A125106 Enumeration of partitions by binary representation: each 1 is a part; the part size is 1 more than the number of 0's in the rest of the number.

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

Offset: 1

Author

Alford Arnold, Dec 10 2006

Keywords

• tabf
• nice
• nonn

Comments

Another way to describe this: starting with the binary representation and a counter set at one, count the 0's from right to left. Write a term equal to the counter for each "1" encountered.

A101211 is a similar sequence, with A005811 elements per row which maps natural numbers to compositions (ordered partitions).

There are two ways to consider this as a table: taking each partition as a row, or taking the partitions generated by 2^(n-1) through 2^n-1 as a row.

Taking the n-th row as multiple partitions, it consists of those partitions with the first hook size (largest part plus number of parts minus 1) equal to n. The number of integers in this n-th row is A001792(n-1), and the row sum is A049611.

Taking each partition as a separate row, the row lengths are A000120, and the row sums are A161511.

Heinz numbers of the rows are A005940. - Gus Wiseman, Jan 17 2023

Examples

			Row 4:
1000 [4]
1001 [3,1]
1010 [3,2]
1011 [2,1,1]
1100 [3,3]
1101 [2,2,1]
1110 [2,2,2]
1111 [1,1,1,1]
		

Links

• Alois P. Heinz, Rows n = 1..12, flattened

Crossrefs

Cf. A000041, A005811, A037016, A101211, A001792, A049611, A126411.

Each partition as row: A000120 (row widths), A161511 (row sums), A243499 (row products).

Cf. A005940. - Franklin T. Adams-Watters, Mar 06 2010

Lasts are A001511.

Firsts are A008687.

Cf. A019565, A029837, A029931, A048793, A059893, A066099, A070939, A242628.

Programs

Maple

b:= proc(n) local c, l, m; l:=[][]; m:= n; c:=1;
      while m>0 do if irem(m, 2, 'm')=0 then c:= c+1
         else l:= c, l fi
      od; l
    end:
T:= n-> seq(b(i), i=2^(n-1)..2^n-1):
seq(T(n), n=1..7);  # Alois P. Heinz, Sep 25 2015

Mathematica

f[k_] := (bits = IntegerDigits[k, 2]; zerosCount = Reverse[ Accumulate[ 1-Reverse[bits] ] ] + 1; Select[ Transpose[ {bits, zerosCount} ], First[#] == 1 & ][[All, 2]]); row[n_] := Table[ f[k], {k, 2^(n-1), 2^n-1}]; Flatten[ Table[ row[n], {n, 1, 5}]] (* Jean-François Alcover, Jan 24 2012 *)
scc[n_]:=Join@@Position[Reverse[IntegerDigits[n,2]],1];
Table[Reverse[scc[n]-Range[Length[scc[n]]]+1],{n,0,20}] (* Gus Wiseman, Jan 17 2023 *)

Formula

Partition 2n is partition n with every part size increased by 1; partition 2n+1 is partition n with an additional part of size 1.

T(n,k) = A272020(n,k) - A000120(n) + k. - Gus Wiseman, Jan 17 2023

Extensions

Edited by Franklin T. Adams-Watters, Jun 11 2009