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%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
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