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a(n) = (n+6)*(n+1)*(n^2 + 7*n + 16)/4.
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%I #31 Oct 09 2024 07:35:36

%S 0,24,84,204,414,750,1254,1974,2964,4284,6000,8184,10914,14274,18354,

%T 23250,29064,35904,43884,53124,63750,75894,89694,105294,122844,142500,

%U 164424,188784,215754,245514,278250,314154,353424,396264,442884,493500,548334

%N a(n) = (n+6)*(n+1)*(n^2 + 7*n + 16)/4.

%C In general, a sequence of the form a(n) = Sum_{k=1..n} (k+x+3)!/(k+x)! will have a closed form of n/4 * (n + 5 + 2*x)*(n^2 + 5*n + 2*x*n + 10 + 2*x^2 + 10*x). - _Gary Detlefs_, Aug 10 2010

%H G. C. Greubel, <a href="/A168538/b168538.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).

%F From _R. J. Mathar_, Mar 21 2010: (Start)

%F a(n) = 6*A063258(n).

%F a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).

%F G.f.: 6*(x-2)*(x^2 - 2*x + 2)/(x-1)^5. (End)

%F From _Gary Detlefs_, Aug 10 2010: (Start)

%F a(n) = Sum_{k=1..n} (k+3)*(k+2)*(k+1), with offset 0.

%F a(n) = (n/4)*(n+5)*(n^2 + 5*n + 10), with offset 0. (End)

%F E.g.f.: (1/4)*(96 + 240*x + 120*x^2 + 20*x^3 + x^4)*exp(x). - _G. C. Greubel_, Jul 25 2016

%t LinearRecurrence[{5,-10,10,-5,1},{0, 24, 84, 204, 414}, 25] (* _G. C. Greubel_, Jul 25 2016 *)

%t Table[1/4 (n+6)(n+1)(n^2+7n+16),{n,-1,40}] (* _Harvey P. Dale_, Jul 07 2019 *)

%o (PARI) a(n) = (n+6)*(n+1)*(n^2 + 7*n + 16)/4; \\ _Michel Marcus_, Jan 10 2015

%o (Magma) [(n+6)*(n+1)*(n^2+7*n+16)/4: n in [-1..35]]; // _Vincenzo Librandi_, Jul 26 2016

%Y Cf. A063258.

%K nonn,easy

%O -1,2

%A Kristo Jorgenson (kristoj(AT)me.com), Nov 29 2009

%E Edited by _N. J. A. Sloane_, Nov 29 2009