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A138467 a(1)=1, then for n>=2 a(n) = n - floor((1/3)*a(a(n-1))). 1
1, 2, 3, 3, 4, 5, 6, 7, 7, 8, 9, 10, 11, 11, 12, 13, 14, 15, 15, 16, 17, 18, 18, 19, 20, 21, 22, 22, 23, 24, 25, 26, 26, 27, 28, 29, 30, 30, 31, 32, 33, 34, 34, 35, 36, 37, 37, 38, 39, 40, 41, 41, 42, 43, 44, 45, 45, 46, 47, 48, 49, 49, 50, 51, 52, 53, 53, 54, 55, 56, 56, 57, 58 (list; graph; refs; listen; history; text; internal format)
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
Sequence in context: A326694 A053207 A334714 * A208746 A230490 A247983
KEYWORD
nonn
AUTHOR
Benoit Cloitre, May 09 2008
EXTENSIONS
More terms from Yalcin Aktar, Jul 13 2008
STATUS
approved

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Last modified March 19 04:58 EDT 2024. Contains 370952 sequences. (Running on oeis4.)