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%I #8 Aug 18 2012 14:09:46
%S 1,5,0,765,41261,0,1175731456,804611664045,0,4133434158867578125,
%T 36792671310208420147421,0,33666995638445382179718361163901,
%U 3930778415673723952392425569428439040,0,637350736211692642266912139961455499346709367565
%N From Wendt's determinant compute (-A048954(2*n)/3)^(1/3).
%C It is known that 3 divides A048954(2*n). It is conjectured that the quotient is a perfect cube.
%C See A048954 for additional comments, references, links, and cross-references.
%H Gerard P. Michon, <a href="http://www.numericana.com/data/wendt.htm">Factorization of Wendt's Determinant</a>(see Remarks and Conjectures).
%F a(n) = (-A048954(2*n)/3)^(1/3).
%F a(n) = 0 if and only if n is divisible by 3.
%t w[n_] := Resultant[x^n - 1, (1 + x)^n - 1, x]; Table[(-w[2 n]/3)^(1/3), {n, 19}]
%Y Cf. A048954, A215615.
%K nonn
%O 1,2
%A _Jonathan Sondow_, Aug 17 2012
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