

A320509


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


2



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
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OFFSET

0,3


COMMENTS

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


LINKS

Seiichi Manyama, Table of n, a(n) for n = 0..300


EXAMPLE

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]


PROG

(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)


CROSSREFS

Cf. A240026, A240027, A320466, A320470, A320510.
Cf. A320387 (distinct parts, nonincreasing, and first difference <= first part).
Sequence in context: A200763 A203291 A220053 * A159999 A003977 A003971
Adjacent sequences: A320506 A320507 A320508 * A320510 A320511 A320512


KEYWORD

nonn


AUTHOR

Seiichi Manyama, Oct 14 2018


STATUS

approved



