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A320388 Number of partitions of n into distinct parts such that the successive differences of consecutive parts are decreasing.

%I A320388 #36 Jan 08 2021 15:23:22
%S A320388 1,1,1,2,2,3,3,4,5,5,6,8,7,9,11,10,12,15,14,16,19,18,21,25,23,26,31,
%T A320388 29,33,38,36,40,46,44,49,56,53,58,66,64,70,77,76,82,92,89,96,106,104,
%U A320388 113,123,120,130,142,141,149,162,160,172,186,184,195,211,210,223,238
%N A320388 Number of partitions of n into distinct parts such that the successive differences of consecutive parts are decreasing.
%C A320388 Partitions are usually written with parts in descending order, but the conditions are easier to check "visually" if written in ascending order.
%C A320388 Partitions into distinct parts (p(1), p(2), ..., p(m)) such that p(k-1) - p(k-2) >  p(k) - p(k-1) for all k >= 3.
%H A320388 Fausto A. C. Cariboni, <a href="/A320388/b320388.txt">Table of n, a(n) for n = 0..2000</a> (terms 0..100 from Seiichi Manyama)
%e A320388 There are a(17) = 15 such partitions of 17:
%e A320388   01: [17]
%e A320388   02: [1, 16]
%e A320388   03: [2, 15]
%e A320388   04: [3, 14]
%e A320388   05: [4, 13]
%e A320388   06: [5, 12]
%e A320388   07: [6, 11]
%e A320388   08: [7, 10]
%e A320388   09: [1, 6, 10]
%e A320388   10: [8, 9]
%e A320388   11: [1, 7, 9]
%e A320388   12: [2, 6, 9]
%e A320388   13: [2, 7, 8]
%e A320388   14: [3, 6, 8]
%e A320388   15: [4, 6, 7]
%e A320388 There are a(18) = 14 such partitions of 18:
%e A320388   01: [18]
%e A320388   02: [1, 17]
%e A320388   03: [2, 16]
%e A320388   04: [3, 15]
%e A320388   05: [4, 14]
%e A320388   06: [5, 13]
%e A320388   07: [6, 12]
%e A320388   08: [7, 11]
%e A320388   09: [8, 10]
%e A320388   10: [1, 7, 10]
%e A320388   11: [1, 8, 9]
%e A320388   12: [2, 7, 9]
%e A320388   13: [3, 7, 8]
%e A320388   14: [1, 4, 6, 7]
%o A320388 (Ruby)
%o A320388 def partition(n, min, max)
%o A320388   return [[]] if n == 0
%o A320388   [max, n].min.downto(min).flat_map{|i| partition(n - i, min, i - 1).map{|rest| [i, *rest]}}
%o A320388 end
%o A320388 def f(n)
%o A320388   return 1 if n == 0
%o A320388   cnt = 0
%o A320388   partition(n, 1, n).each{|ary|
%o A320388     ary0 = (1..ary.size - 1).map{|i| ary[i - 1] - ary[i]}
%o A320388     cnt += 1 if ary0.sort == ary0 && ary0.uniq == ary0
%o A320388   }
%o A320388   cnt
%o A320388 end
%o A320388 def A320388(n)
%o A320388   (0..n).map{|i| f(i)}
%o A320388 end
%o A320388 p A320388(50)
%Y A320388 Cf. A007294, A179254, A179255, A179269, A320382, A320385, A320387.
%Y A320388 Cf. A081489.
%K A320388 nonn
%O A320388 0,4
%A A320388 _Seiichi Manyama_, Oct 12 2018