A227189 -id:A227189 - cp's OEIS frontend

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.

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)).

A227183 a(n) is the sum of parts of the unique unordered partition encoded in the run lengths of the binary expansion of n; row sums of A227739 for n >= 1.

Original entry on oeis.org

0, 1, 2, 2, 4, 3, 3, 3, 6, 5, 4, 6, 5, 4, 4, 4, 8, 7, 6, 8, 8, 5, 7, 9, 7, 6, 5, 7, 6, 5, 5, 5, 10, 9, 8, 10, 10, 7, 9, 11, 12, 9, 6, 10, 11, 8, 10, 12, 9, 8, 7, 9, 9, 6, 8, 10, 8, 7, 6, 8, 7, 6, 6, 6, 12, 11, 10, 12, 12, 9, 11, 13, 14, 11, 8, 12, 13, 10, 12, 14

Offset: 0

Antti Karttunen, Jul 05 2013

Like A129594 this sequence utilizes the fact that compositions (i.e., ordered partitions) can be bijectively mapped to (unordered) partitions by taking the partial sums of the list of composants after one has been subtracted from each except the first one. Compositions in turn are mapped to nonnegative integers via the runlength encoding, where the lengths of maximum runs of 0's or 1's in binary representation of n give the composants. See the OEIS Wiki page and the example below.

Each n occurs A000041(n) times in total and occurs for the first time at A227368(n) and for the last time at position A000225(n). See further comments and conjectures at A227368 and A227370.

			19 has binary expansion "10011", thus the maximal runs of identical bits (scanned from right to left) are [2,2,1]. We subtract one from each after the first one, to get [2,1,0] and then form their partial sums as [2,2+1,2+1+0], which thus maps to unordered partition {2+3+3} which adds to 8. Thus a(19)=8.

Antti Karttunen, Table of n, a(n) for n = 0..8192
A. Karttunen et al., Run-length encoding, OEIS Wiki.
Index entries for sequences related to partitions

Row sums of A227189 and A227739. Cf. A227184 (corresponding products), A227185, A227189, A227192, A129594, A226062, A227368.

Analogous sum sequences computed for other encoding schemes of unordered partitions: A036042, A056239, A161511, A243503. Cf. also A229119, A003188, A075157, A243353 (associated permutations mapping between these schemes).

Table[Function[b, Total@ Accumulate@ Prepend[If[Length@ b > 1, Rest[b] - 1, {}], First@ b] - Boole[n == 0]]@ Map[Length, Split@ Reverse@ IntegerDigits[n, 2]], {n, 0, 79}] // Flatten (* Michael De Vlieger, May 09 2017 *)

def A227183(n):
  '''Sum of parts of the unique unordered partition encoded in the run lengths of the binary expansion of n.'''
  s = 0
  b = n%2
  i = 1
  while (n != 0):
    n >>= 1
    if ((n%2) == b): # Staying in the same run of bits?
      i += 1
    else: # The run changes.
      b = n%2
      s += i
  return(s)

a(n) = Sum_{i=0..A005811(n)-1} A227189(n,i). [The defining formula]
Equivalently, for n>=1, a(n) = Sum_{i=(A173318(n-1)+1)..A173318(n)} A227739(i).
a(n) = A227192(n) - A000217(A005811(n)-1).
Other identities:
a(A129594(n)) = a(n). [This follows from the fact that conjugating a partition doesn't change its total sum]
a(A226062(n)) = a(n). [Which is also true for the "Bulgarian operation"]
From Antti Karttunen, Mar 08 2015: (Start)
Can be also obtained by mapping with an appropriate permutation from the sequences giving sizes of each partition (i.e., sum of their parts) computed for other enumerations similar to A227739:
a(n) = A036042(A229119(n)).
a(n) = A161511(A003188(n)).
a(n) = A056239(A243353(n)).
a(n) = A243503(1+A075157(n)).
(End)

A129594 Involution of nonnegative integers induced by the conjugation of the partition encoded in the run lengths of binary expansion of n.

