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Ruud H.G. van Tol on Tue, 30 Jan 2024 12:50:41 +0100
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- To: pari-users <pari-users@pari.math.u-bordeaux.fr>
- Subject: eval bernpol
- From: "Ruud H.G. van Tol" <rvtol@isolution.nl>
- Date: Tue, 30 Jan 2024 12:50:33 +0100
- Delivery-date: Tue, 30 Jan 2024 12:50:41 +0100
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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