A249160 Smallest number of iterations k such that A068527^(k)(n)=A068527^(k+1)(n).

1, 0, 2, 1, 2, 3, 1, 2, 1, 4, 3, 2, 3, 1, 2, 1, 3, 2, 4, 3, 2, 3, 1, 2, 1, 5, 2, 3, 2, 4, 3, 2, 3, 1, 2, 1, 3, 4, 5, 2, 3, 2, 4, 3, 2, 3, 1, 2, 1, 2, 4, 3, 4, 5, 2, 3, 2, 4, 3, 2, 3, 1, 2, 1, 2, 3, 2, 4, 3, 4, 5, 2, 3, 2, 4, 3, 2, 3, 1, 2, 1, 3, 4, 2, 3, 2, 4, 3, 4, 5, 2, 3, 2, 4, 3, 2, 3, 1, 2, 1

Offset

1,3

Comments

Given a number n, denote its distance from next perfect square >= n as R(n), sequence A068527. The function R(n) has two fixed points, 0 and 2, and for all n>=3, R(n)<n. Thus for any n>=0, there exists a k>=0 such that R^(k)(n)=R^(k+1)(n)=0 or 2. This sequence gives the number of iterations needed to reach the fixed point starting at n.

This sequence is unbounded, but grows very slowly, reaching records of 1, 2, 3, 4, 6 etc at n=1, 3, 6, 10, 26, 170, 7226, etc.

Links

Table of n, a(n) for n=1..100.

Example

R(10) = 6, R(6) = 3, R(3) = 1, R(1) = 0, R(0) = 0. Thus a(10) = 4.

Maple

A249160 := proc(n)
    local k, prev, this;
    prev := n ;
    for k from 1 do
        this := A068527(prev) ;
        if this = prev then
            return k-1;
        end if;
        prev := this ;
    end do:
end proc:
seq(A249160(n), n=1..80) ; # R. J. Mathar, Nov 17 2014

Mathematica

r[n_]:=Ceiling[Sqrt[n]]^2-n; Table[Length[FixedPointList[r, n]]-2, {n, 1, 100}]

Prog

(PARI) r(n)=if(issquare(n), 0, (sqrtint(n)+1)^2-n);
le(n)=b=0; while(n!=0&&n!=2, b=b+1; n=r(n)); return(b);
range(n) = c=List(); for(a = 1, n, listput(c, a)); return(c);
apply(le, range(100))

Crossrefs

Cf. A068527.

Sequence in context: A082691 A280052 A183198 * A346913 A269970 A333518

Adjacent sequences: A249157 A249158 A249159 * A249161 A249162 A249163

Keyword

nonn,easy

Author

Valtteri Raiko, Oct 22 2014

Status

approved