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%I #18 Sep 08 2022 08:45:13
%S 1,600,4983,19468,53505,119676,233695,414408,683793,1066960,1592151,
%T 2290740,3197233,4349268,5787615,7556176,9701985,12275208,15329143,
%U 18920220,23108001,27955180,33527583,39894168,47127025,55301376
%N Figurate numbers based on the 120-cell (4-D polytope with Schlaefli symbol {5,3,3}).
%C This is the 4-dimensional regular convex polytope called the 120-cell, hecatonicosachoron or hyperdodecahedron.
%H Vincenzo Librandi, <a href="/A092183/b092183.txt">Table of n, a(n) for n = 1..1000</a>
%H Hyun Kwang Kim, <a href="http://dx.doi.org/10.1090/S0002-9939-02-06710-2">On Regular Polytope Numbers</a>, Proc. Amer. Math. Soc., 131 (2003), 65-75.
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/120-Cell.html">120-Cell</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1). [_R. J. Mathar_, Jun 21 2010]
%F a(n) = n*((261*n^3)-(504*n^2)+(283*n)-38)/2.
%F a(n) = C(n+3,4) + 595 C(n+2,4) + 1993 C(n+1,4) + 543 C(n,4).
%F a(n) = +5*a(n-1) -10*a(n-2) +10*a(n-3) -5*a(n-4) +a(n-5). G.f.: x*(1+595*x+1993*x^2+543*x^3)/(1-x)^5. [_R. J. Mathar_, Jun 21 2010]
%e a(3) = 3*((261*3^3)-(504*3^2)+(283*3)-38)/2 = 3*(7047-4536+849-38)/2 = 1.5*3322 = 4983
%t Table[SeriesCoefficient[x (1 + 595 x + 1993 x^2 + 543 x^3)/(1 - x)^5, {x, 0, n}], {n, 26}] (* _Michael De Vlieger_, Dec 14 2015 *)
%o (Magma) [n*((261*n^3)-(504*n^2)+(283*n)-38)/2: n in [1..40]]; // _Vincenzo Librandi_, May 22 2011
%o (PARI) a(n) = n*(261*n^3 - 504*n^2 + 283*n - 38)/2; \\ _Michel Marcus_, Dec 14 2015
%Y Cf. A000332, A000583, A014820, A092181, A092182.
%K easy,nonn
%O 1,2
%A Michael J. Welch (mjw1(AT)ntlworld.com), Mar 31 2004
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