A327007 a(n) = number of iterations of f(x)=floor((x^2+n)/(2x)) starting at x=n to reach the value floor(sqrt(n)) (=A000196(n)).

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

Offset: 1

Author

Max Alekseyev, Aug 12 2019

Keywords

• nonn

Comments

Also, we have f(x) = floor((x + floor(n/x))/2).

Notice that f(n) = f(1) = floor((n+1)/2), and so the starting value x = 1 gives the same sequence.

Iterations f(f(...f(a))...) reach floor(sqrt(n)) for any starting integer a >= 1. They either stabilize to floor(sqrt(n)) or alternate between floor(sqrt(n)) and ceiling(sqrt(n)).

Crossrefs

Cf. A000196, A327008, A327009, A327010.

Programs

PARI

{ A327007(n,a=n) = my(k = 0); while(1, my(b = (a+n\a)\2); if(b >= a,break); a = b; k++); k; }