|
|
A110919
|
|
Number of consecutive 1's in the continued fraction for floor(n*phi)/n where phi = (1+sqrt(5))/2 (A001622).
|
|
1
|
|
|
1, 1, 1, 1, 3, 1, 3, 1, 3, 3, 3, 3, 5, 3, 3, 3, 3, 5, 3, 3, 3, 3, 5, 3, 3, 5, 3, 5, 3, 3, 5, 3, 5, 7, 3, 5, 3, 5, 5, 3, 5, 3, 5, 5, 3, 5, 7, 5, 5, 3, 5, 5, 5, 5, 3, 5, 5, 5, 5, 7, 5, 5, 5, 5, 5, 5, 5, 7, 5, 5, 5, 5, 7, 5, 5, 5, 5, 5, 5, 5, 7, 5, 5, 5, 5, 7, 5, 5, 9, 5, 5, 5, 5, 7, 5, 5, 5, 5, 7, 5, 5, 7, 5, 5, 5
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,5
|
|
COMMENTS
|
Terms are always odd.
|
|
LINKS
|
|
|
FORMULA
|
Sum_{k=1..n} a(k) seems to be asymptotic to c*n*log(n) with c around 1.
|
|
EXAMPLE
|
The continued fraction for floor(128*phi)/128 is [1, 1, 1, 1, 1, 1, 1, 2, 1, 2] with 7 consecutive 1's, thus a(128) = 7.
|
|
MATHEMATICA
|
a[1] = 1; a[n_] := -1 + FirstPosition[ContinuedFraction[Floor[n*GoldenRatio]/n], _?(# > 1 &)][[1]]; Array[a, 100] (* Amiram Eldar, May 08 2022 *)
|
|
PROG
|
(PARI) a(n)=if(n<2, 1, s=1; while(component(contfrac(floor(n*(1+sqrt(5))/2)/n), s)==1, s++); s-1)
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|