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Number of integer partitions of n satisfying (length) = (mean). Partitions of n into sqrt(n) parts.
1

%I #9 Jul 07 2023 23:08:56

%S 1,1,0,0,2,0,0,0,0,7,0,0,0,0,0,0,34,0,0,0,0,0,0,0,0,192,0,0,0,0,0,0,0,

%T 0,0,0,1206,0,0,0,0,0,0,0,0,0,0,0,0,8033,0,0,0,0,0,0,0,0,0,0,0,0,0,0,

%U 55974,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0

%N Number of integer partitions of n satisfying (length) = (mean). Partitions of n into sqrt(n) parts.

%F a(n^2) = A206240(n).

%e The a(0) = 1 through a(9) = 7 partitions:

%e () (1) . . (22) . . . . (333)

%e (31) (432)

%e (441)

%e (522)

%e (531)

%e (621)

%e (711)

%t Table[Length[If[n==0,{{}},Select[IntegerPartitions[n],Mean[#]==Length[#]&]]],{n,0,30}]

%Y The strict case is A107379(sqrt(n)).

%Y Without zeros we have A206240.

%Y These partitions have ranks A363951.

%Y A008284 counts partitions by length, A058398 by mean.

%Y A067538 counts partitions with integer mean, ranks A316413.

%Y Cf. A025065, A026905, A237984, A327472, A327482, A363944, A363949.

%K nonn

%O 0,5

%A _Gus Wiseman_, Jul 07 2023