login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

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)).
4
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)).
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
KEYWORD
nonn
AUTHOR
Max Alekseyev, Aug 12 2019
STATUS
approved