A110006 a(n) = n-F(F(n)) where F(x)=A120613(x)=floor(phi*floor(x/phi)) and phi=(1+sqrt(5))/2.

1, 2, 3, 3, 2, 3, 3, 4, 3, 2, 3, 3, 2, 3, 3, 4, 3, 2, 3, 3, 4, 3, 2, 3, 3, 2, 3, 3, 4, 3, 2, 3, 3, 2, 3, 3, 4, 3, 2, 3, 3, 4, 3, 2, 3, 3, 2, 3, 3, 4, 3, 2, 3, 3, 4, 3, 2, 3, 3, 2, 3, 3, 4, 3, 2, 3, 3, 2, 3, 3, 4, 3, 2, 3, 3, 4, 3, 2, 3, 3, 2, 3, 3, 4, 3, 2, 3, 3, 2, 3, 3, 4, 3, 2, 3, 3, 4, 3, 2, 3, 3, 2, 3, 3, 4

Offset

1,2

Comments

To built the sequence start from the infinite Fibonacci word : b(n)=floor(n/phi)-floor((n-1)/phi) for n>=1 giving 0,1,0,1,1,0,1,0,1,1,0,1,1,0,1,0,1,1,..... Then replace each 0 by the block {2,3,3} and each 1 by the block {2,3,3,4,3}. Append an initial 1.

References

Benoit Cloitre, On properties of irrational numbers related to the floor function, in preparation, 2005.

Links

Table of n, a(n) for n=1..105.

Prog

(PARI) a(n)=n-floor((1+sqrt(5))/2*floor((-1+sqrt(5))/2*floor((1+sqrt(5))/2*floor((-1+sqrt(5))/2*n))))

Crossrefs

Cf. A005614 (infinite Fibonacci binary word), A120613.

Cf. sequences for a(n) = n-F^k(n): A003842 (k=1), A110007 (k=3), A110010 (k=4), A110011 (k=5).

Sequence in context: A242285 A238756 A025076 * A289831 A335358 A296611

Adjacent sequences: A110003 A110004 A110005 * A110007 A110008 A110009

Keyword

nonn,easy,changed

Author

Benoit Cloitre, Sep 02 2005

Status

approved