,1,1]    [2,0,0]     2, 2, 2;       6
  12    "1100"      [2,2]      [2,1]     2, 3;          5
  13    "1101"    [1,1,2]    [1,0,1]     1, 1, 2;       4
  14    "1110"      [1,3]      [1,2]     1, 3;          4
  15    "1111"        [4]        [4]     4;             4
  16   "10000"      [4,1]      [4,0]     4, 4;          8

Antti Karttunen, The rows 1..1023 of the table, flattened

Row sums: A227183, row products: A227184, the initial (smallest) term of each row: A136480, the last (largest) term: A227185.

Cf. also A227189, A227738, A227736.

Table[Function[b, Accumulate@ Prepend[If[Length@ b > 1, Rest[b] - 1, {}], First@ b]]@ Map[Length, Split@ Reverse@ IntegerDigits[n, 2]], {n, 34}] // Flatten (* Michael De Vlieger, May 09 2017 *)

(define (A227739 n) (A227189bi (A227737 n) (A227740 n))) ;; The Scheme-code for A227189bi has been given in A227189.

a(n) = A227189(A227737(n),A227740(n)).

A173318 Partial sums of A005811.

Original entry on oeis.org

0, 1, 3, 4, 6, 9, 11, 12, 14, 17, 21, 24, 26, 29, 31, 32, 34, 37, 41, 44, 48, 53, 57, 60, 62, 65, 69, 72, 74, 77, 79, 80, 82, 85, 89, 92, 96, 101, 105, 108, 112, 117, 123, 128, 132, 137, 141, 144, 146, 149, 153, 156, 160, 165, 169, 172, 174, 177, 181, 184, 186, 189

Offset: 0

Jonathan Vos Post, Feb 16 2010

Partial sums of number of runs in binary expansion of n (n>0). Partial sums of number of 1's in Gray code for n. The subsequence of squares in this partial sum begins 0, 1, 4, 9, 144, 169, 256, 289, 324 (since we also have 32 and 128 I wonder about why so many powers). The subsequence of primes in this partial sum begins: 3, 11, 17, 29, 31, 37, 41, 53, 79, 89, 101, 137, 149, 181, 191, 197, 229, 271.

Note: A227744 now gives the squares, which occur at positions given by A227743. - Antti Karttunen, Jul 27 2013

			1 has 1 run in its binary representation "1".
2 has 2 runs in its binary representation "10".
3 has 1 run in its binary representation "11".
4 has 2 runs in its binary representation "100".
5 has 3 runs in its binary representation "101".
Thus a(1) = 1, a(2) = 1+2 = 3, a(3) = 1+2+1 = 4, a(4) = 1+2+1+2 = 6, a(5) = 1+2+1+2+3 = 9.

Antti Karttunen, Table of n, a(n) for n = 0..8191
Richard Blecksmith and Purushottam W. Laud, Some Exact Number Theory Computations via Probability Mechanisms, American Mathematical Monthly, volume 102, number 10, December 1995, pages 893-903, with a(n) = Sum_{j=0..n} b_j as calculated in section 2.
Hsien-Kuei Hwang, Svante Janson and Tsung-Hsi Tsai, Exact and Asymptotic Solutions of a Divide-and-Conquer Recurrence Dividing at Half: Theory and Applications, ACM Transactions on Algorithms, volume 13, number 4, December 2017, article number 47, pages 1-43. Also first authors' copy, 2016. See example 5.5.
Kevin Ryde, Iterations of the Dragon Curve, see index "DirCumul".

Cf. A005811, A000788, A056539, A014707, A014577, A082410, A000975, A034947.
Cf. also A227737, A227741, A227742.
Cf. A227744 (squares occurring), A227743 (indices of squares).

