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a(n) = n*(n-1)^4/2.
4

%I #39 Feb 13 2023 02:51:35

%S 0,0,1,24,162,640,1875,4536,9604,18432,32805,55000,87846,134784,

%T 199927,288120,405000,557056,751689,997272,1303210,1680000,2139291,

%U 2693944,3358092,4147200,5078125,6169176,7440174,8912512,10609215,12555000,14776336,17301504,20160657

%N a(n) = n*(n-1)^4/2.

%C a(n) = n(n-1)^4/2 is half the number of colorings of 5 points on a line with n colors. - _R. H. Hardin_, Feb 23 2002

%H Vincenzo Librandi, <a href="/A019583/b019583.txt">Table of n, a(n) for n = 0..1000</a>

%H Milan Janjic and Boris Petkovic, <a href="http://arxiv.org/abs/1301.4550">A Counting Function</a>, arXiv 1301.4550 [math.CO], 2013.

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).

%F Sum_{j>=2} 1/a(j) = hypergeom([1, 1, 1, 1, 1], [ 2, 2, 2, 3], 1) = -2 + 2*zeta(2) - 2*zeta(3) + 2*zeta(4). - _Stephen Crowley_, Jun 28 2009

%F G.f.: x^2*(1 + 18*x + 33*x^2 + 8*x^3)/(1 - x)^6. - _Colin Barker_, Feb 23 2012

%F From _Amiram Eldar_, Feb 13 2023: (Start)

%F a(n) = A101362(n-1)/2.

%F Sum_{n>=2} (-1)^n/a(n) = 2 + Pi^2/6 + 7*Pi^4/360 - 4*log(2) - 3*zeta(3)/2. (End)

%t CoefficientList[Series[x^2*(1+18*x+33*x^2+8*x^3)/(1-x)^6,{x,0,40}],x] (* _Vincenzo Librandi_, Apr 20 2012 *)

%t a[n_] := n*(n - 1)^4/2; Array[a, 30, 0] (* _Amiram Eldar_, Feb 13 2023 *)

%o (Magma) [n*(n-1)^4/2: n in [0..30]]; // _Vincenzo Librandi_, Apr 20 2012

%Y Cf. A101362.

%K nonn,easy

%O 0,4

%A _N. J. A. Sloane_