A320387 Number of partitions of n into distinct parts such that the successive differences of consecutive parts are nonincreasing, and first difference <= first part.

1, 1, 1, 2, 1, 2, 3, 2, 2, 4, 3, 4, 5, 3, 5, 7, 4, 7, 8, 6, 8, 11, 7, 9, 13, 9, 11, 16, 12, 15, 18, 13, 17, 20, 17, 21, 24, 19, 24, 30, 22, 28, 34, 26, 34, 38, 30, 37, 43, 37, 42, 48, 41, 50, 58, 48, 55, 64, 53, 64, 71, 59, 73, 81, 69, 79, 89, 79, 90, 101, 87, 100, 111

Offset: 0

Author

Seiichi Manyama, Oct 12 2018

Keywords

• nonn

Comments

Partitions are usually written with parts in descending order, but the conditions are easier to check "visually" if written in ascending order.

Generating function of the "second integrals" of partitions: given a partition (p_1, ..., p_s) written in weakly decreasing order, write the sequence B = (b_1, b_2, ..., b_s) = (p_1, p_1 + p_2, ..., p_1 + ... + p_s). The sequence gives the coefficients of the generating function summing q^(b_1 + ... + b_s) over all partitions of all nonnegative integers. - William J. Keith, Apr 23 2022

From Gus Wiseman, Jan 17 2023: (Start)

Equivalently, a(n) is the number of multisets (weakly increasing sequences of positive integers) with weighted sum n. For example, the Heinz numbers of the a(0) = 1 through a(15) = 7 multisets are:

1 2 3 4 7 6 8 10 15 12 16 18 20 26 24 28

5 11 9 17 19 14 21 22 27 41 30 32

13 23 29 31 33 55 39 34

25 35 37 43 45

49 77 47

65

121

These multisets are counted by A264034. The reverse version is A007294. The zero-based version is A359678.

(End)

Examples

			There are a(29) = 15 such partitions of 29:
  01: [29]
  02: [10, 19]
  03: [11, 18]
  04: [12, 17]
  05: [13, 16]
  06: [14, 15]
  07: [5, 10, 14]
  08: [6, 10, 13]
  09: [6, 11, 12]
  10: [7, 10, 12]
  11: [8, 10, 11]
  12: [3, 6, 9, 11]
  13: [5, 7, 8, 9]
  14: [2, 4, 6, 8, 9]
  15: [3, 5, 6, 7, 8]
There are a(30) = 18 such partitions of 30:
  01: [30]
  02: [10, 20]
  03: [11, 19]
  04: [12, 18]
  05: [13, 17]
  06: [14, 16]
  07: [5, 10, 15]
  08: [6, 10, 14]
  09: [6, 11, 13]
  10: [7, 10, 13]
  11: [7, 11, 12]
  12: [8, 10, 12]
  13: [3, 6, 9, 12]
  14: [9, 10, 11]
  15: [4, 7, 9, 10]
  16: [2, 4, 6, 8, 10]
  17: [6, 7, 8, 9]
  18: [4, 5, 6, 7, 8]
		

Links

• Fausto A. C. Cariboni, Table of n, a(n) for n = 0..2000 (terms 0..300 from Seiichi Manyama)

Crossrefs

Cf. A007294, A179254, A179255, A179269, A320382, A320385, A320388.

Number of appearances of n > 0 in A304818, reverse A318283.

A053632 counts compositions by weighted sum.

A358194 counts partitions by weighted sum, reverse A264034.

Weighted sum of prime indices: A359497, A359676, A359682, A359754, A359755.

Cf. A029931, A359361, A359397, A359674, A359677, A359678.

Programs

Mathematica

prix[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
ots[y_]:=Sum[i*y[[i]],{i,Length[y]}];
Table[Length[Select[Range[2^n],ots[prix[#]]==n&]],{n,10}] (* Gus Wiseman, Jan 17 2023 *)

PARI

seq(n)={Vec(sum(k=1, (sqrtint(8*n+1)+1)\2, my(t=binomial(k,2)); x^t/prod(j=1, k-1, 1 - x^(t-binomial(j,2)) + O(x^(n-t+1)))))} \\ Andrew Howroyd, Jan 22 2023

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
  }
  cnt
end
def A320387(n)
  (0..n).map{|i| f(i)}
end
p A320387(50)

Formula

G.f.: Sum_{k>=1} x^binomial(k,2)/Product_{j=1..k-1} (1 - x^(binomial(k,2)-binomial(j,2))). - Andrew Howroyd, Jan 22 2023