A320385 Number of partitions of n into distinct parts such that the successive differences of consecutive parts are decreasing, and first difference < first part.

1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 3, 2, 3, 4, 3, 3, 5, 3, 5, 6, 4, 5, 7, 5, 7, 8, 6, 7, 10, 8, 9, 11, 8, 11, 13, 9, 13, 15, 12, 14, 17, 13, 16, 20, 15, 18, 22, 18, 21, 25, 20, 23, 27, 23, 28, 30, 26, 30, 34, 30, 33, 38, 31, 38, 43, 36, 42, 46, 42, 47, 50, 45, 50, 58, 51, 55

Offset: 0

Author

Seiichi Manyama, Oct 12 2018

Keywords

• nonn

Examples

			There are a(29) = 10 such partitions of 29:
01: [29]
02: [10, 19]
03: [11, 18]
04: [12, 17]
05: [13, 16]
06: [14, 15]
07: [6, 10, 13]
08: [6, 11, 12]
09: [7, 10, 12]
10: [8, 10, 11]
There are a(30) = 8 such partitions of 30:
01: [30]
02: [11, 19]
03: [12, 18]
04: [13, 17]
05: [14, 16]
06: [6, 11, 13]
07: [7, 11, 12]
08: [4, 7, 9, 10]
		

Links

• Fausto A. C. Cariboni, Table of n, a(n) for n = 0..2000 (terms 0..300 from Seiichi Manyama)

Crossrefs

Cf. A007294, A179254, A179255, A179269, A320382, A320387.

Programs

Ruby

def partition(n, min, max)
  return [[]] if n == 0
  [max, n].min.downto(min).flat_map{|i| partition(n - i, min, i - 1).map{|rest| [i, *rest]}}
end
def f(n)
  return 1 if n == 0
  cnt = 0
  partition(n, 1, n).each{|ary|
    ary << 0
    ary0 = (1..ary.size - 1).map{|i| ary[i - 1] - ary[i]}
    cnt += 1 if ary0.sort == ary0 && ary0.uniq == ary0
  }
  cnt
end
def A320385(n)
  (0..n).map{|i| f(i)}
end
p A320385(50)