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A182669 Floor-sum sequence of r, with r = golden ratio = (1+sqrt(5))/2 and a(1)=1, a(2)=3.

%I A182669 #23 Dec 13 2024 13:12:10
%S A182669 1,3,6,11,14,19,22,24,27,32,35,37,40,43,45,48,53,56,58,61,64,66,69,71,
%T A182669 74,77,79,82,87,90,92,95,98,100,103,105,108,111,113,116,119,121,124,
%U A182669 126,129,132,134,137,139,142,145,147,150,153,155,158,160,163,166,168,171,173,174,176,179,181,184
%N A182669 Floor-sum sequence of r, with r = golden ratio = (1+sqrt(5))/2 and a(1)=1, a(2)=3.
%C A182669 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 A182669 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, A182669 is a subsequence of the lower Wythoff sequence, A000201.
%H A182669 Iain Fox, <a href="/A182669/b182669.txt">Table of n, a(n) for n = 1..3000</a>
%e A182669 a(3)=floor(r+3r)=6.
%o A182669 (QBasic) r=(1+5^(1/2))/2: s(1)=1: s(2)=3: s(5)=6
%o A182669 For h=2 to 200: c(h)=h+c(h-1): next h
%o A182669 For h=1 to 100: c=c(h): d=0
%o A182669   For i=1 to h+1: d=d+1: s(c+d)=int(s(i)+s(h+2)*r)
%o A182669   Next i
%o A182669 Next h
%o A182669 For i=1 to 1000: for j=i+1 to 1001
%o A182669 if s(i)>=s(j) then swap s(i),s(j)
%o A182669 next j,i
%o A182669 For i=1 to 120: if s(i+1)<>s(i) then print s(i);
%o A182669 next i
%o A182669 (PARI) lista(nn) = my(S=[1, 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 A182669 Cf. A000201, A182653, A182570.
%K A182669 nonn
%O A182669 1,2
%A A182669 _Clark Kimberling_, Nov 27 2010
%E A182669 139 (generated by m=22, n=64) added by _R. J. Mathar_, Nov 28 2010