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a(n) = n^4 - n^3 - n^2 - n - 1.
6

%I #20 Sep 08 2022 08:45:28

%S -1,-3,1,41,171,469,1037,2001,3511,5741,8889,13177,18851,26181,35461,

%T 47009,61167,78301,98801,123081,151579,184757,223101,267121,317351,

%U 374349,438697,511001,591891,682021,782069,892737,1014751,1148861,1295841,1456489,1631627,1822101,2028781,2252561,2494359

%N a(n) = n^4 - n^3 - n^2 - n - 1.

%C For odd n > 1, a(n) * (n-1) / 2 represents the first integer in a sum of (2 * n^4) consecutive integers that equals n^9. - _Patrick J. McNab_, Dec 25 2016

%H Vincenzo Librandi, <a href="/A125082/b125082.txt">Table of n, a(n) for n = 0..660</a>

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

%F O.g.f.: -60/(-1+x)^3-66/(-1+x)^4-1/(-1+x)-24/(-1+x)^5-20/(-1+x)^2. - _R. J. Mathar_, Feb 26 2008

%F a(0)=-1, a(1)=-3, a(2)=1, a(3)=41, a(4)=171, a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). - _Harvey P. Dale_, Dec 30 2011

%t Table[n^4 - n^3 - n^2 - n - 1, {n, 0, 41}]

%t LinearRecurrence[{5,-10,10,-5,1},{-1,-3,1,41,171},40] (* _Harvey P. Dale_, Dec 30 2011 *)

%o (Magma) [n^4-n^3-n^2-n-1: n in [0..60]]; // _Vincenzo Librandi_, Apr 26 2011

%o (PARI) a(n)=n^4-n^3-n^2-n-1 \\ _Charles R Greathouse IV_, Oct 07 2015

%Y Cf. A083074.

%K sign,easy

%O 0,2

%A _Artur Jasinski_, Nov 19 2006