

A182655


Floorsum sequence of r, with r=(3+sqrt(5))/2 and a(1)=1, a(2)=2.


3



1, 2, 7, 20, 23, 54, 57, 62, 65, 70, 78, 112, 143, 146, 151, 154, 159, 164, 167, 172, 175, 180, 185, 188, 193, 201, 206, 209, 214, 222, 230, 235, 243, 256, 264, 290, 295, 298, 303, 311, 319, 324, 332, 345, 353, 366
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OFFSET

1,2


COMMENTS

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 floorsum sequence determined by (1) and (2) results by arranging the elements of S in strictly increasing order.
Let B be the Beatty sequence of r. Then a floorsum sequence of r is a subsequence of B if and only if a(1) and a(2) are terms of B.


LINKS



EXAMPLE

7 is in the sequence because floor(r*a(1)+r*a(2))=floor(r+2r)=7
57 is in the sequence because floor(r*a(2)+r*a(4))=floor(r*22)=57
61 is not in the sequence because 23*r=60.21... and 24*r=62.83... so there are no integers x,y with floor(r*(x+y))=61
60 is not in the sequence because floor(r*(x+y))=60 requires x+y=23, and no pair of elements of the sequence sum to 23


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



