OFFSET
1,2
COMMENTS
For k >= 1, a(1)=1; a(n) = n - floor((1/k)*a(a(n-1))). - Yalcin Aktar, Jul 13 2008
Conjecture: a(n) = floor(r(p)*(n+1)) with r(p)=(1/2)*(sqrt(p*(p+4))-p). - Yalcin Aktar, Jul 13 2008
From Michel Dekking, Jul 13 2023: (Start)
Here is a correction of these two comments from July 13, 2008:
Consider the following generalization of (a(n)).
Let p>2 be a natural number, and define the sequence b_p by
b_p = 1, b_p(n) = n - floor((1/p)*b_p (b_p (n-1)) for n>1.
Conjecture: b_p(n) = floor(r(p)*(n+1)) where r(p) =(1/2)*(sqrt(p*(p+4))-p).(End)
LINKS
Michel Dekking, Proof by Benoit Cloitre
FORMULA
For n>=1, a(n) = floor(r*(n+1)) where r=(3/2)*(sqrt(7/3)-1).
PROG
(PARI) a(n)=floor((3/2)*(sqrt(7/3)-1)*(n+1))
CROSSREFS
KEYWORD
nonn
AUTHOR
Benoit Cloitre, May 09 2008
EXTENSIONS
More terms from Yalcin Aktar, Jul 13 2008
STATUS
approved