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a(n) = n*(4*n^2 + 2*n + 1).
2

%I #16 Sep 08 2022 08:45:20

%S 0,7,42,129,292,555,942,1477,2184,3087,4210,5577,7212,9139,11382,

%T 13965,16912,20247,23994,28177,32820,37947,43582,49749,56472,63775,

%U 71682,80217,89404,99267,109830,121117,133152,145959,159562,173985,189252

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

%C a(n) = A110449(2*n,n), central terms in triangle A110449.

%C 2*a(n) is the sum of the consecutive integers from A000384(n)+1 to A000384(n+1)-1. - _Bruno Berselli_, Jun 27 2018

%H G. C. Greubel, <a href="/A110451/b110451.txt">Table of n, a(n) for n = 0..5000</a>

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).

%F From _G. C. Greubel_, Aug 24 2017: (Start)

%F a(n) = 4*a(n-1) - 6*a(n-1) + 4*a(n-2) - a(n-4).

%F G.f.: (7*x + 14*x^2 + 3*x^3)/(1 - x)^4.

%F E.g.f.: x*(7 + 14*x + 4*x^2)*exp(x). (End)

%p seq(n*(4*n^2+2*n+1),n=0..40); # _Muniru A Asiru_, Jun 27 2018

%t Table[n*(4*n^2 + 2*n + 1), {n, 0, 50}] (* or *) LinearRecurrence[{4,-6,4,-1}, {0,7,42,129}, 50] (* _G. C. Greubel_, Aug 24 2017 *)

%o (Magma)[n*(4*n^2+2*n+1): n in [0..40]]; // _Vincenzo Librandi_, Dec 26 2010

%o (PARI) x='x+O('x^50); Vec((7*x + 14*x^2 + 3*x^3)/(1 - x)^4) \\ _G. C. Greubel_, Aug 24 2017

%o (GAP) List([0..40],n->n*(4*n^2+2*n+1)); # _Muniru A Asiru_, Jun 27 2018

%Y Cf. A000384, A110449.

%K nonn,easy

%O 0,2

%A _Reinhard Zumkeller_, Jul 21 2005