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%I #15 Aug 27 2019 08:20:57
%S 1,1,1,1,5,1,1,1,7,1,1,1,13,19,5,1,221,1,1,1,1,1,13,17,5,1,47,4913,29,
%T 7,11,53,1,47,325,13,1147,41,1,1,41,1081,11,1,5,1,1,83,1,1,133,1,2491,
%U 97,5,103,61,1,1,19,226493,1,1,1,5,31,1,1,1,1271,289
%N a(n) = (2*n)!/A309205(n).
%C _Bill Gosper_ points out that this is a better fingerprint for the series than A309205.
%t F[n_] := Module[{p}, p = 1 + O[x]; For[k=2, k <= n, k++, p = Cos[x p]]; p];
%t seq[n_] := Module[{v}, v = CoefficientList[F[n], x]; Table[(2(k - 1))!/ Denominator[v[[2k - 1]]], {k, 1, n}]];
%t seq[71] (* _Jean-François Alcover_, Aug 27 2019, from PARI *)
%o (PARI) \\ here F(n) gives n terms of power series.
%o F(n)={my(p=1+O(x)); for(k=2, n, p=cos(x*p)); p}
%o seq(n)={my(v=Vec(F(n))); vector(n, k, (2*(k-1))!/denominator(v[2*k-1]))} \\ _Andrew Howroyd_, Aug 17 2019
%Y Cf. A309204, A309205.
%K nonn,frac
%O 0,5
%A _N. J. A. Sloane_, Jul 28 2019, following a suggestion from _Bill Gosper_.
%E Terms a(31) and beyond from _Andrew Howroyd_, Aug 17 2019