A240690 Number of partitions p of n such that p contains fewer 1s than its conjugate.

0, 1, 1, 2, 3, 4, 7, 8, 14, 16, 26, 30, 47, 54, 81, 95, 136, 161, 224, 266, 361, 431, 571, 684, 891, 1067, 1369, 1641, 2077, 2488, 3116, 3726, 4623, 5520, 6790, 8093, 9884, 11753, 14262, 16923, 20415, 24168, 29006, 34255, 40920, 48214, 57344, 67410, 79863

Offset: 1

Author

Clark Kimberling, Apr 11 2014

Keywords

• nonn
• easy

Comments

a(n+1) = number of partitions p of n such that (# 1s in p) <= (#1s in conjugate(p)).

Examples

			a(6) counts these 4 partitions: 6, 51, 42, 411, of which the respective conjugates are 111111, 21111, 2211, 3111.
G.f. = x^2 + x^3 + 2*x^4 + 3*x^5 + 4*x^6 + 7*x^7 + 8*x^8 + 14*x^9 + 16*x^10 + ...
		

Links

• G. C. Greubel, Table of n, a(n) for n = 1..2500

Crossrefs

Cf. A240691, 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 *)
a[ n_] := SeriesCoefficient[ (-1 + 1 / QPochhammer[ x]) x / (1 + x), {x, 0, n}]; (* Michael Somos, Mar 16 2015 *)

PARI

q='q+O('q^60); concat([0], Vec((-1 + 1/eta(q))*q/(1+q))) \\ G. C. Greubel, Aug 07 2018

Formula

2*a(n) + A240691(n) = A000041(n) for n >= 1.

a(n) + a(n+1) = A000041(n). - Omar E. Pol, Mar 07 2015

G.f.: (-1 + Product_{k>0} (1 - x^k)^(-1)) * x / (1 + x). - Michael Somos, Mar 16 2015

a(n) ~ exp(Pi*sqrt(2*n/3)) / (8*n*sqrt(3)). - Vaclav Kotesovec, Jun 02 2018