%I #40 May 31 2023 09:51:00
%S 1,1,12,8,1,72,528,704,180,1,324,8760,53792,98124,56160,8064,1,1344,
%T 103040,1759520,9936360,21676144,19083456,6356160,604800,1,5436,
%U 1054056,41312704,539233128,2901894144,7118351104,8247838464,4418632656,988952832,68428800,1,21816,10106736,823376896,21574613676,235937470944,1230387808384,3281254260864,4608240745104,3390175943424,1247151098880,204083712000,10897286400
%N Irregular triangle read by rows: coefficients in order of decreasing exponents of polynomials P_g(x) related to Hultman numbers.
%C Row n contains 2*n-1 terms.
%C Evaluating the polynomials at 1 gives A035319.
%H Gheorghe Coserea, <a href="/A185259/b185259.txt">Rows n=1..101, flattened</a>
%H Nikita Alexeev and Peter Zograf, <a href="http://arxiv.org/abs/1111.3061">Hultman numbers, polygon gluings and matrix integrals</a>, arXiv preprint arXiv:1111.3061 [math.PR], 2011.
%H Robert Coquereaux and Jean-Bernard Zuber, <a href="https://arxiv.org/abs/2305.01100">Counting partitions by genus. II. A compendium of results</a>, arXiv:2305.01100 [math.CO], 2023. See p. 9.
%e Triangle begins:
%e [1] 1
%e [2] 1 12 8
%e [3] 1 72 528 704 180
%e [4] 1 324 8760 53792 98124 56160 8064
%e [5] 1 1344 103040 1759520 9936360 21676144 19083456 6356160 604800
%e [6] ...
%t P[n_, x_] := (f = (1-x)^(4n+1); s = Sum[-StirlingS1[2n+2+k, k+1]/ Binomial[2n+2+k, 2] x^k, {k, 0, 2n-2}]; f s + O[x]^(2n-1) // Normal);
%t row[n_] := CoefficientList[P[n, x], x] // Reverse;
%t Array[row, 7] // Flatten (* _Jean-François Alcover_, Sep 05 2018, after _Gheorghe Coserea_ *)
%o (PARI)
%o P(n, v='x) = {
%o my(x='x+O('x^(2*n-1)), f=(1-x)^(4*n+1),
%o s=sum(k=0, 2*n-2, -stirling(2*n+2+k, k+1, 1)/binomial(2*n+2+k,2)*x^k));
%o subst(Pol(f*s, 'x), 'x, v);
%o };
%o concat(vector(7, n, Vec(P(n))))
%o \\ test: N=50; vector(N, n, P(n,1)) == vector(N, n, (4*n)!/((2*n+1)!*4^n))
%o \\ _Gheorghe Coserea_, Jan 30 2018
%Y Cf. A035319, A185263.
%K nonn,tabf
%O 1,3
%A _N. J. A. Sloane_, Jan 21 2012
%E More terms from _Gheorghe Coserea_, Jan 30 2018