Original entry on oeis.org

0, 1, 3, 2, 4, 7, 6, 5, 11, 12, 15, 8, 9, 14, 13, 10, 20, 27, 28, 19, 16, 31, 24, 23, 22, 25, 30, 17, 18, 29, 26, 21, 43, 52, 59, 36, 35, 60, 51, 44, 47, 48, 63, 32, 39, 56, 55, 40, 41, 54, 57, 38, 33, 62, 49, 46, 45, 50, 61, 34, 37, 58, 53, 42, 84, 107, 116, 75, 68, 123

Offset: 0

Antti Karttunen, May 01 2007

This sequence is based on the fact that compositions (i.e. ordered partitions) can be mapped 1-to-1 to partitions by taking the partial sums of the list where one is subtracted from each composant except the first. (See table A227189 where the parts for each partition are listed).
The inverse process, from partitions to compositions, occurs by inserting the first (i.e. smallest) element of a partition sorted into ascending order to the front of the list obtained by adding one to the first differences of the elements.
Compositions map bijectively to nonnegative integers by assigning each run of k consecutive 1's (or 0's) in binary expansion of n with summand k in the composition.
The graph of this sequence is quite interesting.

			a(8) = 11, as 8 is 1000 in binary, mapping to composition 3+1 (we scan the binary expansion from the least to the most significant end), which maps to partition 3+3, whose conjugate-partition is 2+2+2, yielding composition 2+1+1, which maps to binary 1011, 11 in decimal. a(13) = 14, as 13 is 1101 in binary, mapping to composition 1+1+2, which maps to the partition 1+1+2, whose conjugate-partition is 1+3, yielding composition 1+3, which maps to binary 1110, 14 in decimal. a(11) = 8 and a(14) = 13, as taking the conjugate of a partition is a self-inverse operation.

Antti Karttunen, Table of n, a(n) for n = 0..1023
Antti Karttunen, Python-program for computing this and related sequences.
A. Karttunen et al., Run-length encoding, OEIS Wiki.
Index entries for sequences that are permutations of the natural numbers

a(n) = A075158(A122111(1+A075157(n)) - 1). See A129595 for another kind of encoding of integer partitions.
Sequences related to partitions encoded in this way:
Cf. A227189 (parts of partitions listed on separate rows of the array).
Cf. A005811 (number of parts in the partition).
Cf. A136480 (for n>= 1, the smallest part).
Cf. A227185 (the largest part).
Cf. A227183 (sum of parts).
Cf. A227184 (product of parts).
Note that this permutation maps between A005811 and A227185 as follows: A005811(n) = A227185(a(n)) and vice versa: A227185(n) = A005811(a(n)). On the other hand, it keeps A227183 fixed, as A227183(n) = A227183(a(n)).
Cf. also A226062.

A227184 a(n) = product of parts of the unordered partition encoded with the runlengths of binary expansion of n.

Original entry on oeis.org

1, 1, 1, 2, 4, 1, 2, 3, 9, 4, 1, 8, 6, 2, 3, 4, 16, 9, 4, 18, 16, 1, 8, 27, 12, 6, 2, 12, 8, 3, 4, 5, 25, 16, 9, 32, 36, 4, 18, 48, 81, 16, 1, 32, 54, 8, 27, 64, 20, 12, 6, 24, 24, 2, 12, 36, 15, 8, 3, 16, 10, 4, 5, 6, 36, 25, 16, 50, 64, 9, 32, 75, 144, 36, 4, 72

Offset: 0

Antti Karttunen, Jul 04 2013

a(0) = 1, as 0 is here considered to encode an empty partition {}, and the empty product is one.
Like A129594, this sequence is based on the fact that compositions (i.e., ordered partitions) can be mapped 1-to-1 to partitions by taking the partial sums of the list where one is subtracted from each composant except the first (originally explained by Marc LeBrun in his Jan 11 2006 post on SeqFan mailing list, with an additional twist involving factorization and prime exponents, cf. A129595). The example below show how this works.
Compare the scatterplot of this sequence to those of A002487, A243353, A243499 and A253552.

