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

1,4

Comments

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

Links

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

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

Example

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 + ...

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 *)

Prog

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

Crossrefs

Cf. A240691, A000041.

Sequence in context: A215914 A006049 A084541 * A339593 A113050 A278180

Adjacent sequences: A240687 A240688 A240689 * A240691 A240692 A240693

Keyword

nonn,easy

Author

Clark Kimberling, Apr 11 2014

Status

approved