A227739 -id:A227739 - cp's OEIS frontend

A227737 n occurs as many times as there are runs in binary representation of n.

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

Offset: 1

Antti Karttunen, Jul 25 2013

a(n) = the least integer k such that A173318(k) >= n, which implies that each n occurs A005811(n) times.

Although as such quite uninteresting, this sequence is useful for computing irregular tables like A101211, A227736, A227738 and A227739.

			1 has one run in its binary representation "1", thus 1 occurs once.
2 has two runs in its binary representation "10", thus 2 occurs twice.
3 has one run in its binary representation "11", thus 3 occurs only once.
4 has two runs in its binary representation "100", thus 4 occurs twice.
5 has three runs in its binary representation "101", thus 5 occurs three times.
The sequence thus begins as 1, 2,2, 3, 4,4, 5,5,5, ...

Antti Karttunen, Table of n, a(n) for n = 1..5120

Cf. A227740, A227741, A005811, A173318.

Table[ConstantArray[n, Length@ Split@ IntegerDigits[n, 2]], {n, 26}] // Flatten (* Michael De Vlieger, May 09 2017 *)
Table[PadRight[{},Length[Split[IntegerDigits[n,2]]],n],{n,40}]//Flatten (* Harvey P. Dale, Jul 23 2021 *)

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)).

A227762 Numbers in whose minimally runlength-encoded unordered partition all parts are equal; positions of zeros in A227761.

Original entry on oeis.org

0, 1, 2, 3, 4, 6, 8, 9, 10, 12, 15, 16, 20, 24, 25, 28, 30, 35, 36, 42, 48, 49, 54, 56, 63, 64, 72, 80, 81, 88, 90, 99, 100, 110, 120, 121, 130, 132

Offset: 1

Antti Karttunen, Jul 26 2013

After 3 no more primes. First missing composites are: 14, 18, 21, 22, 26, 27, 32, 33, 34, 38, 39, 40, ...

			The first row in A227739 (please see its Example section) that sums to 6 occurs as its row 8 (= A227368(6)). The corresponding partition {3+3} contains only equal parts, thus 6 is a member of this sequence. The first row in A227739 that sums to 5 occurs as its row 9 (= A227368(5)). The corresponding partition {1+2+2} contains more than just one kind of summands, thus 5 do not occur in this sequence.

Cf. A227761, A227739, A227368.

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A227737 n occurs as many times as there are runs in binary representation of n.

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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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A227762 Numbers in whose minimally runlength-encoded unordered partition all parts are equal; positions of zeros in A227761.

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