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a(n) = (n + 4)*(n + 3)*(n + 2)*(n + 1)*n / 5 = 24*A000389(n+4).
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%I #23 Jul 02 2023 02:18:19

%S 0,24,144,504,1344,3024,6048,11088,19008,30888,48048,72072,104832,

%T 148512,205632,279072,372096,488376,632016,807576,1020096,1275120,

%U 1578720,1937520,2358720,2850120,3420144,4077864,4833024,5696064,6678144,7791168,9047808

%N a(n) = (n + 4)*(n + 3)*(n + 2)*(n + 1)*n / 5 = 24*A000389(n+4).

%D L. B. W. Jolley, Summation of Series, Dover, 1961, eq. (48), page 8.

%H Vincenzo Librandi, <a href="/A158874/b158874.txt">Table of n, a(n) for n = 0..10000</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (6,-15,20,-15,6,-1).

%F G.f.: 24*x / (x-1)^6 . - _R. J. Mathar_, Oct 03 2011

%F E.g.f.: x*(x^4 + 20*x^3 + 120*x^2 + 240*x + 120)*exp(x)/5. - _G. C. Greubel_, Nov 21 2017

%F From _Amiram Eldar_, Jul 02 2023: (Start)

%F Sum_{n>=1} 1/a(n) = 5/96.

%F Sum_{n>=1} (-1)^(n+1)/a(n) = 10*log(2)/3 - 655/288. (End)

%t Table[(n + 4)*(n + 3)*(n + 2)*(n + 1)*n/5, {n,0,50}] (* _G. C. Greubel_, Nov 21 2017 *)

%o (Magma) [n*(n^4+10*n^3+35*n^2+50*n+24)/5: n in [0..30]]; // _Vincenzo Librandi_, Oct 05 2011

%o (PARI) for(n=0,30, print1((n + 4)*(n + 3)*(n + 2)*(n + 1)*n/5, ", ")) \\ _G. C. Greubel_, Nov 21 2017

%Y Partial sums of A052762.

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_, Nov 29 2009