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%I #54 Sep 21 2023 21:03:58
%S 1,2,2,4,2,8,2,11,9,14,2,46,2,24,51,66,2,126,2,202,144,69,2,632,194,
%T 116,381,756,2,1707,2,1417,956,316,2043,5295,2,511,2293,9151,2,10278,
%U 2,8409,14671,1280,2,36901,8035,21524,11614,25639,2,53138,39810,85004
%N Number of partitions of n in which the number of parts divides n.
%C Also sum of p(n,d) over the divisors d of n, where p(n,m) is the count of partitions of n in exactly m parts. - _Wouter Meeussen_, Jun 07 2009
%C From _Gus Wiseman_, Sep 24 2019: (Start)
%C Also the number of integer partitions of n whose maximum part divides n. The Heinz numbers of these partitions are given by A326836. For example, the a(1) = 1 through a(8) = 11 partitions are:
%C (1) (2) (3) (4) (5) (6) (7) (8)
%C (11) (111) (22) (11111) (33) (1111111) (44)
%C (211) (222) (422)
%C (1111) (321) (431)
%C (2211) (2222)
%C (3111) (4211)
%C (21111) (22211)
%C (111111) (41111)
%C (221111)
%C (2111111)
%C (11111111)
%C (End)
%H Chai Wah Wu, <a href="/A067538/b067538.txt">Table of n, a(n) for n = 1..10000</a> (n = 1..500 from Wouter Meeussen, n = 501..1000 from Alois P. Heinz, n = 1001..5000 from David A. Corneth)
%H Eric W. Weisstein, <a href="http://mathworld.wolfram.com/PartitionFunctionP.html">Partition Function P</a>
%H Wikipedia, <a href="https://www.wikipedia.org/wiki/integer_partition">Integer Partition</a>
%F a(p) = 2 for all primes p.
%e a(3)=2 because 3 is a prime; a(4)=4 because the five partitions of 4 are {4}, {3, 1}, {2, 2}, {2, 1, 1}, {1, 1, 1, 1}, and the number of parts in each of them divides 4 except for {2, 1, 1}.
%e From _Gus Wiseman_, Sep 24 2019: (Start)
%e The a(1) = 1 through a(8) = 11 partitions whose length divides their sum are the following. The Heinz numbers of these partitions are given by A316413.
%e (1) (2) (3) (4) (5) (6) (7) (8)
%e (11) (111) (22) (11111) (33) (1111111) (44)
%e (31) (42) (53)
%e (1111) (51) (62)
%e (222) (71)
%e (321) (2222)
%e (411) (3221)
%e (111111) (3311)
%e (4211)
%e (5111)
%e (11111111)
%e (End)
%t Do[p = IntegerPartitions[n]; l = Length[p]; c = 0; k = 1; While[k < l + 1, If[ IntegerQ[ n/Length[ p[[k]] ]], c++ ]; k++ ]; Print[c], {n, 1, 57}, All]
%t p[n_,k_]:=p[n,k]=p[n-1,k-1]+p[n-k,k];p[n_,k_]:=0/;k>n;p[n_,n_]:=1;p[n_,0]:=0
%t Table[Plus @@ (p[n,# ]&/ @ Divisors[n]),{n,36}] (* _Wouter Meeussen_, Jun 07 2009 *)
%t Table[Count[IntegerPartitions[n], q_ /; IntegerQ[Mean[q]]], {n, 50}] (*_Clark Kimberling_, Apr 23 2019 *)
%o (PARI) a(n) = {my(nb = 0); forpart(p=n, if ((vecsum(Vec(p)) % #p) == 0, nb++);); nb;} \\ _Michel Marcus_, Jul 03 2018
%o (Python)
%o # uses A008284_T
%o from sympy import divisors
%o def A067538(n): return sum(A008284_T(n,d) for d in divisors(n,generator=True)) # _Chai Wah Wu_, Sep 21 2023
%Y Cf. A000005, A000041, A143773, A298422, A298423, A298426.
%Y The strict case is A102627.
%Y Partitions with integer geometric mean are A067539.
%Y Cf. A018818, A102627, A316413, A326622, A326836, A326843, A326850.
%K easy,nonn
%O 1,2
%A _Naohiro Nomoto_, Jan 27 2002
%E Extended by _Robert G. Wilson v_, Oct 16 2002
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