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
Keywords
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.
Programs
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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)).
Comments