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a(n) = 2^(n-4)*(n+2)*(n+3)*(n+4)*(n+5)*(n+6)/15.
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%I #37 Sep 08 2022 08:45:09

%S 3,21,112,504,2016,7392,25344,82368,256256,768768,2236416,6336512,

%T 17547264,47628288,127008768,333398016,862912512,2205220864,

%U 5571084288,13927710720,34487664640,84651540480,206108098560,498094571520

%N a(n) = 2^(n-4)*(n+2)*(n+3)*(n+4)*(n+5)*(n+6)/15.

%C Old definition was "Sequence associated with recurrence a(n) = 2*a(n-1) + k*(k+2)*a(n-2)". See the first comment in A080928.

%C The sixth column of A080928 (after 0) is 2*a(n).

%H G. C. Greubel, <a href="/A080952/b080952.txt">Table of n, a(n) for n = 0..1000</a> (terms 0..300 from Vincenzo Librandi)

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (12,-60,160,-240,192,-64).

%F G.f.: (1-x)*(4*x^2-2*x+1)*(4*x^2-6*x+3)/(1-2x)^6.

%F a(n) = 12*a(n-1) - 60*a(n-2) + 160*a(n-3) - 240*a(n-4) + 192*a(n-5) - 64*a(n-6), n>=6. - _Harvey P. Dale_, Jun 11 2011

%F Let b(n) = A000292(n+1)+n+1+A000389(n+5) = (n+1)*(n^4+14*n^3+91*n^2+254*n+360)/120 = 3, 12, 34, 80, 166, 314,.. Then a(n) = 2^n*b(n) - 2^(n-1)*b(n-1). - _R. J. Mathar_, Jun 11 2011

%F From _Amiram Eldar_, Jan 07 2022: (Start)

%F Sum_{n>=0} 1/a(n) = 40*log(2) - 82/3.

%F Sum_{n>=0} (-1)^n/a(n) = 1314 - 3240*log(3/2). (End)

%t LinearRecurrence[{12, -60, 160, -240, 192, -64}, {3, 21, 112, 504, 2016, 7392}, 30] (* or *) CoefficientList[Series[(1-x) (3 - 12 x + 28 x^2 - 32 x^3 + 16 x^4)/ (1 - 2 x)^6, {x, 0, 30}], x] (* _Harvey P. Dale_, Jun 11 2011 *)

%o (Magma) I:=[3,21,112,504,2016,7392]; [n le 6 select I[n] else 12*Self(n-1)-60*Self(n-2)+160*Self(n-3)-240*Self(n-4)+192*Self(n-5)-64*Self(n-6): n in [1..30]]; // _Vincenzo Librandi_, Aug 06 2013

%o (PARI) my(x='x+O('x^50)); Vec((1-x)*(4*x^2-2*x+1)*(4*x^2-6*x+3)/(1-2*x)^6) \\ _G. C. Greubel_, Nov 24 2017

%Y Cf. A000292, A000389, A080928.

%K nonn,easy

%O 0,1

%A _Paul Barry_, Feb 26 2003

%E Replaced the previous definition with the closed form from _Bruno Berselli_, Aug 06 2013