A179269 Number of partitions of n into distinct parts such that the successive differences of consecutive parts are increasing, and first difference > first part.

1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 5, 5, 5, 7, 7, 7, 10, 10, 10, 13, 14, 14, 18, 19, 19, 23, 25, 25, 30, 32, 33, 38, 41, 42, 48, 52, 54, 60, 65, 67, 75, 81, 84, 92, 99, 103, 113, 121, 126, 136, 147, 153, 165, 177, 184, 197, 213, 221, 236, 253, 264, 280, 301, 313, 331, 355, 371, 390, 418, 435, 458

Offset: 0

Author

Joerg Arndt, Jan 05 2011

Keywords

• nonn

Comments

Conditions as in A179254; additionally, if more than 1 part, first difference > first part.

Equivalently, number of partitions for which the sequence of part counts by decreasing part size is 1, 2, 3, ... - Olivier Gérard, Jul 28 2017

Examples

			a(10) = 5 as there are 5 such partitions of 10: 1 + 3 + 6 = 1 + 9 = 2 + 8 = 3 + 7 = 10.
a(10) = 5 as there are 5 such partitions of 10: 10, 8 + 1 + 1, 6 + 2 + 2, 4 + 3 + 3, 3 + 2 + 2 + 1 + 1 + 1 (second definition).
From _Gus Wiseman_, May 04 2019: (Start)
The a(3) = 1 through a(13) = 7 partitions whose differences are strictly increasing (with the last part taken to be 0) are the following (A = 10, B = 11, C = 12, D = 13). The Heinz numbers of these partitions are given by A325460.
  (3)  (4)   (5)   (6)   (7)   (8)   (9)   (A)    (B)    (C)    (D)
       (31)  (41)  (51)  (52)  (62)  (72)  (73)   (83)   (93)   (94)
                         (61)  (71)  (81)  (82)   (92)   (A2)   (A3)
                                           (91)   (A1)   (B1)   (B2)
                                           (631)  (731)  (831)  (C1)
                                                                (841)
                                                                (931)
The a(3) = 1 through a(11) = 5 partitions whose multiplicities form an initial interval of positive integers are the following (A = 10, B = 11). The Heinz numbers of these partitions are given by A307895.
  (3)  (4)    (5)    (6)    (7)    (8)    (9)    (A)       (B)
       (211)  (311)  (411)  (322)  (422)  (522)  (433)     (533)
                            (511)  (611)  (711)  (622)     (722)
                                                 (811)     (911)
                                                 (322111)  (422111)
(End)
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 0..10000 (terms 0..200 from Seiichi Manyama)
• Gus Wiseman, Sequences counting and ranking integer partitions by the differences of their successive parts.

Crossrefs

Cf. A179254 (condition only on differences), A007294 (nondecreasing instead of strictly increasing), A179255, A320382, A320385, A320387, A320388.

Cf. A007862, A240027, A307895, A320509, A320510, A325324, A325357, A325391, A325460.

Programs

Mathematica

Table[Length@
  Select[IntegerPartitions[n],
   And @@ Equal[Range[Length[Split[#]]], Length /@ Split[#]] &], {n,
0, 40}]   (* Olivier Gérard, Jul 28 2017 *)
Table[Length[Select[IntegerPartitions[n],Less@@Differences[Append[#,0]]&]],{n,0,30}] (* Gus Wiseman, May 04 2019 *)

PARI

R(n)={my(L=List(), v=vectorv(n, i, 1), w=1, t=1); while(v, listput(L,v); w++; t+=w; v=vectorv(n, i, sum(k=1, (i-1)\t, L[w-1][i-k*t]))); Mat(L)}
seq(n)={my(M=R(n)); concat([1], vector(n, i, vecsum(M[i,])))} \\ Andrew Howroyd, Aug 27 2019

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.reverse && ary0.uniq == ary0
  }
  cnt
end
def A179269(n)
  (0..n).map{|i| f(i)}
end
p A179269(50) # Seiichi Manyama, Oct 12 2018

Sage

def A179269(n):
    has_increasing_diffs = lambda x: min(differences(x,2)) >= 1
    special = lambda x: (x[1]-x[0]) > x[0]
    allowed = lambda x: (len(x) < 2 or special(x)) and (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

Formula

G.f.: Sum_{k>=0} x^(k*(k+1)*(k+2)/6) / Product_{j=1..k} (1 - x^(j*(j+1)/2)) (conjecture). - Ilya Gutkovskiy, Apr 25 2019