eval bernpol

["Ruud H.G. van Tol" <rvtol@isolution.nl>]

Is this a canonical and effective way to use bernpol?

A285068(n) = my(x=1/3); denominator(3^n * eval(bernpol(n)));

? [ A285068(n) |n<-[0..20] ]
% [1, 2, 2, 1, 10, 1, 14, 1, 10, 1, 22, 1, 910, 1, 2, 1, 170, 1, 266, 1, 
110]

If not, what are alternatives?

-- Ruud