A240056 Number of partitions of n such that m(1) > m(2), where m = multiplicity.

0, 1, 1, 1, 3, 4, 5, 9, 12, 16, 24, 32, 42, 59, 77, 100, 134, 173, 221, 288, 366, 463, 590, 741, 926, 1163, 1444, 1787, 2215, 2726, 3342, 4101, 5003, 6087, 7402, 8964, 10827, 13069, 15718, 18865, 22617, 27041, 32263, 38453, 45719, 54264, 64326, 76102, 89884

Offset: 0

Author

Clark Kimberling, Mar 31 2014

Keywords

• nonn
• easy

Examples

			a(7) counts these 9 partitions: 61, 511, 4111, 331, 3211, 31111, 22111, 211111, 1111111.
		

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..10000 (terms 0..1000 from Alois P. Heinz)

Crossrefs

Cf. A182713, A174455, A000041.

Programs

Mathematica

z = 60; f[n_] := f[n] = IntegerPartitions[n]; t1 = Table[Count[f[n], p_ /; Count[p, 1] < Count[p, 2]], {n, 0, z}]  (* A182713 *)
t2 = Table[Count[f[n], p_ /; Count[p, 1] <= Count[p, 2]], {n, 0, z}] (* A182713(n+2) *)
t3 = Table[Count[f[n], p_ /; Count[p, 1] == Count[p, 2]], {n, 0, z}] (* A174455 *)
t4 = Table[Count[f[n], p_ /; Count[p, 1] > Count[p, 2]], {n, 0, z}]  (* A240056 *)
t5 = Table[Count[f[n], p_ /; Count[p, 1] >= Count[p, 2]], {n, 0, z}] (* A240056(n+1) *)

Formula

a(n) = A000041(n) - A182713(n+2) = a(n+1) - A174455(n) for n >= 0.

a(n) ~ exp(sqrt(2*n/3)*Pi) / (2 * 3^(3/2) * n). - Vaclav Kotesovec, Jan 15 2022