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%I #17 Jun 03 2015 18:27:55
%S 0,0,1,2,3,5,7,10,13,16,20,24,28,33,38,44,50,56,63,70,78,86,94,103,
%T 112,121,131,141,152,163,174,186,198,210,223,236,250,264,278,293,308,
%U 324,340,356,373,390,407,425,443,462,481,500,520,540,561,582,603,625
%N [1/s]+[2/s]+...+[n/s], where s=(golden ratio)^2 and []=floor.
%C A183136(n) + a(n) = A000217(n) (the triangular numbers).
%F floor(1/s)+floor(2/s)+...+floor(n/s), where s = (3+sqrt(5))/2 = (golden mean)^2.
%F a(n+1) = a(n) + n - A005206(n). - _John Furey_, Jun 03 2015
%e a(7)=0+0+1+1+1+2+2.
%t Accumulate[With[{c=GoldenRatio^2},Floor[Range[60]/c]]] (* _Harvey P. Dale_, Apr 20 2011 *)
%Y Cf. A183136, A000217, A005206, A060143, A060144.
%K nonn
%O 1,4
%A _Clark Kimberling_, Dec 26 2010
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