			8 has binary expansion "1000", whose runs have lengths [3,1] when arranged from the least significant to the most significant end. Taking partial sums of 3 and 0, we get 3 and 3, whose product is 9, thus a(8) = 9.
For 44, in binary "101100", the run lengths are [2,2,1,1] (from the least significant end), and subtracting one from all terms except the first one, we get [2,1,0,0], whose partial sums are [2,3,3,3], and 2*3*3*3 = 54, thus a(44)=54.

Antti Karttunen, Table of n, a(n) for n = 0..8192

For n>=1, a(n) gives the product of nonzero terms on row n of table A227189/A227739.
Cf. A227183 (gives the corresponding sums).
See also A167489 for a similar sequence, which gives the product of parts of the compositions (ordered partitions).
Cf. A243499, A003963, A243504 (other such product sequences) and A003188, A243353, A075157 (associated permutations mapping between these schemes).
Cf. also A002487, A243353, A253552.

Table[Function[b, Times @@ Accumulate@ Prepend[If[Length@ b > 1, Rest[b] - 1, {}], First@ b]]@ Map[Length, Split@ Reverse@ IntegerDigits[n, 2]], {n, 0, 75}] // Flatten (* Michael De Vlieger, May 09 2017 *)

def A227184(n):
  '''Product of parts of the unique unordered partition encoded in the run lengths of the binary expansion of n.'''
  p = 1
  b = n%2
  i = 1
  while (n != 0):
    n >>= 1
    if ((n%2) == b): i += 1
    else:
      b = n%2
      p *= i
  return(p)

(define (A227184 n) (if (zero? n) 1 (apply * (binexp_to_ascpart n))))
(define (binexp_to_ascpart n) (let ((runlist (reverse! (binexp->runcount1list n)))) (PARTSUMS (cons (car runlist) (map -1+ (cdr runlist))))))
(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)))))))
(define (PARTSUMS a) (cdr (reverse! (fold-left (lambda (psums n) (cons (+ n (car psums)) psums)) (list 0) a))))

Can be also obtained by mapping with an appropriate permutation from the products of parts of each partition computed for other enumerations similar to A227739:
a(n) = A243499(A003188(n)).
a(n) = A003963(A243353(n)).
a(n) = A243504(1+A075157(n)).

A227368 a(n) = Index k where A227183(k) for the first time gets value n; the runlength binary code for minimally runlength-encoded unordered partition of size n.

Original entry on oeis.org

0, 1, 2, 5, 4, 9, 8, 17, 16, 23, 32, 39, 40, 71, 72, 87, 80, 151, 144, 167, 160, 295, 288, 327, 320, 351, 576, 607, 640, 671, 672, 1183, 1184, 1311, 1312, 1375, 1344, 2399, 2368, 2655, 2624, 2719, 2688, 4767, 4736, 5279, 5248, 5407, 5376, 5503, 9472, 9599, 10496

Offset: 0

Antti Karttunen, Jul 08 2013

nonn

The word "minimally" in the description means that the integer in whose binary representation some unordered partition of n is encoded should be as small as possible. This sequence gives such a minimal integer for each n, which encodes an unordered partition whose sum is n. The details of the encoding system are explained in A227183.
Also, a(n) gives the index of the first row of A227189/A227739 which sums to n.
Project: Find an algorithm which computes a(n) with a more sophisticated method than just by a blind search. This is a kind of an optimization problem for representing n as a special "bit-packed" sum: the smallest summand of size x costs x bits, and its any subsequent usage in the sum costs just one bit each time. Using any additional summand y > x costs (y-x)+1 bits when used first time, and then again additional usages cost only one bit each. Goal: minimize the number of bits needed. If multiple candidates with the same number of bits are found, then the one which results the smallest integer (when interpreted as a binary number) wins.
For any composite n = t*u, the upper bound for the size of a(n) is t+u-1 bits.
A000267(n) seems to give the binary width of a(n+1). Compare to the conjecture given at A227370.

