A227192 -id:A227192 - cp's OEIS frontend

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.

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)

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]

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.

A360259 a(0) = 0, and for any n > 0, let k > 0 be as small as possible and such that F(2) + ... + F(1+k) >= n (where F(m) denotes A000045(m), the m-th Fibonacci number); a(n) = k + a(F(2) + ... + F(1+k) - n).

Original entry on oeis.org

0, 1, 3, 2, 6, 4, 3, 10, 6, 7, 5, 4, 15, 8, 9, 11, 7, 8, 6, 5, 21, 10, 11, 13, 12, 16, 9, 10, 12, 8, 9, 7, 6, 28, 12, 13, 15, 14, 18, 16, 15, 22, 11, 12, 14, 13, 17, 10, 11, 13, 9, 10, 8, 7, 36, 14, 15, 17, 16, 20, 18, 17, 24, 20, 21, 19, 18, 29, 13, 14, 16

Offset: 0

Rémy Sigrist, Jan 31 2023

See A095791 for the corresponding k's.

This sequence has similarities with A227192; here we use Fibonacci numbers, there powers of 2.

			The first terms, alongside the corresponding k's, are:
  n      a(n)  k
  -----  ----  ---
      0     0  N/A
      1     1    1
      2     3    2
      3     2    2
      4     6    3
      5     4    3
      6     3    3
      7    10    4
      8     6    4
      9     7    4
     10     5    4
     11     4    4
     12    15    5

Rémy Sigrist, Table of n, a(n) for n = 0..10946

See A095791, A360260 and A360265 for similar sequences.

Cf. A000045, A001911, A227192.

{ t = k = 0; print1 (0); for (n = 1, #a = vector(70), if (n > t, t += fibonacci(1+k++);); print1 (", "a[n] = k+if (t==n, 0, a[t-n]));); }

a(A001911(n)) = n.

cp's OEIS Frontend

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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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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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.

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A360259 a(0) = 0, and for any n > 0, let k > 0 be as small as possible and such that F(2) + ... + F(1+k) >= n (where F(m) denotes A000045(m), the m-th Fibonacci number); a(n) = k + a(F(2) + ... + F(1+k) - n).

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