A320509 Number of partitions of n such that the successive differences of consecutive parts are nonincreasing, and first difference <= first part.
1, 1, 2, 3, 3, 4, 6, 4, 6, 8, 7, 8, 11, 7, 12, 14, 10, 13, 19, 12, 18, 21, 16, 19, 27, 19, 25, 30, 25, 30, 37, 25, 35, 40, 35, 42, 49, 35, 49, 56, 46, 54, 66, 50, 65, 72, 60, 70, 83, 68, 84, 90, 80, 94, 110, 86, 107, 116, 98, 119, 137, 111, 134, 146, 130, 148, 165, 141, 169
Offset: 0
Keywords
Examples
There are a(11) = 8 such partitions of 11: 01: [11] 02: [4, 7] 03: [5, 6] 04: [2, 4, 5] 05: [3, 4, 4] 06: [2, 3, 3, 3] 07: [1, 2, 2, 2, 2, 2] 08: [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1] There are a(12) = 11 such partitions of 12: 01: [12] 02: [4, 8] 03: [5, 7] 04: [6, 6] 05: [2, 4, 6] 06: [3, 4, 5] 07: [4, 4, 4] 08: [3, 3, 3, 3] 09: [1, 2, 3, 3, 3] 10: [2, 2, 2, 2, 2, 2] 11: [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
Links
- Fausto A. C. Cariboni, Table of n, a(n) for n = 0..2000 (terms 0..300 from Seiichi Manyama)
- Gus Wiseman, Sequences counting and ranking integer partitions by the differences of their successive parts.
Crossrefs
Programs
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Mathematica
Table[Length[Select[IntegerPartitions[n],GreaterEqual@@Differences[Append[#,0]]&]],{n,0,30}] (* Gus Wiseman, May 03 2019 *)
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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| ary << 0 ary0 = (1..ary.size - 1).map{|i| ary[i - 1] - ary[i]} cnt += 1 if ary0.sort == ary0 } cnt end def A320509(n) (0..n).map{|i| f(i)} end p A320509(50)
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