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%I #31 Jun 18 2021 15:25:08
%S 0,1,10,14,30,55,104,140,204,285,406,506,650,819,1044,1240,1496,1785,
%T 2146,2470,2870,3311,3840,4324,4900,5525,6254,6930,7714,8555,9516,
%U 10416,11440,12529,13754,14910,16206,17575,19096,20540,22140,23821,25670,27434,29370
%N a(0)=0; then a(4*k+1)=a(4*k)+(4*k+1)^2, a(4*k+2)=a(4*k+1)+(4*k+3)^2, a(4*k+3)=a(4*k+2)+(4*k+2)^2, a(4*k+4)=a(4*k+3)+(4*k+4)^2.
%C Probable answer to the riddle in A115603.
%C Partial sums of the squares of the terms of A116966.
%H Harvey P. Dale, <a href="/A115391/b115391.txt">Table of n, a(n) for n = 0..1000</a> [a(0)=0 prepended by _Georg Fischer_, Jun 18 2021]
%H <a href="/index/Rec#order_10">Index entries for linear recurrences with constant coefficients</a>, signature (2,-1,0,2,-4,2,0,-1,2,-1).
%F G.f.: x*(4*x^7-3*x^6+8*x^5+7*x^4+12*x^3-5*x^2+8*x+1) / ((x-1)^4*(x+1)^2*(x^2+1)^2). - _Colin Barker_, Jul 18 2013
%F a(n) = (2*n+1)*(2*n*(n+1)+3*(1+cos(n*Pi)-2*cos(n*Pi/2)))/12. - _Luce ETIENNE_, Feb 01 2017
%t LinearRecurrence[{2,-1,0,2,-4,2,0,-1,2,-1},{0,1,10,14,30,55,104,140,204,285,406},50] (* _Harvey P. Dale_, Jul 01 2020 *)
%Y Cf A000330, A004770
%K nonn,easy
%O 0,3
%A _Pierre CAMI_, Mar 15 2006
%E More terms from _Stefan Steinerberger_, Mar 31 2006
%E Entry revised by _Don Reble_, Apr 06 2006
%E More terms from _Colin Barker_, Jul 18 2013
%E Offset adapted to definition by _Georg Fischer_, Jun 18 2021
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