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Binomial(n-1,3)+3*binomial(n-1,4)+6*binomial(n-1,5)+5*binomial(n-1,6).
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%I #23 Jul 27 2022 18:57:11

%S 0,0,0,1,7,31,106,301,742,1638,3312,6237,11077,18733,30394,47593,

%T 72268,106828,154224,218025,302499,412699,554554,734965,961906,

%U 1244530,1593280,2020005,2538081,3162537,3910186,4799761,5852056,7090072,8539168,10227217,12184767,14445207,17044938,20023549,23423998,27292798

%N Binomial(n-1,3)+3*binomial(n-1,4)+6*binomial(n-1,5)+5*binomial(n-1,6).

%C Coefficient of q^3 in the polynomial NT_{n,mu}(q).

%H M. Jones, S. Kitaev, J. Remmel, <a href="http://arxiv.org/abs/1311.3332">Frame patterns in n-cycles</a>, arXiv preprint arXiv:1311.3332 [math.CO], 2013.

%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 From _Colin Barker_, Jan 16 2014: (Start)

%F a(n) = (720-2136*n+2450*n^2-1395*n^3+425*n^4-69*n^5+5*n^6)/720.

%F G.f.: -x^4*(x^3+3*x^2+1) / (x-1)^7.

%F (End)

%F E.g.f.: (1/720)*exp(x)*x^3*(120 + 90*x + 36*x^2 + 5*x^3). - _Stefano Spezia_, Jan 09 2019

%p b:=binomial;

%p f:=n->b(n-1,3)+3*b(n-1,4)+6*b(n-1,5)+5*b(n-1,6);

%p [seq(f(n),n=1..50)];

%t a[n_] := 1/720 (n-1)(n-2)(n-3)(-120 + 136n - 39n^2 + 5n^3); Array[a, 42] (* _Jean-François Alcover_, Jan 09 2019 *)

%t LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,0,0,1,7,31,106},50] (* _Harvey P. Dale_, Jul 27 2022 *)

%o (PARI) Vec(-x^4*(x^3+3*x^2+1)/(x-1)^7 + O(x^100)) \\ _Colin Barker_, Jan 16 2014

%Y Cf. A235594.

%K nonn,easy

%O 1,5

%A _N. J. A. Sloane_, Jan 13 2014