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

Author

Valtteri Raiko, Oct 22 2014

Keywords

• nonn
• easy

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)=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.

Examples

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

Crossrefs

Cf. A068527.

Programs

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}]

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))