cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A348339 a(n) is the number of distinct numbers of steps required for the last n digits of integers to repeat themselves by iterating the map m -> m^2.

Original entry on oeis.org

3, 6, 9, 12, 19, 28, 39, 52, 67, 84
Offset: 1

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Author

Ya-Ping Lu, Oct 13 2021

Keywords

Comments

Conjecture: For n >= 5, a(n) = a(n-1) + 2*n - 3 - ceiling(log_5 ((n-1)/16)), or a(n) = (n-1)^2 + 3 - Sum_{5..n} ceiling(log_5 ((i-1)/16)). The largest number of steps required is 4*5^(n-2) + (n-2) for n >= 4.

Examples

			a(1) = 3. Integers ending with 0, 1, 5 or 6 take 1 step to repeat the last digit. Integers ending with 4 or 9 require 2 steps and 2, 3, 7 or 8 require 3 steps to repeat their last digit. Thus, the distinct numbers of steps for n = 1 are {1, 2, 3} and a(1) = 3.
a(2) = 6 because the distinct steps are: {1, 2, 3, 4, 5, 6}.
a(3) = 9: {1, 2, 3, 4, 5, 6, 20, 21, 22}.
a(4) = 12: {1, 2, 3, 4, 5, 6, 20, 21, 22, 100, 101, 102}.
a(5) = 19: {1, 2, 3, 4, 5, 6, 7, 20, 21, 22, 23, 100, 101, 102, 103, 500, 501, 502, 503}.
a(6) = 28: {1, 2, 3, 4, 5, 6, 7, 8, 20, 21, 22, 23, 24, 100, 101, 102, 103, 104, 500, 501, 502, 503, 504, 2500, 2501, 2502, 2503, 2504}.
The paths of the last 1, 2, and 3 digits of integers resulted from iterating the map, m -> m*m, are shown in the Links.
		

Crossrefs

Programs

  • Python
    def tail(m):
        global n; s = str(m)
        return m if len(s) <= n else int(s[-n:])
    for n in range(1, 10):
        M = []
        for i in range(10**n):
            t = i; L = [t]
            while i >= 0:
                t = tail(t*t)
                if t not in L: L.append(t)
                else: break
            d = len(L)
            if d not in M: M.append(d)
        print(len(M), end = ', ')
    
  • Python
    def A348339(n):
        m, s = 10**n, set()
        for k in range(m):
            c, k2, kset = 0, k, set()
            while k2 not in kset:
                kset.add(k2)
                c += 1
                k2 = k2*k2 % m
            s.add(c)
        return len(s) # Chai Wah Wu, Oct 19 2021

Extensions

a(10) from Martin Ehrenstein, Oct 20 2021