OFFSET
2,1
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 if g=0 (non-crossing partitions), by [Zuber] when g=1 (this sequence) and g=2, see A371251. One has C(n=6,[3^2],1) = 6, C(n=9,[3^3],1) = 102, etc.
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. 19; preprint, arXiv:2305.01100 [math.CO], 2023. See p. 19.
Jean-Bernard Zuber, Counting partitions by genus. I. Genus 0 to 2, Enumer. Comb. Appl. 4 (2) (2024) #S2R13. See pp. 14-17; preprint, arXiv:2303.05875 [math.CO], 2023. See pp. 16-19.
FORMULA
The g.f. Z for C(3n,[3^n],1) is given [Zuber] by Z = (-6*x^6*z^6)/((-1 + 3*x^3*z^2)*(1 - 2*x^3*z^3)^4) 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 latter obeys z = 1 + (x*z)^3, therefore Z obeys the cubic equation Z = (-216*x^12-6*x^3(-4+27*x^3)^3*Z^2-(-4+27*x^3)^5*Z^3)/(36*x^6*(-1+81*(x^3+9*x^6))).
The expression of Z given in [Zuber] is
(1152*x^3*sin((1/3)*arcsin((3*sqrt(3*x^3))/2))^6)/((2*cos((1/3)*arccos(1-(27 x^3/2)))-1)*(9*sqrt(x^3)-4*sqrt(3)*sin((1/3)*arcsin((3*sqrt(3*x^3))/2)))^4). Its expansion starts as 6*x^6 + 102*x^9 + 1212*x^12 + ...
As a function of X = x^3, this g.f. can be simplified as
(32*sin(t)^2*sin(3*t)^2)/(27(1+2*sin(t))^5*(1-2*sin(t))^5) where t = (1/3)*arcsin((3/2)*sqrt(3*X)). See the Mathematica program below.
D-finite with recurrence +104*(n-1)*(2*n-3)*a(n) +6*(-2052*n^2+9531*n-11920)*a(n-1) +81*(2133*n^2-11313*n+15580)*a(n-2) +4374*(-207*n^2+1233*n-1930)*a(n-3) +177147*(3*n-10)*(3*n-11)*a(n-4)=0. - R. J. Mathar, Mar 25 2024
EXAMPLE
a(2) = 6.
G.f. = 6*X^2 + 102*X^3 + 1212*X^4 + ...
MATHEMATICA
Table[SeriesCoefficient[(32 Sin[t]^2 Sin[3 t]^2)/( 27 (1 + 2 Sin[t])^5 (1 - 2 Sin[t])^5) /. {t -> 1/3 ArcSin[(3*Sqrt[3*X])/2]}, {X, 0, p}], {p, 2, 21}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Robert Coquereaux, Mar 16 2024
STATUS
approved