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Number of partitions p of n such that (# 1s in p) = (#1s in conjugate(p)).
2

%I #12 Jun 02 2018 11:01:19

%S 1,0,1,1,1,3,1,6,2,10,4,17,7,27,14,41,25,63,42,95,70,140,113,207,176,

%T 302,272,436,411,628,610,897,897,1270,1303,1791,1869,2509,2661,3492,

%U 3753,4838,5249,6665,7294,9130,10066,12453,13799,16902,18815,22831,25511

%N Number of partitions p of n such that (# 1s in p) = (#1s in conjugate(p)).

%F a(n) = A240690(n+1) - A240690(n) for n >= 1.

%F a(n) + 2*A240690(n) = A000041(n) for n >= 1.

%F a(n) + a(n-1) = A002865(n) = A187219(n) = A085811(n+3), in all cases for n >= 2. - _Omar E. Pol_, Mar 07 2015

%F a(n) ~ exp(Pi*sqrt(2*n/3)) * Pi / (24*sqrt(2)*n^(3/2)). - _Vaclav Kotesovec_, Jun 02 2018

%e a(6) counts these 3 partitions: 33, 321, 222, of which the respective conjugates are 222, 321, 33.

%t z = 53; f[n_] := f[n] = IntegerPartitions[n]; c[p_] := Table[Count[#, _?(# >= i &)], {i, First[#]}] &[p]; (* conjugate of partition p *)

%t Table[Count[f[n], p_ /; Count[p, 1] < Count[c[p], 1]], {n, 1, z}] (* A240690 *)

%t Table[Count[f[n], p_ /; Count[p, 1] <= Count[c[p], 1]], {n, 1, z}] (* A240690(n+1) *)

%t Table[Count[f[n], p_ /; Count[p, 1] == Count[c[p], 1]], {n, 1, z}] (* A240691 *)

%Y Cf. A240690, A000041.

%K nonn,easy

%O 1,6

%A _Clark Kimberling_, Apr 11 2014