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Floor-sum sequence of r with r = golden ratio = (1+sqrt(5))/2 and a(1)=1, a(2)=2.
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%I #17 Apr 25 2019 03:19:18

%S 1,2,4,8,9,14,16,17,19,21,24,25,27,29,30,32,33,35,37,38,40,42,43,45,

%T 46,48,50,51,53,55,56,58,59,61,63,64,66,67,69,71,72,74,76,77,79,80,82,

%U 84,85,87,88,90,92,93,95,97,98,100,101,103,105

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

%C Let S be the set generated by these rules:

%C (1) if m and n are in S and m<n, then floor(mr+nr) is in S;

%C (2) one or more specific numbers are in S by decree.

%C The floor-sum sequence determined by (1) and (2) results by arranging the elements of S in strictly increasing order.

%H Iain Fox, <a href="/A182653/b182653.txt">Table of n, a(n) for n = 1..3000</a>

%e Viewing the floor-sum as a binary operation o, we create S in successive generations:

%e 1, 2 (0th generation);

%e 1o2=4 (1st generation);

%e 1o4=8, 2o4=9 (2nd generation);

%e 1o8=14, 2o8=16, 4o8=19 and four others (3rd generation).

%p A182653 := proc(amax)

%p a := {1,2} ;r := (1+sqrt(5))/2 ;

%p while true do

%p anew := {} ;

%p for i in a do

%p for j in a do

%p if i <> j then

%p S := floor(r*(i+j)) ;

%p if is(S <= amax) then

%p anew := anew union { S };

%p 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 A182653(106) ;

%o (PARI) lista(nn) = my(S=[1, 2], 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 24 2019

%Y Cf. A182654, A182655, A182656.

%K nonn

%O 1,2

%A _Clark Kimberling_, Nov 26 2010