login
Number of partitions of n for which (number of occurrences of the least part) = (number of occurrences of greatest part).
3

%I #10 Jun 24 2015 00:44:57

%S 1,2,3,4,4,8,6,11,12,18,16,32,27,44,51,70,73,114,116,169,192,250,282,

%T 391,432,559,657,831,952,1223,1395,1747,2040,2492,2910,3580,4130,5007,

%U 5857,7040,8171,9813,11372,13550,15771,18625,21611,25549,29540,34706

%N Number of partitions of n for which (number of occurrences of the least part) = (number of occurrences of greatest part).

%C The partitions of n are partitioned by the partitions counted by A236543, A236544, A236545 (see Example); consequently, A000041(n) = A236543(n) + A236544(n) + A236545(n) for n >= 1.

%e Among the 15 partitions of 7, the following 6 have #(occurrences of least part) = #(occurrences of greatest part): 7, 61, 52, 43, 421, 111111; the following 7 have " > " in place of " = ": 511, 4111, 322, 3211, 31111, 22111, 211111; and the remaining 2, have " < ": 331, 221.

%t z = 65; s = Map[Map[Length, {Select[#, First[#] == Last[#] &], Select[#, First[#] > Last[#] &], Select[#, First[#] < Last[#] &]} &[Map[{Count[#, Min[#]], Count[#, Max[#]]} &, IntegerPartitions[#]]]] &, Range[z]]; t = Flatten[s];

%t t1 = Table[t[[3 k - 2]], {k, 1, z}] (* A236543 *)

%t t2 = Table[t[[3 k - 1]], {k, 1, z}] (* A236544 *)

%t t3 = Table[t[[3 k]], {k, 1, z}] (* A236545 *)

%t (* _Peter J. C. Moses_, Jan 28 2014 *)

%Y Cf. A236544, A236545, A000041.

%K nonn,easy

%O 1,2

%A _Clark Kimberling_, Jan 28 2014