A215616 From Wendt's determinant compute (-A048954(2*n)/3)^(1/3).

1, 5, 0, 765, 41261, 0, 1175731456, 804611664045, 0, 4133434158867578125, 36792671310208420147421, 0, 33666995638445382179718361163901, 3930778415673723952392425569428439040, 0, 637350736211692642266912139961455499346709367565

Offset: 1

Author

Jonathan Sondow, Aug 17 2012

Keywords

• nonn

Comments

It is known that 3 divides A048954(2*n). It is conjectured that the quotient is a perfect cube.

See A048954 for additional comments, references, links, and cross-references.

Links

• Gerard P. Michon, Factorization of Wendt's Determinant(see Remarks and Conjectures).

Crossrefs

Cf. A048954, A215615.

Programs

Mathematica

w[n_] := Resultant[x^n - 1, (1 + x)^n - 1, x]; Table[(-w[2 n]/3)^(1/3), {n, 19}]

Formula

a(n) = (-A048954(2*n)/3)^(1/3).

a(n) = 0 if and only if n is divisible by 3.