A363623 Irregular triangle read by rows where T(n,k) is the number of integer partitions of n with reverse-weighted alternating sum k (leading and trailing 0's omitted).

1, 1, 1, 1, 1, 2, 2, 0, 1, 2, 2, 1, 1, 1, 1, 1, 3, 1, 0, 3, 0, 1, 1, 1, 1, 3, 2, 0, 3, 1, 2, 0, 1, 0, 1, 2, 5, 1, 0, 3, 1, 2, 2, 2, 1, 1, 0, 1, 0, 1, 2, 5, 3, 0, 4, 2, 2, 0, 3, 2, 1, 3, 0, 0, 1, 0, 1, 1, 1, 1, 7, 2, 0, 4, 1, 5, 2, 3, 1, 3, 0, 2, 3, 1, 2, 1, 0, 0, 1, 0, 1, 1, 1, 1

Offset: 0

Author

Gus Wiseman, Jun 15 2023

Keywords

• nonn
• tabf

Comments

We define the reverse-weighted alternating sum of a sequence (y_1,...,y_k) to be Sum_{i=1..k} (-1)^(k-i) i * y_{k-i+1}. For example:

- (3,3,2,1,1) has reverse-weighted alternating sum 1*1 - 2*1 + 3*2 - 4*3 + 5*3 = 8.

- (1,2,2,3) has reverse-weighted alternating sum -1*3 + 2*2 - 3*2 + 4*1 = -1.

Examples

			Triangle begins:
  1
  1
  1  1
  1  2
  2  0  1  2
  2  1  1  1  1  1
  3  1  0  3  0  1  1  1  1
  3  2  0  3  1  2  0  1  0  1  2
  5  1  0  3  1  2  2  2  1  1  0  1  0  1  2
  5  3  0  4  2  2  0  3  2  1  3  0  0  1  0  1  1  1  1
Row n = 6 counts the following partitions:
  k=3       k=4       k=6       k=8      k=9   k=10    k=11
--------------------------------------------------------------
  (33)      (222)  .  (6)    .  (21111)  (51)  (3111)  (411)
  (2211)              (42)
  (111111)            (321)
		

Crossrefs

Row sums are A000041.

Column k = floor((n+1)/2) is A119620.

The unweighted version is A344612 aerated, reverse A103919.

The corresponding rank statistic is A363620, reverse A363619.

The reverse version is A363622.

A053632 counts compositions by weighted sum.

A264034 counts partitions by weighted sum, reverse A358194.

A316524 gives alternating sum of prime indices, reverse A344616.

A363624 gives weighted alternating sum of Heinz partition, reverse A363625.

Cf. A008284, A067538, A222855, A222970, A318283, A320387, A360672, A360675, A362559, A363532, A363621, A363626.

Programs

Mathematica

revaltwtsum[y_]:=Sum[(-1)^(Length[y]-k)*k*y[[-k]],{k,1,Length[y]}];
Table[Length[Select[IntegerPartitions[n],revaltwtsum[#]==k&]],{n,0,15},{k,Floor[(n+1)/2],Ceiling[n*(n+1)/4]}]