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Number of ways to choose three points on a centered hexagonal grid of size n.
2

%I #40 Feb 17 2024 04:04:23

%S 0,35,969,7770,35990,121485,333375,790244,1679580,3280455,5989445,

%T 10349790,17083794,27128465,41674395,62207880,90556280,128936619,

%U 180007425,246923810,333395790,443749845,582993719,756884460,971999700,1235812175,1556767485,1944365094

%N Number of ways to choose three points on a centered hexagonal grid of size n.

%C A centered hexagonal grid of size n is a grid with A003215(n-1) points forming a hexagonal lattice.

%H Vincenzo Librandi, <a href="/A240826/b240826.txt">Table of n, a(n) for n = 1..1000</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/HexNumber.html">Hex Number</a>.

%H <a href="/index/Rec#order_07">Index entries for linear recurrences with constant coefficients</a>, signature (7,-21,35,-35,21,-7,1).

%F a(n) = binomial(A003215(n-1), 3)

%F = binomial(3*n^2-3*n+1, 3)

%F = 1/2*n*(n-1)*(3*n^2-3*n+1)*(3*n^2-3*n-1)

%F = 9/2*n^6-27/2*n^5+27/2*n^4-9/2*n^3-1/2*n^2+1/2*n.

%F G.f.: -x^2*(35*x^4+724*x^3+1722*x^2+724*x+35) / (x-1)^7. - _Colin Barker_, Apr 18 2014

%F Sum_{n>=2} 1/a(n) = sqrt(3/7)*Pi*tan(sqrt(7/3)*Pi/2) + sqrt(3)*Pi*tanh(Pi/(2*sqrt(3))) - 2. - _Amiram Eldar_, Feb 17 2024

%p seq(binomial(3*n^2-3*n+1, 3), n=1..28); # _Martin Renner_, May 31 2014

%p op(PolynomialTools[CoefficientList](convert(series(-x^2*(35*x^4+724*x^3+1722*x^2+724*x+35)/(x-1)^7, x=0, 29), polynom), x)[2..29]); # _Martin Renner_, May 31 2014

%t CoefficientList[Series[- x(35 x^4 + 724 x^3 + 1722 x^2 + 724 x + 35)/(x - 1)^7, {x, 0, 50}], x] (* _Vincenzo Librandi_, Apr 19 2014 *)

%t LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,35,969,7770,35990,121485,333375},40] (* _Harvey P. Dale_, Sep 12 2019 *)

%Y Cf. A003215, A178208, A241219.

%K nonn,easy

%O 1,2

%A _Martin Renner_, Apr 17 2014