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a(n) = 1 - n^4.
5

%I #21 Sep 08 2022 08:44:48

%S 1,0,-15,-80,-255,-624,-1295,-2400,-4095,-6560,-9999,-14640,-20735,

%T -28560,-38415,-50624,-65535,-83520,-104975,-130320,-159999,-194480,

%U -234255,-279840,-331775,-390624,-456975,-531440,-614655,-707280,-809999,-923520,-1048575,-1185920,-1336335,-1500624

%N a(n) = 1 - n^4.

%H Vincenzo Librandi, <a href="/A024002/b024002.txt">Table of n, a(n) for n = 0..630</a>

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

%F a(n) = -A123865(n) for n>0.

%F From _G. C. Greubel_, May 11 2017: (Start)

%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.: (1 - 5*x - 5*x^2 - 15*x^3)/(1 - x)^5.

%F E.g.f.: (1 - x - 7*x^2 - 6*x^3 - x^4)*exp(x). (End)

%F Sum_{k>=2} -1/a(k) = A256919 = 7/8 - Pi*coth(Pi)/4. - _Vaclav Kotesovec_, Dec 08 2020

%t Table[1 - n^4, {n, 0, 50}] (* _Bruno Berselli_, Jun 12 2015 *)

%t CoefficientList[Series[(1 - 5*x - 5*x^2 - 15*x^3)/(1 - x)^5, {x, 0, 50}], x] (* _G. C. Greubel_, May 11 2017 *)

%o (Magma) [1-n^4: n in [0..50]]; // _Vincenzo Librandi_, Apr 29 2011

%o (PARI) x='x+O('x^50); Vec((1 - 5*x - 5*x^2 - 15*x^3)/(1 - x)^5) \\ _G. C. Greubel_, May 11 2017

%Y Cf. A123865.

%K sign,easy

%O 0,3

%A _N. J. A. Sloane_

%E Corrected by _T. D. Noe_, Nov 08 2006