%I #12 Nov 24 2016 09:31:26
%S 1,12,144,864,27648,248832,11943936,35831808,1528823808,1719926784,
%T 220150628352,2972033482752,1141260857376768,106993205379072,
%U 54780521154084864,13695130288521216,42071440246337175552,739537035580145664,6058287395472553279488
%N a(n) is denominator of rational z(n) associated with the non-orientable map asymptotics constant p((n+1)/2).
%H Gheorghe Coserea, <a href="/A278121/b278121.txt">Table of n, a(n) for n = 0..201</a>
%H S. R. Carrell, <a href="https://arxiv.org/abs/1406.1760v2">The Non-Orientable Map Asymptotics Constant pg</a>, arXiv:1406.1760 [math.CO], 2014.
%H Stavros Garoufalidis, Marcos Marino, <a href="https://arxiv.org/abs/0812.1195v4">Universality and asymptotics of graph counting problems in nonorientable surfaces</a>, arXiv:0812.1195 [math.CO], 2008.
%F a(n) = denominator(z(n)), where z(n) = 1/2 * (y(n/2)/3^(n/2) + (5*n-6)/6 * z(n-1) + Sum {k=1..n-1} z(k)*z(n-k)), with z(0) = -1, y(n) = A269418(n)/A269419(n) and y(n+1/2) = 0 for all n.
%F p((n+1)/2) = 4 * (A278120(n)/A278121(n)) * (3/2)^((n+1)/2) / gamma((5*n-1)/4), where p((n+1)/2) is the non-orientable map asymptotics constant for type g=(n+1)/2 and gamma is the Gamma function.
%e For n=3 we have p(2) = 4 * (25/864) * (3/2)^2 / gamma(7/2) = 5/(36*sqrt(Pi)).
%e For n=5 we have p(3) = 4 * (15745/248832)*(3/2)^3/gamma(6) = 3149/442368.
%e n z(n) p((n+1)/2)
%e 0 -1 3/(sqrt(6)*gamma(3/4))
%e 1 1/12 1/2
%e 2 5/144 2/(sqrt(6)*gamma(1/4))
%e 3 25/864 5/(36*sqrt(Pi))
%e 4 1033/27684 1033/(13860*sqrt(6)*gamma(3/4))
%e 5 15745/248832 3149/442368
%e 6 1599895/11943936 319979/(18796050*sqrt(6)*gamma(1/4))
%e 7 12116675/35831808 484667/(560431872*sqrt(Pi))
%e 8 1519810267/1528823808 1519810267/(4258429005600*sqrt(6)*gamma(3/4))
%e 9 5730215335/1719926784 1146043067/41094783959040
%e ...
%o (PARI)
%o A269418_seq(N) = {
%o my(y = vector(N)); y[1] = 1/48;
%o for (n = 2, N,
%o y[n] = (25*(n-1)^2-1)/48 * y[n-1] + 1/2*sum(k = 1, n-1, y[k]*y[n-k]));
%o concat(-1, y);
%o };
%o seq(N) = {
%o my(y = A269418_seq(N), z = vector(N)); z[1] = 1/12;
%o for (n = 2, N,
%o my(t1 = if(n%2, 0, y[1+n\2]/3^(n\2)),
%o t2 = sum(k=1, n-1, z[k]*z[n-k]));
%o z[n] = (t1 + (5*n-6)/6 * z[n-1] + t2)/2);
%o concat(-1, z);
%o };
%o apply(denominator, seq(18))
%Y Cf. A269418, A269419, A278120 (numerator).
%K nonn,frac
%O 0,2
%A _Gheorghe Coserea_, Nov 12 2016