			   n   a(n)   binary   corresponding partition    sum = n
                       (cf. A227183 for details)
   0     0         0   (0)                          0
   1     1         1   (1)                          1
   2     2        10   (1 + 1)                      2
   3     5       101   (1 + 1 + 1)                  3
   4     4       100   (2 + 2)                      4
   5     9      1001   (1 + 2 + 2)                  5
   6     8      1000   (3 + 3)                      6
   7    17     10001   (1 + 3 + 3)                  7
   8    16     10000   (4 + 4)                      8
   9    23     10111   (3 + 3 + 3)                  9
  10    32    100000   (5 + 5)                     10
  11    39    100111   (3 + 4 + 4)                 11
  12    40    101000   (3 + 3 + 3 + 3)             12
  13    71   1000111   (3 + 5 + 5)                 13
  14    72   1001000   (3 + 3 + 4 + 4)             14
  15    87   1010111   (3 + 3 + 3 + 3 + 3)         15
  16    80   1010000   (4 + 4 + 4 + 4)             16
  17   151  10010111   (3 + 3 + 3 + 4 + 4)         17
  18   144  10010000   (4 + 4 + 5 + 5)             18
  19   167  10100111   (3 + 4 + 4 + 4 + 4)         19
  20   160  10100000   (5 + 5 + 5 + 5)             20
a(5) = 9, because 5 occurs for the first time in A227183 as A227183(9).
Note that for 20, there is for example also a code 175, "10101111" in binary, which results a partition (4 + 4 + 4 + 4 + 4) (= 20), but as 160 < 175, and there are no other partitions of 20 which would result even smaller code number, 160 is the winner (the minimal code), and thus a(20)=160.
A227761 gives the maximum difference between successive parts that occurs in these partitions.

Antti Karttunen, Table of n, a(n) for n = 0..132
Index entries for sequences related to partitions

Same sequence sorted into ascending order: A227369.

Cf. also A227183, A227761, A227762.

def A227368(n):
  '''Index k where A227183(k) for the first time gets value n. A naive implementation.'''
  k = 0
  while(A227183(k) != n): k += 1
  return(k)

a(n) = A227369(A227370(n)) [See comments and conjecture at A227370]

A227185 The largest part in the unordered partition encoded in the runlengths of the binary expansion of n.

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Jul 05 2013

The bijective encoding of nonordered partitions via compositions (ordered partitions) present in the binary expansion of n is explained in A227184.

It appears that a(4n+2) = a(2n+1). - Ralf Stephan, Jul 20 2013

			12 has binary expansion "1100", for which the lengths of runs (consecutive blocks of 0- or 1-bits) are [2,2]. Converting this to a partition in the manner explained in A227184 gives the partition {2+3}. Its largest part is 3, thus a(12)=3, which is actually the first time when this sequence differs from A043276.

Antti Karttunen, Table of n, a(n) for n = 0..8192

For all n, A005811(n) = a(A129594(n)). Cf. also A136480 (for n>= 1, gives the smallest part) and A227183, A227184, A226062, A092339, A227147.

a(n) gives the rightmost nonzero term on the n-th row of A227189.

Table[Function[b, Max@ Accumulate@ Prepend[If[Length@ b > 1, Rest[b] - 1, {}], First@ b] - Boole[n == 0]]@ Map[Length, Split@ Reverse@ IntegerDigits[ n, 2]], {n, 0, 120}] // Flatten (* Michael De Vlieger, May 09 2017 *)

(define (A227185 n) (if (zero? n) n (+ 1 (- (A029837 (+ 1 n)) (A005811 n)))))
(define (A227185v2 n) (if (zero? n) n (car (reverse (binexp_to_ascpart n))))) ;; Alternative definition, using the auxiliary functions given in A227184.

