FUNCTIONAL SMARANDACHE ITERATIONS


1) Functional Smarandache Iteration of First Kind:

   Let f: A ---> A be a function, such that f(x) <= x for all x, 
   and 
       
       min {f(x), x belongs to A} > = m0, different from negative infinity.
   
   Let f have p >= 1 fix points:  m0 <= x1 < x2 < ... < xp.
   [The point x is called fix if f(x) = x.]
   Then

   SI1 (x) = the smallest number of iterations k such that
      f

   f(f(...f(x)...)) = constant.
   iterated k times



Example:
   Let n > 1 be an integer, and d(n) be the number of positive divisors of n,
   d: N ---> N.

   Then SI1 (n) is the smallest number of iterations k
           d

   such that d(d(...d(n)...)) = 2;
             iterated k times

   because d(n) < n for n > 2, and the fix points of the function d are 1
   and 2.

   Thus SI1 (6) = 3, because d(d(d(6))) = d(d(4)) = d(3) = 2 = constant.
           d

        SI1 (5) = 1, because d(5) = 2.
           d




2) Functional Smarandache Iteration of Second Kind:

   Let g: A ---> A be a function, such that g(x) > x for all x, 
   and let b > x.  Then:

   SI2 (x, b) = the smallest number of iterations k such that
      g

   g(g(...g(x)...)) >= b.
   iterated k times



Example:
   Let n > 1 be an integer, and sigma(n) be the sum of positive divisors
   of n (1 and n included),  sigma: N ---> N.

   Then SI2     (n, b) is the smallest number of iterations k such that 
           sigma
   
   sigma(sigma(...sigma(n)...)) >= b,
     iterated k times

   because sigma(n) > n for n > 1.

   Thus SI2     (4, 11) = 3, because sigma(sigma(sigma(4))) = 
           sigma

   sigma(sigma(7)) = sigma(8) = 15 >= 11.



3) Functional Smarandache Iteration of Third Kind:

   Let h: A ---> A be a function, such that h(x) < x for all x, 
   and let b < x.  Then: 

   SI3 (x, b) = the smallest number of iterations k such that
      h

   h(h(...h(x)...)) <= b.
   iterated k times


Example:
  Let n be an integer and gd(n) be the greatest divisor of n, less than n,
  gd: N* ---> N*.  Then  gd(n) < n for n > 1.
  
  SI3  (60, 3) = 4, because gd(gd(gd(gd(60)))) = gd(gd(gd(30))) = 
     gd
  
  gd(gd(15)) = gd(5) = 1 <= 3.
  



References:
  [1] Ibstedt, H., "Smarandache Iterations of First and Second Kinds",
      <Abstracts of Papers Presented to the American Mathematical Society>,
      Vol. 17, No. 4, Issue 106, 1996, p. 680.
  [2] Ibstedt, H., "Surfing on the Ocean of Numbers - A Few Smarandache
      Notions and Similar Topics", Erhus University Press, Vail, 1997;
      pp. 52-58.
  [3] Smarandache, F., "Unsolved Problem: 52", <Only Problems, Not 
      Solutions!>, Xiquan Publishing House, Phoenix, 1993.