%I #6 Feb 15 2026 21:23:06
%S 1,1,1,1,1,1,45,45,315,63,14175,675,93555,66825,14189175,8513505,
%T 638512875,70945875,19538493975,19538493975,9280784638125,40176556875,
%U 714620417135625,54970801318125,147926426347074375,11378955872851875,48076088562799171875,6868012651828453125
%N Denominator of Sum_{k=1..floor((n+1)/2)} binomial(n, 2*k-2) * c(k), where c(k) = A002430(k)/A036279(k) is the k-th positive coefficient in the Taylor series for tan(x).
%H Amiram Eldar, <a href="/A393303/b393303.txt">Table of n, a(n) for n = 1..450</a>
%H John Greene, <a href="https://www.fq.math.ca/Scanned/40-3/greene.pdf">The Burgstahler Coincidence</a>, The Fibonacci Quarterly, Vol. 40, No. 3 (2002), pp. 194-202.
%t c[n_] := (-1)^n * (16^n-4^n) * Zeta[1-2*n]/(2*n-1)!;
%t a[n_] := Denominator[Sum[Binomial[n, 2*k-2] * c[k], {k, 1, Floor[(n+1)/2]}]];
%t Array[a, 28]
%o (PARI) c(n) = (-1)^(n+1) * (16^n-4^n) * bernfrac(2*n)/(2*n)!;
%o a(n) = denominator(sum(k = 1, (n+1)\2, binomial(n, 2*k-2) * c(k)));
%Y Cf. A002430, A036279, A393302 (numerators).
%K nonn,easy,frac
%O 1,7
%A _Amiram Eldar_, Feb 10 2026