Numbers n such that n^2 = (1/3)*(n+floor(sqrt(3)*n*floor(sqrt(3)*n))).

A081065

Numbers n such that n^2 = (1/3)*(n+floor(sqrt(3)*n*floor(sqrt(3)*n))).

Terms

    a(0) =2a(1) =24a(2) =330a(3) =4592a(4) =63954a(5) =890760a(6) =12406682a(7) =172802784a(8) =2406832290a(9) =33522849272a(10) =466913057514

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