A323835 Start with n and repeatedly double it and apply the "delete multiple digits" map m -> A320485(m); a(n) is the number of steps needed to reach either 0 or 1, or -1 if neither 0 nor 1 is ever reached.

0, 0, 27, 12, 26, 41, 11, 31, 25, 4, 40, 1, 10, 37, 30, 43, 24, 35, 3, 42, 39, 15, 1, 20, 9, 2, 36, 26, 29, 13, 42, 32, 23, 1, 34, 44, 2, 18, 41, 21, 38, 45, 14, 15, 1, 45, 19, 2, 8, 30, 1, 20, 35, 2, 25, 1, 28, 27, 12, 26, 41, 1, 31, 43, 22, 34, 5, 20, 33, 30

Offset

0,3

Comments

The first values of k for which a(k) = -1 are 91, 182, 364, 455, 728, 910, 1456, 1729, 1820, 1853, 1879. - Giovanni Resta, Feb 04 2019

From Chai Wah Wu, Feb 04 2019: (Start)

a(n) <= 64 for all n.

Let f(n) = A320486(2*n) and k = 9876543210. If n > k/2, then f(n) <= k. Note that a(n) = a(f(n)) + 1 if a(f(n)) >= 0 and a(n) = -1 if a(f(n)) = -1.

If k/2 < n <= k, then f(n) <= n*198/1000 < k/2. Thus if n > k, f(f(n)) <= k/2.

This means that we only need to study trajectories for 0 <= n <= k. The longest trajectories in this range have 64 steps and are reached by the 9 numbers 1233546907, 1323546907, 1335246907, 1335467407, 1335469072, 1335469207, 1335471907, 1337046907, 2133546907. The first application of f(.) takes all these numbers to the number 26709381, which then follows 63 steps to 1. Since these 9 numbers all have a double digit 3, they are not in the range of f and thus not part of a longer trajectory. Thus for all n > k, a(f(n)) <= 63, and a(n) <= 64.

There are 74801508 numbers in the range 0 <= n <= k such that a(n) = -1.

(End)

All trajectories will reach one of four cycles: 0, 1, 91, or 910. - Chai Wah Wu, Feb 11 2019

Links

David Consiglio, Jr., Table of n, a(n) for n = 0..999

Example

As we can see from A320487, 2 reaches 1 in 27 steps: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 3, 6, 12, 24, 48, 96, 192, 384, 768, 1536, 3072, 61, 1, so a(2)=27.

Prog

(Python)
def A323835(n):
    mset, m, c = set(), n, 0
    while True:
        if m == 0 or m == 1:
            return c
        m = int('0'+''.join(d if str(2*m).count(d) == 1 else '' for d in str(2*m)))
        if m in mset:
            return -1
        mset.add(m)
        c += 1  # Chai Wah Wu, Feb 04 2019

Crossrefs

Cf. A320485, A320487, A323832.

Sequence in context: A069893 A013681 A103948 * A040705 A317106 A033347

Adjacent sequences: A323832 A323833 A323834 * A323836 A323837 A323838

Keyword

nonn,base

Author

N. J. A. Sloane, Feb 03 2019

Extensions

More terms from David Consiglio, Jr., Feb 04 2019

Status

approved