A323832 Start with n and repeatedly double it and apply the "repeatedly delete any run of identical digits" operation described in A323830; a(n) is the number of steps needed to reach one of 0, 1, or 5, or -1 if none of these three numbers is ever reached.

0, 0, 19, 12, 18, 0, 11, 23, 17, 4, 19, 1, 10, 29, 22, 32, 16, 5, 3, 47, 18, 15, 1, 20, 9, 2, 28, 26, 21, 13, 31, 24, 15, 1, 4, 23, 2, 18, 46, 21, 17, 51, 14, 15, 1, 24, 19, 2, 8, 10, 1, 33, 27, 24, 25, 1, 20, 19, 12, 18, 30, 1, 23, 7, 14, 29, 5, 20, 3

Offset

0,3

Comments

Conjecture: every number will eventually reach one of 0, 1, or 5.

From Chai Wah Wu, Feb 04 2019, Feb 13 2019: (Start)

Conjecture is true for n < 10^10.

1604466 takes 115 steps to reach 5 and is the largest value for a(n) for n < 10^7.

91070713 takes 121 steps to reach 5 and is the largest value for a(n) for n < 10^8.

126591463 and 801282051 both take 128 steps to reach 5 and this is the largest value for a(n) for n < 10^9.

The numbers 1582393271, 1582393293, 4645106705 all take 131 steps to reach 5 and this is the largest value for a(n) for n < 10^10.

(End)

Links

Chai Wah Wu, Table of n, a(n) for n = 0..10000

Example

Starting with 2, the trajectory is 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 636, 1272, 25, 50, 1, reaching 1 in 20 steps, so a(2) = 20.
3 reaches 1 in 12 steps: 3, 6, 12, 24, 48, 96, 192, 384, 768, 1536, 3072, 61, 1, so a(3) = 12.
10 reaches 5 in 19 steps: 10, 20, 40, 80, 160, 320, 640, 1280, 2560, 5120, 10240, 20480, 40960, 81920, 163840, 327680, 6360, 12720, 250, 5, so a(10) = 19.

Prog

(Python)
from re import split
def A321801(n):
    return int('0'+''.join(d for d in split('(0+)|(1+)|(2+)|(3+)|(4+)|(5+)|(6+)|(7+)|(8+)|(9+)', str(n)) if d != '' and d != None and len(d) == 1))
def f(n):
    x = 2*n
    y = A321801(x)
    while x != y:
        x, y = y, A321801(y)
    return x
def A323832(n):
    mset, m, c = set(), n, 0
    while True:
        if m == 1 or m == 0 or m == 5:
            return c
        m = f(m)
        if m in mset:
            return -1
        mset.add(m)
        c += 1 # Chai Wah Wu, Feb 04 2019, Feb 11 2019

Crossrefs

Cf. A321801, A321802, A323830, A323830.

Sequence in context: A356570 A040344 A261717 * A306384 A196188 A089294

Adjacent sequences: A323829 A323830 A323831 * A323833 A323834 A323835

Keyword

nonn,base

Author

N. J. A. Sloane, Feb 03 2019

Status

approved