A240691 Number of partitions p of n such that (# 1s in p) = (#1s in conjugate(p)).

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, 302, 272, 436, 411, 628, 610, 897, 897, 1270, 1303, 1791, 1869, 2509, 2661, 3492, 3753, 4838, 5249, 6665, 7294, 9130, 10066, 12453, 13799, 16902, 18815, 22831, 25511

Offset: 1

Author

Clark Kimberling, Apr 11 2014

Keywords

• nonn
• easy

Examples

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

Crossrefs

Cf. A240690, A000041.

Programs

Mathematica

z = 53; f[n_] := f[n] = IntegerPartitions[n]; c[p_] := Table[Count[#, _?(# >= i &)], {i, First[#]}] &[p];  (* conjugate of partition p *)
Table[Count[f[n], p_ /; Count[p, 1] < Count[c[p], 1]], {n, 1, z}]  (* A240690 *)
Table[Count[f[n], p_ /; Count[p, 1] <= Count[c[p], 1]], {n, 1, z}]  (* A240690(n+1) *)
Table[Count[f[n], p_ /; Count[p, 1] == Count[c[p], 1]], {n, 1, z}] (* A240691 *)

Formula

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

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

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

a(n) ~ exp(Pi*sqrt(2*n/3)) * Pi / (24*sqrt(2)*n^(3/2)). - Vaclav Kotesovec, Jun 02 2018