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Number of genus 1 partitions of the set [3n] into n blocks of length 3.
2

%I #20 Mar 25 2024 08:46:13

%S 6,102,1212,12330,114888,1011486,8558712,70324884,564931230,

%T 4457508264,34662068784,266296074408,2025114297696,15267023594670,

%U 114233412701424,849144504823848,6275680692866946,46143888578211414,337737723001251660,2461833584990710434

%N Number of genus 1 partitions of the set [3n] into n blocks of length 3.

%C 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.

%H Robert Coquereaux and Jean-Bernard Zuber, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL27/Coquereaux/coque5.html">Counting partitions by genus: a compendium of results</a>, Journal of Integer Sequences, Vol. 27 (2024), Article 24.2.6. See p. 19; <a href="https://arxiv.org/abs/2305.01100">preprint</a>, arXiv:2305.01100 [math.CO], 2023. See p. 19.

%H Jean-Bernard Zuber, <a href="https://ecajournal.haifa.ac.il/Volume2024/ECA2024_S2A13.pdf">Counting partitions by genus. I. Genus 0 to 2</a>, Enumer. Comb. Appl. 4 (2) (2024) #S2R13. See pp. 14-17; <a href="https://arxiv.org/abs/2303.05875">preprint</a>, arXiv:2303.05875 [math.CO], 2023. See pp. 16-19.

%F 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.

%F 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))).

%F The expression of Z given in [Zuber] is

%F (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 + ...

%F As a function of X = x^3, this g.f. can be simplified as

%F (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.

%F 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

%e a(2) = 6.

%e G.f. = 6*X^2 + 102*X^3 + 1212*X^4 + ...

%t 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}]

%Y Cf. A001764 for C(3n,[3^n],0) and A371251 for C(3n,[3^n],2).

%K nonn

%O 2,1

%A _Robert Coquereaux_, Mar 16 2024