OFFSET
0,11
COMMENTS
Are the partitions counted all of length 4 or 5?
The (one-based) weighted sum of a sequence (y_1,...,y_k) is Sum_{i=1..k} i*y_i. The reverse-weighted sum is the weighted sum of the reverse, also the sum of partial sums. For example, the weighted sum of (4,2,2,1) is 1*4 + 2*2 + 3*2 + 4*1 = 18 and the reverse-weighted sum is 4*4 + 3*2 + 2*2 + 1*1 = 27.
EXAMPLE
The partition (6,4,4,1) has sum 15 and reverse-weighted sum 45 so is counted under a(15).
The a(n) partitions for n = {5, 10, 15, 16, 21, 24}:
(1,1,1,1,1) (4,3,2,1) (6,4,4,1) (6,5,4,1) (8,6,6,1) (9,7,7,1)
(2,2,2,2,2) (6,5,2,2) (6,6,2,2) (8,7,4,2) (9,8,5,2)
(7,3,3,2) (7,4,3,2) (9,5,5,2) (9,9,3,3)
(3,3,3,3,3) (9,6,3,3) (10,6,6,2)
(10,4,4,3) (10,7,4,3)
(11,5,5,3)
(12,4,4,4)
MATHEMATICA
Table[Length[Select[IntegerPartitions[n], Total[Accumulate[#]]==3n&]], {n, 0, 30}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Gus Wiseman, Jun 10 2023
STATUS
approved