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Number of integer partitions of n whose product of parts is >= n.
17

%I #27 May 11 2021 06:16:31

%S 1,1,1,1,2,2,5,7,13,18,28,40,60,80,113,152,205,266,353,454,590,751,

%T 959,1210,1529,1905,2381,2953,3658,4501,5539,6772,8278,10065,12230,

%U 14801,17893,21544,25921,31089,37240,44478,53068,63150,75063,89018,105438,124632

%N Number of integer partitions of n whose product of parts is >= n.

%H Alois P. Heinz, <a href="/A319005/b319005.txt">Table of n, a(n) for n = 0..1000</a>

%H Pankaj Jyoti Mahanta, <a href="https://arxiv.org/abs/2010.07353">On the number of partitions of n whose product of the summands is at most n</a>, arXiv:2010.07353 [math.CO], 2020.

%e The a(1) = 1 through a(9) = 18 partitions:

%e (1) (2) (3) (4) (5) (6) (7) (8) (9)

%e (22) (32) (33) (43) (44) (54)

%e (42) (52) (53) (63)

%e (222) (322) (62) (72)

%e (321) (331) (332) (333)

%e (421) (422) (432)

%e (2221) (431) (441)

%e (521) (522)

%e (2222) (531)

%e (3221) (621)

%e (3311) (3222)

%e (4211) (3321)

%e (22211) (4221)

%e (4311)

%e (5211)

%e (22221)

%e (32211)

%e (33111)

%p b:= proc(n, i, p) option remember; `if`(n=0 or i=1, `if`(p>1,

%p 0, 1), b(n, i-1, p) +b(n-i, min(i, n-i), max(p/i, 1)))

%p end:

%p a:= n-> b(n$3):

%p seq(a(n), n=0..50); # _Alois P. Heinz_, Oct 22 2018

%t Table[Length[Select[IntegerPartitions[n],Times@@#>=n&]],{n,50}]

%t (* Second program: *)

%t b[n_, i_, p_] := b[n, i, p] = If[n == 0 || i == 1, If[p > 1, 0, 1],

%t b[n, i - 1, p] + b[n - i, Min[i, n - i], Max[p/i, 1]]];

%t a[n_] := b[n, n, n];

%t a /@ Range[0, 50] (* _Jean-François Alcover_, May 11 2021, after _Alois P. Heinz_ *)

%Y Column sums of A319000.

%Y Cf. A001055, A002865, A069016, A096276, A301987, A318950, A319057, A319916.

%K nonn

%O 0,5

%A _Gus Wiseman_, Oct 22 2018