A138467 a(1)=1, then for n>=2 a(n) = n - floor((1/3)*a(a(n-1))).

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Offset: 1

Author

Benoit Cloitre, May 09 2008

Keywords

• nonn

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

Crossrefs

Cf. A005206, A135414.

Programs

PARI

a(n)=floor((3/2)*(sqrt(7/3)-1)*(n+1))

Formula

For n>=1, a(n) = floor(r*(n+1)) where r=(3/2)*(sqrt(7/3)-1).

Extensions

More terms from Yalcin Aktar, Jul 13 2008