A227736 - OEIS

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.

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

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,4

COMMENTS

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, A066099, A108244 and A124734.

Each row n (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.

LINKS

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

Mikhail Kurkov, Comments on A227736 [verification needed]

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 Apparently unpublished.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.

FORMULA

a(n) = A227186(A227737(n), A227740(n)).

a(n) = A101211(A227741(n)).

EXAMPLE

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)

MATHEMATICA

Array[Length /@ Reverse@ Split@ IntegerDigits[#, 2] &, 34] // Flatten (* Michael De Vlieger, Dec 11 2020 *)

PROG

(Scheme) (define (A227736 n) (A227186bi (A227737 n) (A227740 n))) ;; The Scheme-function for A227186bi has been given in A227186.

(Haskell)

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

CROSSREFS

Cf. A227738 and also A227739 for similar table for unordered partitions.

Cf. A030308, A245562, A245563.

Sequence in context: A300752 A300751 A070000 * A228528 A366040 A219244

Adjacent sequences: A227733 A227734 A227735 * A227737 A227738 A227739

KEYWORD

nonn,base,tabf

AUTHOR

Antti Karttunen, Jul 25 2013

STATUS

approved