%I #29 Sep 08 2022 08:45:50
%S 0,1,25,124,380,905,1841,3360,5664,8985,13585,19756,27820,38129,51065,
%T 67040,86496,109905,137769,170620,209020,253561,304865,363584,430400,
%U 506025,591201,686700,793324,911905,1043305,1188416,1348160,1523489
%N n*(n+1)*(15*n^2-n-8)/12.
%C This sequence is related to A007587 by a(n) = n*A007587(n)-sum(i=0..n-1, A007587(i)).
%C This is the case d=5 in the general formula n*(n*(n+1)*(2*d*n-2*d+3)/6)-sum(i=0..n-1, i*(i+1)*(2*d*i-2*d+3)/6) = n*(n+1)*(3*d*n^2-d*n+4*n-2*d+2)/12. - _Bruno Berselli_, Dec 07 2010
%C The inverse binomial transform yields 0, 1, 23, 52, 30, 0, 0 (0 continued). - _R. J. Mathar_, Dec 09 2010
%H Vincenzo Librandi, <a href="/A172047/b172047.txt">Table of n, a(n) for n = 0..1000</a>
%H B. Berselli, A description of the recursive method in Comments lines: website <a href="http://www.lanostra-matematica.org/2008/12/sequenze-numeriche-e-procedimenti.html">Matem@ticamente</a> (in Italian).
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F G.f.: -x*(1+20*x+9*x^2)/(x-1)^5. - _R. J. Mathar_, Dec 09 2010
%F a(n)-a(-n) = A063521(n). - _Bruno Berselli_, Aug 26 2011
%t CoefficientList[Series[x (1 + 20 x + 9 x^2)/(1 - x)^5, {x, 0, 40}], x] (* _Vincenzo Librandi_, Jan 01 2014 *)
%o (Magma) [n*(n+1)*(15*n^2-n-8)/12: n in [0..50]]; // _Vincenzo Librandi_, Jan 01 2014
%Y Cf. A007587.
%K nonn,easy
%O 0,3
%A _Vincenzo Librandi_, Jan 24 2010