A179254 Number of partitions of n into distinct parts such that the successive differences of consecutive parts are strictly increasing.
1, 1, 1, 2, 2, 3, 3, 5, 5, 6, 8, 9, 9, 13, 14, 15, 19, 21, 22, 28, 30, 32, 39, 42, 44, 54, 58, 61, 72, 77, 82, 96, 102, 108, 124, 133, 141, 160, 171, 180, 203, 218, 230, 256, 273, 289, 320, 342, 361, 395, 423, 447, 486, 520, 548, 594, 635, 669, 721, 769, 811, 871, 928, 978, 1044, 1114
Offset: 0
Keywords
Examples
There are a(17) = 21 such partitions of 17: 01: [ 1 2 4 10 ] 02: [ 1 2 5 9 ] 03: [ 1 2 14 ] 04: [ 1 3 13 ] 05: [ 1 4 12 ] 06: [ 1 5 11 ] 07: [ 1 16 ] 08: [ 2 3 12 ] 09: [ 2 4 11 ] 10: [ 2 5 10 ] 11: [ 2 15 ] 12: [ 3 4 10 ] 13: [ 3 5 9 ] 14: [ 3 14 ] 15: [ 4 5 8 ] 16: [ 4 13 ] 17: [ 5 12 ] 18: [ 6 11 ] 19: [ 7 10 ] 20: [ 8 9 ] 21: [ 17 ] - _Joerg Arndt_, Mar 31 2014
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..1000 (first 225 terms from Joerg Arndt)
Crossrefs
Programs
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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 - 1).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.reverse && ary0.uniq == ary0 } cnt end def A179254(n) (0..n).map{|i| f(i)} end p A179254(50) # Seiichi Manyama, Oct 12 2018
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Sage
def A179254(n): has_increasing_diffs = lambda x: min(differences(x,2)) >= 1 allowed = lambda x: len(x) < 3 or has_increasing_diffs(x) return len([x for x in Partitions(n,max_slope=-1) if allowed(x[::-1])]) # D. S. McNeil, Jan 06 2011
Comments