%I
%S 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,
%T 39,37,42,48,47,50,59,56,63,70,68,74,83,82,89,97,97,104,116,113,123,
%U 133,133,142,155,153,166,178,178,189,204,204,218,232,235,247,265,265,283,299
%N Number of partitions of n such that the successive differences of consecutive parts are strictly decreasing.
%C Partitions are usually written with parts in descending order, but the conditions are easier to check "visually" if written in ascending order.
%C Partitions (p(1), p(2), ..., p(m)) such that p(k1)  p(k2) > p(k)  p(k1) for all k >= 3.
%C 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
%H Fausto A. C. Cariboni, <a href="/A320470/b320470.txt">Table of n, a(n) for n = 0..2000</a>
%H Gus Wiseman, <a href="/A325325/a325325.txt">Sequences counting and ranking integer partitions by the differences of their successive parts.</a>
%e There are a(10) = 8 such partitions of 10:
%e 01: [10]
%e 02: [1, 9]
%e 03: [2, 8]
%e 04: [3, 7]
%e 05: [4, 6]
%e 06: [5, 5]
%e 07: [1, 4, 5]
%e 08: [2, 4, 4]
%e There are a(11) = 10 such partitions of 11:
%e 01: [11]
%e 02: [1, 10]
%e 03: [2, 9]
%e 04: [3, 8]
%e 05: [4, 7]
%e 06: [5, 6]
%e 07: [1, 4, 6]
%e 08: [1, 5, 5]
%e 09: [2, 4, 5]
%e 10: [3, 4, 4]
%t Table[Length[Select[IntegerPartitions[n],Greater@@Differences[#]&]],{n,0,30}] (* _Gus Wiseman_, May 03 2019 *)
%o (Ruby)
%o def partition(n, min, max)
%o return [[]] if n == 0
%o [max, n].min.downto(min).flat_map{i partition(n  i, min, i).map{rest [i, *rest]}}
%o end
%o def f(n)
%o return 1 if n == 0
%o cnt = 0
%o partition(n, 1, n).each{ary
%o ary0 = (1..ary.size  1).map{i ary[i  1]  ary[i]}
%o cnt += 1 if ary0.sort == ary0 && ary0.uniq == ary0
%o }
%o cnt
%o end
%o def A320470(n)
%o (0..n).map{i f(i)}
%o end
%o p A320470(50)
%Y Cf. A049988, A240026, A240027, A320466, A320510, A325325, A325358, A325393, A325457.
%K nonn
%O 0,3
%A _Seiichi Manyama_, Oct 13 2018