Accumulate[Join[{0},Table[Length[Split[IntegerDigits[n,2]]],{n,110}]]] (* Harvey P. Dale, Jul 29 2013 *)

a(n) = my(v=binary(n+1),d=0,e=4); for(i=1,#v, if(v[i], v[i]=#v-i+d;d+=e;e=0, e=4)); fromdigits(v,2)>>1; \\ Kevin Ryde, Aug 27 2021

a(n) = sum(i=0..n) A005811(i) = sum(i=0..n) (A037834(i)+1) = sum(i=0..n) (A069010(i) + A033264(i)).
a(A000225(n)) = A001787(n) = A000788(A000225(n)). - Antti Karttunen, Jul 27 2013 & Aug 09 2013
a(2n) = 2*a(n) + n - 2*(ceiling(A005811(n)/2) - (n mod 2)), a(2n+1) = 2*a(n) + n + 1. - Ralf Stephan, Aug 11 2013

A227738 Irregular table read by rows: each row n (n>=1) lists the positions where the runs of bits change between 0's and 1's in the binary expansion of n, when scanning it from the least significant to the most significant end.

Original entry on oeis.org

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

Offset: 1

Antti Karttunen, Jul 25 2013

Row n has A005811(n) terms.
As a sequence, seems to have a particular fractal structure, probably allowing additional formulas.
Row n lists the positions of 1-bits in the binary expansion of the Gray code for n, A003188(n), when 1 is the rightmost position. A003188(17) = 25 = 11001_2 gives row 17: 1,4,5. - Alois P. Heinz, Feb 01 2023

			Table begins as:
  Row  n in    Terms on
   n   binary  that row
   1      1    1;
   2     10    1,2;
   3     11    2;
   4    100    2,3;
   5    101    1,2,3;
   6    110    1,3;
   7    111    3;
   8   1000    3,4;
   9   1001    1,3,4;
  10   1010    1,2,3,4;
  11   1011    2,3,4;
  12   1100    2,4;
  13   1101    1,2,4;
  14   1110    1,4;
  15   1111    4;
  16  10000    4,5;
etc.
The terms also give the partial sums of runlengths, when the binary expansion of n is scanned from the least significant to the most significant end.

Antti Karttunen, The rows 1..1023 of the table, flattened
Wikipedia, Gray code

Each row n (n>=1) contains the initial A005811(n) nonzero terms from the beginning of row n of A227188. A227192(n) gives the sum of terms on row n. A136480 gives the first column.
Cf. also A227188, A227736, A227739.
A318926 is a compressed version. If the order is reversed we get A101211 and A318927.
Cf. A003188, A360287.

T:= n-> (l-> seq(`if`(l[i]=1, i, [][]), i=1..nops(l)))(
                 Bits[Split](Bits[Xor](n, iquo(n, 2)))):
seq(T(n), n=1..50);  # Alois P. Heinz, Feb 01 2023

Table[Rest@FoldList[Plus,0,Length/@Split[Reverse[IntegerDigits[n,2]]]],{n,34}]//Flatten (* Wouter Meeussen, Aug 31 2013 *)

a(n) = A227188(A227737(n),A227740(n)).

Alternatively, if A227740(n) is 0, then a(n) = A227736(n), otherwise a(n) = a(n-1) + A227736(n). [Each row gives cumulative sums of the runlengths of binary representation of n]

A227740 Integers from 0 to A037834(n) followed by integers from 0 to A037834(n+1) and so on.

Original entry on oeis.org

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

Offset: 1

Antti Karttunen, Jul 25 2013

nonn

Equivalently, integers from 0 to A005811(n)-1 followed by integers from 0 to A005811(n+1)-1 and so on.

Vincenzo Librandi, Table of n, a(n) for n = 1..2272

Cf. A227737, A227741, A005811, A037834, A173318.

Cf. also A227736, A227738, A227739.

Table[Range[0, #] &@ Total@ Flatten@ Map[Abs@ Differences@ # &,
Partition[IntegerDigits[n, 2], 2, 1]], {n, 34}] // Flatten (* Michael De Vlieger, May 09 2017 *)

(define (A227740 n) (- n (+ 1 (A173318 (- (A227737 n) 1)))))

a(n) = n - (1 + A173318(A227737(n)-1)).

A227741 Simple self-inverse permutation of natural numbers: List each block of A005811(n) numbers from A173318(n-1)+1 to A173318(n) in reverse order.

