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Number of partitions of n such that (greatest part) <= (multiplicity of least part).
3

%I #9 Apr 12 2014 16:21:52

%S 1,1,1,1,3,2,5,4,8,10,13,15,25,25,37,46,61,70,97,112,150,177,224,270,

%T 347,407,508,611,754,895,1106,1304,1594,1892,2283,2708,3262,3835,4595,

%U 5421,6452,7574,8993,10530,12445,14564,17123,19992,23465,27302,31931

%N Number of partitions of n such that (greatest part) <= (multiplicity of least part).

%F a(n) = A240178(n) + A240183(n), for n >= 1.

%F a(n) + A240179(n) = A000041(n) for n >= 0.

%e a(8) counts these 8 partitions: 41111, 32111, 311111, 2222, 22211, 221111, 2111111, 11111111.

%t z = 60; f[n_] := f[n] = IntegerPartitions[n];

%t t1 = Table[Count[f[n], p_ /; Max[p] < Count[p, Min[p]]], {n, 0, z}] (* A240178 except for n=0 *)

%t t2 = Table[Count[f[n], p_ /; Max[p] <= Count[p, Min[p]]], {n, 0, z}] (* A240182 *)

%t t3 = Table[Count[f[n], p_ /; Max[p] == Count[p, Min[p]]], {n, 0, z}] (* A240183 *)

%t t4 = Table[Count[f[n], p_ /; Max[p] > Count[p, Min[p]]], {n, 0, z}] (* A240184 *)

%t t5 = Table[Count[f[n], p_ /; Max[p] >= Count[p, Min[p]]], {n, 0, z}] (* A240179 *)

%Y Cf. A240178, A240183, A240184, A240179, A000041.

%K nonn,easy

%O 0,5

%A _Clark Kimberling_, Apr 02 2014