A327009 -id:A327009 - cp's OEIS frontend

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

Max Alekseyev, Aug 12 2019

nonn

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)).

Cf. A000196, A327008, A327009, A327010.

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

A327008 a(n) = number of iterations of f(x)=floor((x^2+n^2)/(2x)) starting at x=n^2 to reach the value n.

Original entry on oeis.org

0, 1, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9

Offset: 1

Max Alekseyev, Aug 12 2019

nonn

Cf. A000196, A327007, A327009, A327010.

a(n) = A327007(n^2).

A327010 Indices of records in A327008.

Original entry on oeis.org

1, 2, 3, 4, 7, 11, 19, 33, 56, 100, 177, 321, 585, 1073, 1980, 3684, 6886, 12909, 24305, 45886, 86912, 165059, 314181, 599343, 1145657, 2193951, 4208697, 8086122, 15558747, 29977903, 57833191, 111705481, 216002074, 418115774

Offset: 1

Max Alekseyev, Aug 12 2019

nonn

a(n) = smallest m such that A327008(m) = n-1.

Cf. A000196, A327007, A327008, A327009.

For n >= 5, a(n) = sqrt(A327009(n+1)+1).

cp's OEIS Frontend

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)).

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A327008 a(n) = number of iterations of f(x)=floor((x^2+n^2)/(2x)) starting at x=n^2 to reach the value n.

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A327010 Indices of records in A327008.

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