A320470 Number of partitions of n such that the successive differences of consecutive parts are strictly decreasing.

1, 1, 2, 2, 3, 4, 4, 5, 7, 6, 8, 10, 10, 11, 14, 13, 16, 19, 18, 20, 25, 23, 27, 31, 30, 34, 39, 37, 42, 48, 47, 50, 59, 56, 63, 70, 68, 74, 83, 82, 89, 97, 97, 104, 116, 113, 123, 133, 133, 142, 155, 153, 166, 178, 178, 189, 204, 204, 218, 232, 235, 247, 265, 265, 283, 299

Offset: 0

Author

Seiichi Manyama, Oct 13 2018

Keywords

• nonn

Comments

Partitions are usually written with parts in descending order, but the conditions are easier to check "visually" if written in ascending order.

Partitions (p(1), p(2), ..., p(m)) such that p(k-1) - p(k-2) > p(k) - p(k-1) for all k >= 3.

The differences of a sequence are defined as if the sequence were increasing, so for example the differences of (6,3,1) are (-3,-2). Then a(n) is the number of integer partitions of n whose differences are strictly decreasing. The Heinz numbers of these partitions are given by A325457. Of course, the number of such integer partitions of n is also the number of reversed integer partitions of n whose differences are strictly decreasing, which is the author's interpretation. - Gus Wiseman, May 03 2019

Examples

			There are a(10) = 8 such partitions of 10:
01: [10]
02: [1, 9]
03: [2, 8]
04: [3, 7]
05: [4, 6]
06: [5, 5]
07: [1, 4, 5]
08: [2, 4, 4]
There are a(11) = 10 such partitions of 11:
01: [11]
02: [1, 10]
03: [2, 9]
04: [3, 8]
05: [4, 7]
06: [5, 6]
07: [1, 4, 6]
08: [1, 5, 5]
09: [2, 4, 5]
10: [3, 4, 4]
		

Links

• Fausto A. C. Cariboni, Table of n, a(n) for n = 0..2000
• Gus Wiseman, Sequences counting and ranking integer partitions by the differences of their successive parts.

Crossrefs

Cf. A049988, A240026, A240027, A320466, A320510, A325325, A325358, A325393, A325457.

Programs

Mathematica

Table[Length[Select[IntegerPartitions[n],Greater@@Differences[#]&]],{n,0,30}] (* Gus Wiseman, May 03 2019 *)

Ruby

def partition(n, min, max)
  return [[]] if n == 0
  [max, n].min.downto(min).flat_map{|i| partition(n - i, min, i).map{|rest| [i, *rest]}}
end
def f(n)
  return 1 if n == 0
  cnt = 0
  partition(n, 1, n).each{|ary|
    ary0 = (1..ary.size - 1).map{|i| ary[i - 1] - ary[i]}
    cnt += 1 if ary0.sort == ary0 && ary0.uniq == ary0
  }
  cnt
end
def A320470(n)
  (0..n).map{|i| f(i)}
end
p A320470(50)