Defining formula:
a(0)=0; and for n>=1, a(n) = A029837(n+1) - (A005811(n)-1). [Because the largest part in the unordered partition in this encoding scheme is computed as (c_1 + (c_2-1) + (c_3-1) + ... + (c_k-1)) where c_1 .. c_k are the parts of the k-part composition that sum together as c_1 + c_2 + ... + c_k = A029837(n+1) (the binary width of n), so we subtract from the total binary width of n the number of runs (A005811) minus 1.]
Equivalently: a(n) = A092339(n)+1 for n>0.
a(n) = A005811(A129594(n)). [This just states the fact that when conjugating a partition, the largest part of an old partition will be the number of the parts in the new, conjugated partition.]

A227186 Array A(n,k) read by antidiagonals: the length of the (k+1)-th run (k>=0) of binary digits of n, first run starting from the least significant bit of n.

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Jul 06 2013

A(n,k) is set to zero if there are less than k+1 runs.

The irregular table A101211 gives the nonzero terms of each row in reverse order. The terms on row n sum to A029837(n+1). The product of nonzero terms on row n>0 is A167489(n). Number of nonzero terms on each row: A005811.

			The top-left corner of the array:
0, 0, 0, 0, 0, ... (0, in binary 0, has no runs (by convention), thus at first we have all-0 sequence)
1, 0, 0, 0, 0, ... (1, in binary 1, has one run of length 1)
1, 1, 0, 0, 0, ... (2, in binary 10, has two runs of length 1 both)
2, 0, 0, 0, 0, ... (3, in binary 11, has one run of length 2)
2, 1, 0, 0, 0, ... (4, in binary 100, the rightmost run of length 2 given first, then the second run of length 1)
1, 1, 1, 0, 0, ... (5, in binary 101, has three runs of one bit each)
1, 2, 0, 0, 0, ...
3, 0, 0, 0, 0, ...
3, 1, 0, 0, 0, ...
1, 2, 1, 0, 0, ...
1, 1, 1, 1, 0, ...
2, 1, 1, 0, 0, ...
2, 2, 0, 0, 0, ...
1, 1, 2, 0, 0, ...
1, 3, 0, 0, 0, ...
4, 0, 0, 0, 0, ...

Antti Karttunen, The first 141 antidiagonals of the table, flattened

Used to compute A227183. Cf. also A163575, A227188, A227189.

A227186 := proc(n,k)
    local bdgs,ru,i,b,a;
    bdgs := convert(n,base,2) ;
    if nops(bdgs) = 0 then
        return 0 ;
    end if;
    ru := 0 ;
    i := 1 ;
    b := op(i,bdgs) ;
    a := 1 ;
    for i from 2 to nops(bdgs) do
        if op(i,bdgs) <> op(i-1,bdgs) then
            if ru = k then
                return a;
            end if;
            a := 1 ;
            ru := ru+1 ;
        else
            a := a+1 ;
        end if;
    end do:
    if ru =k then
        a ;
    else
        0 ;
    end if;
end proc: # R. J. Mathar, Jul 23 2013

A227186(n,k)=while(k>=0,for(c=1,n,bittest(n,0)==bittest(n\=2,0)&next;k&break;return(c));n||return;k--) \\ To let A(0,0)=1 add "!n||!" in front of while(...). TO DO: add default value k=-1 and implement "flattened" sequence, such that A227186(n) yields a(n). M. Hasler, Jul 21 2013

(define (A227186 n) (A227186bi (A002262 n) (A025581 n)))
(define (A227186bi n k) (cond ((< (A005811 n) (+ 1 k)) 0) ((zero? k) (A136480 n)) (else (A227186bi (A163575 n) (- k 1)))))

A(n,0) = A136480(n), n>0.

