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

1, 3, 6, 11, 14, 19, 22, 24, 27, 32, 35, 37, 40, 43, 45, 48, 53, 56, 58, 61, 64, 66, 69, 71, 74, 77, 79, 82, 87, 90, 92, 95, 98, 100, 103, 105, 108, 111, 113, 116, 119, 121, 124, 126, 129, 132, 134, 137, 139, 142, 145, 147, 150, 153, 155, 158, 160, 163, 166, 168, 171, 173, 174, 176, 179, 181, 184

Offset: 1

Author

Clark Kimberling, Nov 27 2010

Keywords

• nonn

Comments

Let S be the set generated by these rules: (1) if m and n are in S and m

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.

Examples

			a(3)=floor(r+3r)=6.
		

Links

• Iain Fox, Table of n, a(n) for n = 1..3000

Crossrefs

Cf. A000201, A182653, A182570.

Programs

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

QBasic

r=(1+5^(1/2))/2: s(1)=1: s(2)=3: s(5)=6
For h=2 to 200: c(h)=h+c(h-1): next h
For h=1 to 100: c=c(h): d=0
  For i=1 to h+1: d=d+1: s(c+d)=int(s(i)+s(h+2)*r)
  Next i
Next h
For i=1 to 1000: for j=i+1 to 1001
if s(i)>=s(j) then swap s(i),s(j)
next j,i
For i=1 to 120: if s(i+1)<>s(i) then print s(i);
next i

Extensions

139 (generated by m=22, n=64) added by R. J. Mathar, Nov 28 2010

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