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Number of partitions p of n such that floor(mean(p)) or ceiling(mean(p)) is a part.
5

%I #9 Apr 26 2014 21:14:49

%S 0,1,2,3,4,6,8,12,16,22,31,44,50,78,102,125,163,230,271,379,441,575,

%T 760,978,1073,1457,1865,2250,2704,3544,3955,5293,6154,7637,9533,11171,

%U 12702,16718,20215,23926,26949,34725,39187,49415,56914,66105,82244,98231

%N Number of partitions p of n such that floor(mean(p)) or ceiling(mean(p)) is a part.

%e a(6) counts these 8 partitions: 6, 33, 321, 3111, 222, 21111, 111111.

%t z = 30; f[n_] := f[n] = IntegerPartitions[n];

%t t1 = Table[Count[f[n], p_ /; MemberQ[p, Floor[Mean[p]]] && MemberQ[p, Ceiling[Mean[p]]]], {n, 0, z}] (* A241340 *)

%t t2 = Table[Count[f[n], p_ /; ! MemberQ[p, Floor[Mean[p]]] && MemberQ[p, Ceiling[Mean[p]]]], {n, 0, z}] (* A241341 *)

%t t3 = Table[Count[f[n], p_ /; MemberQ[p, Floor[Mean[p]]] && ! MemberQ[p, Ceiling[Mean[p]]]], {n, 0, z}] (* A241342 *)

%t t4 = Table[Count[f[n], p_ /; ! MemberQ[p, Floor[Mean[p]]] && ! MemberQ[p, Ceiling[Mean[p]]]], {n, 0, z}] (* A241343 *)

%t t5 = Table[Count[f[n], p_ /; MemberQ[p, Floor[Mean[p]]] || MemberQ[p, Ceiling[Mean[p]]]], {n, 0, z}] (* A241344 *)

%Y Cf. A241340, A241341, A241342, A241343.

%K nonn,easy

%O 0,3

%A _Clark Kimberling_, Apr 20 2014