This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A297179 #46 Mar 05 2024 14:38:49
%S A297179 1,28,399,4179,36498,282282,1999998,13258674,83417334,503090588,
%T A297179 2929953026,16569715890,91386952020,493234934220,2612295374940,
%U A297179 13607257868820,69841333755270,353777814426960,1770937330172010,8770508370593970,43015147164809820,209104302965011740
%N A297179 Number of genus-2 partitions of [n].
%C A297179 From _Robert Coquereaux_, Feb 12 2024: (Start)
%C A297179 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.
%C A297179 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.
%C A297179 Defining a(n) to be the sum over k of T(n,k) one shows that a(n) obeys the recurrence
%C A297179 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.
%C A297179 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)
%H A297179 Robert Coquereaux and Jean-Bernard Zuber, <a href="https://arxiv.org/abs/2305.01100">Counting partitions by genus. A compendium of results</a>, arXiv:2305.01100 [math.CO], 2023. See p. 5.
%H A297179 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. 9.
%H A297179 Robert Cori and G. Hetyei, <a href="https://arxiv.org/abs/1710.09992">Counting partitions of a fixed genus</a>, arXiv preprint arXiv:1710.09992 [math.CO], 2017.
%F A297179 From _Robert Coquereaux_, Feb 12 2024: (Start)
%F A297179 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!).
%F A297179 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)
%t A297179 a[n_] := (2^(n - 9) (88 n - 39 n^2 + 5 n^3 - 84) (2 n - 9)!!) / (45 (n - 6)!);
%t A297179 Table[a[n], {n, 6, 27}] (* _Peter Luschny_, Feb 13 2024 *)
%Y A297179 Row sums of A297178.
%Y A297179 Column g=2 of A370235.
%K A297179 nonn
%O A297179 6,2
%A A297179 _N. J. A. Sloane_, Dec 27 2017