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

1,3

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)).

Links

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

Prog

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

Crossrefs

Cf. A000196, A327008, A327009, A327010.

Sequence in context: A085033 A230254 A299484 * A336767 A096446 A274701

Adjacent sequences: A327004 A327005 A327006 * A327008 A327009 A327010

Keyword

nonn

Author

Max Alekseyev, Aug 12 2019

Status

approved