Original entry on oeis.org

1, 3, 2, 4, 6, 5, 9, 8, 7, 11, 10, 12, 14, 13, 17, 16, 15, 21, 20, 19, 18, 24, 23, 22, 26, 25, 29, 28, 27, 31, 30, 32, 34, 33, 37, 36, 35, 41, 40, 39, 38, 44, 43, 42, 48, 47, 46, 45, 53, 52, 51, 50, 49, 57, 56, 55, 54, 60, 59, 58, 62, 61, 65, 64, 63, 69, 68, 67, 66

Offset: 1

Antti Karttunen, Jul 25 2013

nonn

This permutation maps between such irregular tables as A101211 and A227736 which are otherwise identical, except for the order in which the lengths of runs have been listed. In other words, A227736(n) = A101211(a(n)) and vice versa, A101211(n) = A227736(a(n)).

Antti Karttunen, Table of n, a(n) for n = 1..1001
Index entries for sequences that are permutations of the natural numbers

Cf. A227742 (gives the fixed points).

Cf. also A227737, A227740, A005811, A173318.

(define (A227741 n) (- (A173318 (A227737 n)) (A227740 n)))

a(n) = A173318(A227737(n)) - A227740(n).

A100922 k appears A000120(k) times (appearances equal number of 1-bits).

Original entry on oeis.org

1, 2, 3, 3, 4, 5, 5, 6, 6, 7, 7, 7, 8, 9, 9, 10, 10, 11, 11, 11, 12, 12, 13, 13, 13, 14, 14, 14, 15, 15, 15, 15, 16, 17, 17, 18, 18, 19, 19, 19, 20, 20, 21, 21, 21, 22, 22, 22, 23, 23, 23, 23, 24, 24, 25, 25, 25, 26, 26, 26, 27, 27, 27, 27, 28, 28, 28, 29, 29, 29, 29, 30, 30, 30, 30

Offset: 0

Rick L. Shepherd, Nov 21 2004

Clearly every positive integer appears at least once in this sequence.

			The binary representation of 16 is 10000, which has one 1-bit (and four 0-bits), hence 16 appears once in this sequence (and four times in A100921).

Antti Karttunen, Table of n, a(n) for n = 0..11264

Cf. A100921 (n's appearances equal its number of 0-bits), A030530 (n's appearances equal its total number of bits), A227737 (n's appearances equal its total number of runs), A000069, A000120, A000788, A163510, A243067.

T:= n-> n$add(i, i=Bits[Split](n)):
seq(T(n), n=1..30);  # Alois P. Heinz, Nov 11 2024

Table[Table[n,DigitCount[n,2,1]],{n,30}]//Flatten (* Harvey P. Dale, Aug 31 2017 *)

def A000788(n): return (n+1)*n.bit_count()+(sum((m:=1<>j)-(r if n<<1>=m*(r:=k<<1|1) else 0)) for j in range(1,n.bit_length()+1))>>1)
def A100922(n):
    if n == 0: return 1
    m, k = 1, 1
    while A000788(m)<=n: m<<=1
    while m-k>1:
        r = m+k>>1
        if A000788(r)>n:
            m = r
        else:
            k = r
    return m # Chai Wah Wu, Nov 11 2024

a(n) = the least k such that A000788(k) > n. - Antti Karttunen, Jun 20 2014

Sum_{n>=1} (-1)^(n+1)/a(n) = Sum_{n>=1} (-1)^(n+1)/A000069(n) = 0.67968268... . - Amiram Eldar, Feb 18 2024

cp's OEIS Frontend

A101211 Triangle read by rows: n-th row is length of run of leftmost 1's, followed by length of run of 0's, followed by length of run of 1's, etc., in the binary representation 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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A227739 Irregular table where row n lists in nondecreasing order the parts of unordered partition encoded in the runlengths of binary expansion of n; nonzero terms of A227189.

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A173318 Partial sums of A005811.

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A227738 Irregular table read by rows: each row n (n>=1) lists the positions where the runs of bits change between 0's and 1's in the binary expansion of n, when scanning it from the least significant to the most significant end.

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A227740 Integers from 0 to A037834(n) followed by integers from 0 to A037834(n+1) and so on.

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A227741 Simple self-inverse permutation of natural numbers: List each block of A005811(n) numbers from A173318(n-1)+1 to A173318(n) in reverse order.