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%I #28 Sep 08 2022 08:14:50
%S 9,96,500,1800,5145,12544,27216,54000,99825,174240,290004,463736,
%T 716625,1075200,1572160,2247264,3148281,4332000,5865300,7826280,
%U 10305449,13406976,17250000,21970000,27720225,34673184,43022196,52983000,64795425,78725120,95065344
%N Fourth column (m=3) of triangle A090447.
%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) = A090447(n,3).
%F a(n) = (n^3*(n-1)^2*(n-2)^1)/(1!*2!*3!) for n >= 3.
%F From _Colin Barker_, Jan 21 2013: (Start)
%F a(n) = (n^6-4*n^5+5*n^4-2*n^3)/12.
%F G.f.: -x^3*(x^3+17*x^2+33*x+9)/(x-1)^7. (End)
%F a(n) = A000330(n-1)^2 - A000292(n-1)^2. - _J. M. Bergot_, May 02 2014
%F a(n) = 7*a(n-1)-21*a(n-2)+35*a(n-3)-35*a(n-4)+21*a(n-5)-7*a(n-6)+a(n-7). - _Wesley Ivan Hurt_, May 04 2021
%F From _Amiram Eldar_, Sep 08 2022: (Start)
%F Sum_{n>=3} 1/a(n) = 207/4 - 9*Pi^2/2 - 6*zeta(3).
%F Sum_{n>=3} (-1)^(n+1)/a(n) = 165/4 - Pi^2/4 - 48*log(2) - 9*zeta(3)/2. (End)
%p seq(mul(binomial(n,k),k=1..3),n=3..30); # _Zerinvary Lajos_, Dec 13 2007
%t a[n_] := Product[Binomial[n, k], {k, 0, 3}]; Array[a, 30, 3] (* _Amiram Eldar_, Sep 08 2022 *)
%Y Cf. A000292, A000330, A090447.
%K nonn,easy
%O 3,1
%A _Wolfdieter Lang_, Dec 23 2003
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