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%I #41 Mar 02 2023 11:25:42
%S 43175,1382000,10495575,44230400,134984375,335890800,726002375,
%T 1415475200,2550752775,4319750000,6957037175,10749024000,16039143575,
%U 23233036400,32803734375,45296844800,61335734375,81626713200,106964218775,138236000000,176428301175
%N a(n) = square root of the square part of discriminant of Brioschi quintic polynomial x^5-10*n*x^3+45*n^2*x-n^2.
%C The squarefree part is always 5.
%H Matthew Moore, <a href="https://web.archive.org/web/20211016014955/https://www.math.arizona.edu/~ura-reports/053/Moore.Matthew/Final.pdf">Theorems and Algorithms Associated with Solving the General Quintic</a> [Appears to give incorrect formula for the Brioschi quintic]
%H Tito Piezas III and Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/BrioschiQuinticForm.html">Brioschi Quintic Form</a>.
%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).
%F a(n) = 25*n^4*(1728*n-1). - _Klaus Brockhaus_, Oct 28 2007
%F G.f.: 25*x*(1729*x^4 + 44938*x^3+114048*x^2+44918*x+1727) / (x-1)^6. - _Colin Barker_, Sep 02 2013
%t Table[25n^4(1728n-1),{n,1,100}]
%o (PARI) a(n) = my(p=poldisc(x^5 - 10*n*x^3 + 45*n^2*x - n^2)); sqrtint(p/core(p)); \\ _Michel Marcus_, Mar 02 2023
%Y Cf. A134448.
%K nonn,easy
%O 1,1
%A _Artur Jasinski_, Oct 26 2007, Oct 28 2007
%E Corrected by _Klaus Brockhaus_, Oct 28 2007
%E More terms from _Colin Barker_, Sep 02 2013
%E Name corrected by _Amiram Eldar_, Mar 02 2023