%I #7 Mar 30 2012 18:57:12
%S 1,2,7,20,23,54,57,62,65,70,78,112,143,146,151,154,159,164,167,172,
%T 175,180,185,188,193,201,206,209,214,222,230,235,243,256,264,290,295,
%U 298,303,311,319,324,332,345,353,366
%N Floorsum sequence of r, with r=(3+sqrt(5))/2 and a(1)=1, a(2)=2.
%C Let S be the set generated by these rules: (1) if m and n are in S and m<n, then floor(mr+nr) is in S; (2) two or more specific numbers are in S. The floorsum sequence determined by (1) and (2) results by arranging the elements of S in strictly increasing order.
%C Let B be the Beatty sequence of r. Then a floorsum sequence of r is a subsequence of B if and only if a(1) and a(2) are terms of B.
%e 7 is in the sequence because floor(r*a(1)+r*a(2))=floor(r+2r)=7
%e 57 is in the sequence because floor(r*a(2)+r*a(4))=floor(r*22)=57
%e 61 is not in the sequence because 23*r=60.21... and 24*r=62.83... so there are no integers x,y with floor(r*(x+y))=61
%e 60 is not in the sequence because floor(r*(x+y))=60 requires x+y=23, and no pair of elements of the sequence sum to 23
%Y Cf. A182653, A182654, A182656. A001950.
%K nonn
%O 1,2
%A _Clark Kimberling_, Nov 26 2010
