A297179 Number of genus-2 partitions of [n].

1, 28, 399, 4179, 36498, 282282, 1999998, 13258674, 83417334, 503090588, 2929953026, 16569715890, 91386952020, 493234934220, 2612295374940, 13607257868820, 69841333755270, 353777814426960, 1770937330172010, 8770508370593970, 43015147164809820, 209104302965011740

Offset: 6

Author

N. J. A. Sloane, Dec 27 2017

Keywords

• nonn

Comments

From Robert Coquereaux, Feb 12 2024: (Start)

Call B(n, g) the number of genus g partitions of a set with n elements (genus-dependent Bell number). Then a(n) = B(n, 2) with B(6, 2) = 1.

The entries of the triangle T(n, k) giving the number of genus 2 partitions of a set with n elements with k parts are known from R. Cori and G. Hetyei A297178.

Defining a(n) to be the sum over k of T(n,k) one shows that a(n) obeys the recurrence

a(n) = a(n-1) * (2*(-9 + 2*n) (-84 + n (88 + n*(-39 + 5*n)))) / ((-6 + n)*(-216 + n*(181 + n*(-54 + 5 n)))) with a(1) = a(2) = a(3) = a(4) = a(5) = 0 and a(6) = 1.

This determines a(n) for all n. One can solve the above recurrence and find an explicit formula, given below, for a(n) as a function of n. (End)

Links

• Robert Coquereaux and Jean-Bernard Zuber, Counting partitions by genus. A compendium of results, arXiv:2305.01100 [math.CO], 2023. See p. 5.
• 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. 9.
• Robert Cori and G. Hetyei, Counting partitions of a fixed genus, arXiv preprint arXiv:1710.09992 [math.CO], 2017.

Crossrefs

Row sums of A297178.

Column g=2 of A370235.

Programs

Mathematica

a[n_] := (2^(n - 9) (88 n - 39 n^2 + 5 n^3 - 84) (2 n - 9)!!) / (45 (n - 6)!);
Table[a[n], {n, 6, 27}]  (* Peter Luschny, Feb 13 2024 *)

Formula

From Robert Coquereaux, Feb 12 2024: (Start)

a(n) = (1/(2^9*3^2*5)) * ((-84 + 88*n - 39*n^2 + 5*n^3) /((2*n - 1) * (2*n - 3) * (2*n - 5) * (2*n - 7))) * (1/(n - 6)!) * ((2*n)!/n!).

E.g.f.: (1/720) * exp(2*x) *(x^2*(-6 + 6*x - 9*x^2 + 5*x^3)*BesselI(0, 2*x) + x*(6 - 6*x + 12*x^2 - 8*x^3 + 5*x^4)*BesselI(1, 2*x)). (End)