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From Wendt's determinant compute (-A048954(2*n)/3)^(1/3).
3

%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