|
%I #9 Sep 08 2017 09:52:01
%S 1,3,5,7,10,12,14,16,19,21,23,25,27,30,32,34,36,39,41,43,45,48,50,52,
%T 54,57,59,61,63,65,68,70,72,74,77,79,81,83,86,88,90,92,95,97,99,101,
%U 103,106,108,110,112,115,117,119,121,124,126,128,130,133,135,137,139,141,144,146,148,150,153,155,157,159,162,164,166,168,171,173,175,177,180,182,184,186,188,191,193,195,197,200,202,204,206,209,211,213,215,218,220,222,224,226,229,231,233,235,238,240,242,244,247,249,251,253,256,258,260,262,264,267
%N a(n) = floor((n-1/2)*r), where r=sqrt(5); complement of A184587.
%C r = sqrt(5) and s = (5+sqrt(5))/4 form a Beatty pair. This yields the pair of complementary homogeneous Beatty sequences A022839 and A108598. From a theorem of Thoralf Skolem follows that (a(n)) and A184587 are complementary inhomogeneous Beatty sequences. - _Michel Dekking_, Sep 08 2017
%F a(n)=floor[(n-1/2)r], where r=sqrt(5).
%t r=5^(1/2); c=1/2; s=r/(r-1);
%t Table[Floor[n*r-c*r],{n,1,120}] (* A184586 *)
%t Table[Floor[n*s+c*s],{n,1,120}] (* A184587 *)
%Y Cf. A184587.
%K nonn
%O 1,2
%A _Clark Kimberling_, Jan 17 2011
%E Name and formula corrected by _Michel Dekking_, Sep 08 2017
|
