A371251 Number of genus 2 partitions of the set [3n] into n blocks of length 3.

1, 144, 6046, 149674, 2771028, 42679084, 578872364, 7153349724, 82324041285, 895669007200, 9311524010712, 93235420275816, 904560813228072, 8543205886920516, 78838778199275032, 713005588584772184, 6334935141516816267, 55407394283320881984

Offset: 2

Author

Robert Coquereaux, Mar 16 2024

Keywords

• nonn

Comments

Call C(p, [alpha], g) the number of partitions of the cyclically ordered set [p], of cyclic type [alpha], and of genus g (genus g Faa di Bruno coefficients of type [alpha]). The number C(3n, [3^n], g) of genus g partitions of the set [3n] into n blocks of length 3 is given by A001764 when g=0 (non-crossing partitions), by [Zuber] when g = 1 (see A371250) and g=2 (this sequence). One has C(n=6,[3^2],2) = 1, C(n=9, [3^3], 2) = 144, etc.

Examples

			a(2) = 1.
G.f. = X^2 + 144*X^3 + 6046*X^4 + 149674*X^5 + 2771028*X^6 + ...
		

Links

• Robert Coquereaux and Jean-Bernard Zuber, Counting partitions by genus: a compendium of results, Journal of Integer Sequences, Vol. 27 (2024), Article 24.2.6. See p. 27; preprint, arXiv:2305.01100 [math.CO], 2023. See p. 27.
• Jean-Bernard Zuber, Counting partitions by genus. I. Genus 0 to 2, Enumer. Comb. Appl. 4 (2) (2024) #S2R13. See pp. 16-19; preprint, arXiv:2303.05875 [math.CO], 2023. See pp. 16-19.

Crossrefs

Cf. A001764 for C(3n,[3^n],0) and A371250 for C(3n,[2^n],1).

Programs

Mathematica

Table[SeriesCoefficient[(( 16 (16321 - 21668 Cos[2 t] + 5054 Cos[4 t] + 578 Cos[6 t] -  276 Cos[8 t]) Sin[3 t]^4)/( 6561 (1 - 4 Sin[t]^2)^11) /. {t -> ArcSin[u]/3}) /. {u -> (3 Sqrt[3 X])/2}, {X, 0, p}], {p, 2, 19}]

Formula

The g.f. Z for C(3n,[3^n],2) obeys the equation [Zuber, sect 4.4.2]

Z = (x^6*(1+113*x^3*z^3+(1610*x^6-72*x^9)*z^6 - 16*x^9*(308+9*x^3)*z^9 + 34016*x^12*z^12 - 90880*x^15*z^15 + 56832*x^18*z^18))/((-1+3*x^3*z^2)*(-1+2*x^3*z^3)^11) where z = (2*sin(arcsin((3*sqrt(3)*sqrt(x^3))/2)/3))/ (sqrt(3)*sqrt(x^3)) = 1+x^3+3x^6+... is the g.f. for the sequence C(3n,[3^n],0), given by A001764. The expansion of Z starts as x^6 + 144*x^9 + 6046*x^12 + 149674*x^15 + ...

As a function of X = x^3, this g.f. can be simplified as (16*(16321 - 21668*cos(2*t) + 5054*cos(4*t) + 578*cos(6*t) - 276*cos(8*t))*sin(3*t)^4)/ (6561(1-4*sin(t)^2)^11) where t = (1/3)*arcsin((3/2)*sqrt(3*X)).