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Floor-sum sequence of r, where r = golden ratio = (1+sqrt(5))/2 and a(1)=2, a(2)=3.
1

%I #17 Apr 25 2019 03:19:43

%S 2,3,8,16,17,29,30,32,38,40,50,51,53,55,56,59,61,64,66,67,69,72,74,76,

%T 77,79,84,85,87,88,90,92,93,95,98,100,101,103,106,108,110,111,113,114,

%U 116,118,119,121,122,124,126,127,129,131,132,134,135,137,139,140

%N Floor-sum sequence of r, where r = golden ratio = (1+sqrt(5))/2 and a(1)=2, a(2)=3.

%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 floor-sum 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 floor-sum sequence of r is a subsequence of B if and only if a(1) and a(2) are terms of B. Thus, A182670 is not a subsequence of the lower Wythoff sequence, A000201.

%H Iain Fox, <a href="/A182670/b182670.txt">Table of n, a(n) for n = 1..3000</a>

%e a(3) = floor(2r+3r) = 8.

%p A182670 := proc(amax)

%p a := {2,3} ;

%p r := (1+sqrt(5))/2 ;

%p while true do

%p anew := {} ;

%p for i in a do for j in a do

%p if i <> j then S := floor(r*(i+j)) ; if is(S <= amax) then anew := anew union { S }; end if;

%p end if;

%p end do:

%p end do:

%p if a union anew = a then

%p return sort(a) ;

%p end if;

%p a := a union anew ;

%p end do:

%p end proc:

%p A182670(140) ;

%o (PARI) lista(nn) = my(S=[2, 3], r=(1+sqrt(5))/2, new, k); while(1, new=[]; for(m=1, #S, for(n=m+1, #S, k=floor(r*(S[m]+S[n])); if(k<=nn, new=setunion(new, [k])))); if(S==setunion(S, new), return(S)); S=setunion(S, new)) \\ _Iain Fox_, Apr 25 2019

%Y Cf. A000201, A182653, A182669.

%K nonn

%O 1,1

%A _Clark Kimberling_, Nov 27 2010