A227188 Square array A(n,k) read by antidiagonals: the one-based bit-index where the (k+1)-st run in the binary expansion of n ends, as read from the least significant end.

Original entry on oeis.org

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

Offset: 0

Antti Karttunen, Jul 06 2013

A(n,k) is set to zero if there are fewer runs in n than k+1.
Equally, when A005811(n) > 1, A(n,k) gives the zero-based bit-index where the (k+2)-th run in the binary expansion of n starts, counted from the least significant end.
Each row gives the partial sums of the terms on the corresponding row in A227186, up to the first zero.

			The top-left corner of the array:
row #  row starts as
   0    0, 0, 0, 0, 0, ...
   1    1, 0, 0, 0, 0, ...
   2    1, 2, 0, 0, 0, ...
   3    2, 0, 0, 0, 0, ...
   4    2, 3, 0, 0, 0, ...
   5    1, 2, 3, 0, 0, ...
   6    1, 3, 0, 0, 0, ...
   7    3, 0, 0, 0, 0, ...
   8    3, 4, 0, 0, 0, ...
   9    1, 3, 4, 0, 0, ...
  10    1, 2, 3, 4, 0, ...
  11    2, 3, 4, 0, 0, ...
  12    2, 4, 0, 0, 0, ...
  13    1, 2, 4, 0, 0, ...
  14    1, 4, 0, 0, 0, ...
  15    4, 0, 0, 0, 0, ...
  16    4, 5, 0, 0, 0, ...
etc.
For example, for n = 8, whose binary expansion is "1000", we get the run lengths 3 and 1 (scanning from the right), partial sums of which are 3 and 4, thus row 8 begins as A(8,0)=3, A(8,1)=4, A(8,2)=0, ...

Antti Karttunen, The first 141 antidiagonals of the table, flattened

Cf. A227192 (row sums). Number of nonzero terms on each row: A005811.

Cf. also A227186, A227189, A163575.

A227188 := proc(n,k)
    local bdgs,ru,i,b,a;
    bdgs := convert(n,base,2) ;
    if nops(bdgs) = 0 then
        return 0 ;
    end if;
    ru := 0 ;
    i := 1 ;
    b := op(i,bdgs) ;
    for i from 2 to nops(bdgs) do
        if op(i,bdgs) <> op(i-1,bdgs) then
            if ru = k then
                return i-1;
            end if;
            ru := ru+1 ;
        end if;
    end do:
    if ru =k then
        nops(bdgs) ;
    else
        0 ;
    end if;
end proc: # R. J. Mathar, Jul 23 2013

Table[PadRight[Rest@FoldList[Plus,0,Length/@Split[Reverse[IntegerDigits[j,2]]]],i+1-j][[i+1-j]],{i,0,12},{j,0,i}] (* Wouter Meeussen, Aug 31 2013 *)

(define (A227188 n) (A227188bi (A002262 n) (A025581 n)))
(define (A227188bi n k) (cond ((< (A005811 n) (+ 1 k)) 0) ((zero? k) (A136480 n)) (else (+ (A136480 n) (A227188bi (A163575 n) (- k 1))))))

A(n,0) = A136480(n), n>0.

cp's OEIS Frontend

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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A227183 a(n) is the sum of parts of the unique unordered partition encoded in the run lengths of the binary expansion of n; row sums of A227739 for n >= 1.

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A129594 Involution of nonnegative integers induced by the conjugation of the partition encoded in the run lengths of binary expansion of n.

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A227184 a(n) = product of parts of the unordered partition encoded with the runlengths of binary expansion of n.

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A227368 a(n) = Index k where A227183(k) for the first time gets value n; the runlength binary code for minimally runlength-encoded unordered partition of size n.

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A227185 The largest part in the unordered partition encoded in the runlengths of the binary expansion of n.

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A227186 Array A(n,k) read by antidiagonals: the length of the (k+1)-th run (k>=0) of binary digits of n, first run starting from the least significant bit of n.

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Maple

PARI