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%I #45 Sep 08 2022 08:45:32
%S 0,0,24,216,960,3000,7560,16464,32256,58320,99000,159720,247104,
%T 369096,535080,756000,1044480,1414944,1883736,2469240,3192000,4074840,
%U 5142984,6424176,7948800,9750000,11863800,14329224,17188416,20486760,24273000,28599360,33521664
%N a(n) = n^5 - n^3.
%H Vincenzo Librandi, <a href="/A133754/b133754.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 a(n) = 12*n*(2*binomial(n+2,4)- binomial(n+1,3)). - _Gary Detlefs_, Mar 25 2012
%F Sum_{n>=2} 1/a(n) = 5/4 - zeta(3). - _Daniel Suteu_, Feb 06 2017
%F From _G. C. Greubel_, Sep 02 2019: (Start)
%F G.f.: 24*x^2*(1 + 3*x + x^2)/(1-x)^6.
%F E.g.f.: x^2*(12 + 24*x + 10*x^2 + x^3)*exp(x). (End)
%F Sum_{n>=2} (-1)^n/a(n) = 3*zeta(3)/4 + 2*log(2) - 9/4. - _Amiram Eldar_, Jan 09 2021
%p seq(n^5 - n^3, n=0..50); # _G. C. Greubel_, Sep 02 2019
%t Table[n^5-n^3, {n,0,50}] (* _Vladimir Joseph Stephan Orlovsky_, Apr 18 2011 *)
%o (Magma) [n^5-n^3: n in [0..50]]; // _Vincenzo Librandi_, Feb 20 2012
%o (PARI) a(n)=n^5-n^3 \\ _Charles R Greathouse IV_, Feb 20 2012
%o (Sage) [n^5 - n^3 for n in (0..50)] # _G. C. Greubel_, Sep 02 2019
%o (GAP) List([0..50], n-> n^5 - n^3); # _G. C. Greubel_, Sep 02 2019
%Y Cf. A000578, A000584, A155977.
%K easy,nonn
%O 0,3
%A _Rolf Pleisch_, Mar 16 2008
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